Solutions for Calculus 11th Us Edition by Larson by 6alsm

Chapter 2 Differentiation

Chapter Comments

The material presented in Chapter 2 forms the basis for the remainder of calculus. Much of it needs to be memorized, beginning with the definition of a derivative of a function found on page 103. Students need to have a thorough understanding of the tangent line problem and they need to be able to find an equation of a tangent line. Frequently, students will use the function f ′(x) as the slope of the tangent line. They need to understand that f ′(x) is the formula for the slope and the actual value of the slope can be found by substituting into f ′(x) the appropriate value for x On pages 105–106 of Section 2.1, you will find a discussion of situations where the derivative fails to exist. These examples (or similar ones) should be discussed in class.

As you teach this chapter, vary your notations for the derivative. One time write y ′; another time write dy dx or f ′(x). Terminology is also important. Instead of saying “find the derivative,” sometimes say, “differentiate.” This would be an appropriate time, also, to talk a little about Leibnitz and Newton and the discovery of calculus.

Sections 2.2, 2.3, and 2.4 present a number of rules for differentiation. Have your students memorize the Product Rule and the Quotient Rule (Theorems 2.7 and 2.8) in words rather than symbols. Students tend to be lazy when it comes to trigonometry and therefore, you need to impress upon them that the formulas for the derivatives of the six trigonometric functions need to be memorized also. You will probably not have enough time in class to prove every one of these differentiation rules, so choose several to do in class and perhaps assign a few of the other proofs as homework.

The Chain Rule, in Section 2.4, will require two days of your class time. Students need a lot of practice with this and the algebra involved in these problems. Many students can find the derivative of f (x) = x2√1 x2 without much trouble, but simplifying the answer is often difficult for them. Insist that they learn to factor and write the answer without negative exponents. Strive to get the answer in the form given in the back of the book. This will help them later on when the derivative is set equal to zero.

Implicit differentiation is often difficult for students. Have students think of y as a function of x and therefore y 3 is [ f (x)] 3 This way they can relate implicit differentiation to the Chain Rule studied in the previous section.

Try to get your students to see that related rates, discussed in Section 2.6, are another use of the Chain Rule.

Section 2.1 The Derivative and the Tangent Line Problem

Section Comments

2.1 The Derivative and the Tangent Line Problem—Find the slope of the tangent line to a curve at a point. Use the limit definition to find the derivative of a function. Understand the relationship between differentiability and continuity.

Teaching Tips

Ask students what they think “the line tangent to a curve” means. Draw a curve with tangent lines to show a visual picture of tangent lines. For example:

= f (x)

(x)

When talking about the tangent line problem, use the suggested example of finding the equation of the tangent line to the parabola y = x2 at the point (1, 1)

Compute an approximation of the slope m by choosing a nearby point Q(x, x2) on the parabola and computing the slope mPQ of the secant line PQ.

After going over Examples 1–3, return to Example 2 where f (x) = x2 + 1 and note that f ′(x) = 2x. How can we find the equation of the line tangent to f and parallel to 4x y = 0?

Because the slope of the line is 4,

2x = 4

x = 2

So, at the point (2, 5), the tangent line is parallel to 4x y = 0 The equation of the tangent line is y 5 = 4(x 2) or y = 4x 3

Be sure to find the derivatives of various types of functions to show students the different types of techniques for finding derivatives. Some suggested problems are f (x) = 4x3 3x2 , g(x) = 2 (x 1), and h(x) = √2x + 5.

How Do You See It? Exercise

Page 108, Exercise 64 The figure shows the graph of g ′

(a) g′(0) =

(b) g′(3) =

(d) What can you conclude about the graph of g knowing that g′( 4) = 7 3 ?

(e) Is g(6) g(4) positive or negative? Explain.

(f) Is it possible to find g(2) from the graph? Explain.

Solution

(a) g′(0) = 3

(b) g′(3) = 0

(d) Because g′( 4) = 7 3 , g is increasing (rising) at x = 4

(e) Because g′(4) and g′(6) are both positive, g(6) is greater than g(4) and g(6) g(4) > 0

(f) No, it is not possible. All you can say is that g is decreasing (falling) at x = 2

Suggested Homework Assignment

Pages 107–109: 1, 3, 7, 11, 21–27 odd, 37, 43–47 odd, 53, 57, 61, 77, 87, 93, and 95.

Section 2.2 Basic Differentiation Rules and Rates of Change

Section Comments

2.2 Basic Differentiation Rules and Rates of Change—Find the derivative of a function using the Constant Rule. Find the derivative of a function using the Power Rule. Find the derivative of a function using the Constant Multiple Rule. Find the derivative of a function using the Sum and Difference Rules. Find the derivatives of the sine function and of the cosine function. Use derivatives to find rates of change.

Teaching Tips

Start by showing proofs of the Constant Rule and the Power Rule. Students who are mathematics majors need to start seeing proofs early on in their college careers as they will be taking Functions of a Real Variable at some point.

Go over an example in class like f (x) = 5x2 + x x . Show students that before differentiating they can rewrite the function as f (x) = 5x + 1 Then they can differentiate to obtain f ′(x) = 5

Use this example to emphasize the prudence of examining the function first before differentiating. Rewriting the function in a simpler, equivalent form can expedite the differentiating process.

Give mixed examples of finding derivatives. Some suggested examples are:

This will test students’ understanding of the various differentiation rules of this section.

How Do You See It? Exercise

Page 119, Exercise 76 Use the graph of f to answer each question. To print an enlarged copy of the graph, go to MathGraphs.com.

(a) Between which two consecutive points is the average rate of change of the function greatest?

(b) Is the average rate of change of the function between A and B greater than or less than the instantaneous rate of change at B?

(c) Sketch a tangent line to the graph between C and D such that the slope of the tangent line is the same as the average rate of change of the function between C and D.

Solution

(a) The slope appears to be steepest between A and B

(b) The average rate of change between A and B is greater than the instantaneous rate of change at B.

(c)

Suggested Homework Assignment

Pages 118–120: 1, 3, 5, 7–29 odd, 35, 39–53 odd, 55, 59, 65, 75, 85–89 odd, 91, 95, and 97.

Section 2.3

Product and Quotient Rules and Higher-Order Derivatives

Section Comments

2.3 Product and Quotient Rules and Higher-Order Derivatives—Find the derivative of a function using the Product Rule. Find the derivative of a function using the Quotient Rule. Find the derivative of a trigonometric function. Find a higher-order derivative of a function.

Teaching Tips

Some students have difficulty simplifying polynomial and rational expressions. Students should review these concepts by studying Appendices A.2–A.4 and A.7 in Precalculus, 10th edition, by Larson.

When teaching the Product and Quotient Rules, give proofs of each rule so that students can see where the rules come from. This will provide mathematics majors a tool for writing proofs, as each proof requires subtracting and adding the same quantity to achieve the desired results. For the Project Rule, emphasize that there are many ways to write the solution. Remind students that there must be one derivative in each term of the solution. Also, the Product Rule can be extended to more that just the product of two functions. Simplification is up to the discretion of the instructor. Examples such as f (x) = (2x2 3x)(5x3 + 6) can be done with or without the Product Rule. Show the class both ways.

After the Quotient Rule has been proved to the class, give students the memorization tool of LO d HI – HI d LO. This will give students a way to memorize what goes in the numerator of the Quotient Rule.

Some examples to use are f (x) = 2x

x and g(

) = 4 (

x) 3 x

Save f (x) for the next section as this will be a good example for the Chain Rule. g(x) is a good example for first finding the least common denominator.

How Do You See It? Exercise

Page 132, Exercise 120 The figure shows the graphs of the position, velocity, and acceleration functions of a particle.

(a) Copy the graphs of the functions shown. Identify each graph. Explain your reasoning. To print an enlarged copy of the graph, go to MathGraphs.com

(b) On your sketch, identify when the particle speeds up and when it slows down. Explain your reasoning.

s position

v velocity function a acceleration function

(b) The speed of the particle is the absolute value of its velocity. So, the particle’s speed is slowing down on the intervals (0, 4 3), and (8 3, 4) and it speeds up on the intervals (4 3, 8 3) and (4, 6)

Suggested Homework Assignment

Pages 129–132: 1, 3, 9, 13, 19, 23, 29–55 odd, 59, 61, 63, 75, 77, 91–107 odd, 111, 113, 117, and 131–135 odd.

Section 2.4 The Chain Rule

Section Comments

2.4 The Chain Rule—Find the derivative of a composite function using the Chain Rule. Find the derivative of a function using the General Power Rule. Simplify the derivative of a function using algebra. Find the derivative of a trigonometric function using the Chain Rule.

Teaching Tips

Begin this section by asking students to consider finding the derivative of F(x) = √x2 + 1 F is a composite function. Letting y = f (u) = √u and u = g(x) = x2 + 1, then y = F(x) = f (g(x)) or F = f ∘ g When stating the Chain Rule, be sure to state it using function notation and using Leibniz notation as students will see both forms when studying other courses with other texts. Following the definition, be sure to prove the Chain Rule as done on page 134.

Be sure to give examples that involve all rules discussed so far. Some examples include:

f (x) = (sin(6x))4 , g(x) = ( 3 + sin(2x) 3 √x + 3 )2 , and h(x) = (√x 2 x ) ∙ [8x + cos(x2 + 1)]3

You can use Exercise 98 on page 141 to review the following concepts:

• Product Rule

• Chain Rule

• Quotient Rule

• General Power Rule

Students need to understand these rules because they are the foundation of the study of differentiation.

Use the solution to show students how to solve each problem. As you apply each rule, give the definition of the rule verbally. Note that part (b) is not possible because we are not given g′(3)

Solution

(a)

f (x) = g(x)h(x)

f ′(x) = g(x)h′(x) + g′(x)h(x)

f ′(5) = ( 3)( 2) + (6)(3) = 24

(b) f (x) = g(h(x))

f ′(x) = g′(h(x))h′(x)

f ′(5) = g′(3)( 2) = 2g′(3)

Not possible. You need g′(3) to find f ′(5).

f ′(x) = h(x)g′(x) g(x)h′(x) [h(x)]2

f ′(x) = (3)(6) ( 3)( 2) (3)2 = 12 9 = 4 3

(d) f (x) = [g(x)]3

f ′(x) = 3[g(x)]2g′(x)

f ′(5) = 3( 3)2(6) = 162

How Do You See It? Exercise

Page 142, Exercise 106 The cost C (in dollars) of producing x units of a product is C = 60x + 1350 For one week, management determined that the number of units produced x at the end of t hours can be modeled by x = 1 6t3 + 19t2 0 5t 1 The graph shows the cost C in terms of the time t 1 2345 Time (in hours)

of Producing a Product

C t

(a) Using the graph, which is greater, the rate of change of the cost after 1 hour or the rate of change of the cost after 4 hours?

(b) Explain why the cost function is not increasing at a constant rate during the eight-hour shift.

Solution

(a) According to the graph, C ′(4) > C ′(1).

(b) Answers will vary.

Suggested Homework Assignment

Pages 140–143: 1–53 odd, 63, 67, 75, 81, 83, 91, 97, 121, and 123.

Section 2.5 Implicit Differentiation

Section Comments

2.5 Implicit Differentiation—Distinguish between functions written in implicit form and explicit form. Use implicit differentiation to find the derivative of a function.

Teaching Tips

Material learned in this section will be vital for students to have for related rates. Be sure to ask students to find dy dx when x = c.

You can use the exercise below to review the following concepts:

• Finding derivatives when the variables agree and when they disagree

• Using implicit differentiation to find the derivative of a function

Determine if the statement is true. If it is false, explain why and correct it. For each statement, assume y is a function of x

(a) d dx cos(x 2) = 2x sin(x 2)

(b) d dy cos( y2) = 2y sin( y2)

Implicit differentiation is often difficult for students, so as you review this concept remind students to think of y as a function of x Part (a) is true, and part (b) can be corrected as shown below. Part (c) requires implicit differentiation. Note that the result can also be written as 2y sin( y2) dy dx .

Solution

(a) True

(b) False. d dy cos( y2) = 2y sin( y2).

A good way to teach students how to understand the differentiation of a mix of variables in part (c) is to let g = y Then g ′ = y ′ So, d dx cos( y2) = d dx cos( g2) = sin ( g2) ∙ 2gg ′ = sin( y2) ∙ 2y y ′

How Do You See It? Exercise

Page 151, Exercise 70 Use the graph to answer the questions.

(a) Which is greater, the slope of the tangent line at x = 3 or the slope of the tangent line at x = 1?

(b) Estimate the point(s) where the graph has a vertical tangent line.

Solution

(a) The slope is greater at x = 3.

(b) The graph has vertical tangent lines at about ( 2, 3) and (2, 3)

Suggested Homework Assignment

Pages 149–150: 1–17 odd, 25–35 odd, 53, and 61.

Section 2.6 Related Rates

Section Comments

2.6 Related Rates—Find a related rate. Use related rates to solve real-life problems.

Teaching Tips

Begin this lesson with a quick review of implicit differentiation with an implicit function in terms of x and y differentiated with respect to time. Follow this with an example similar to Example 1 on page 152, outlining the step-by-step procedure at the top of page 153 along with the guidelines at the bottom of page 153. Be sure to tell students, that for every related rate problem, to write down the given information, the equation needed, and the unknown quantity. A suggested problem to work out with the students is as follows:

A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 foot per second, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall?

Be sure to go over a related rate problem similar to Example 5 on page 155 so that students are exposed to working with related rate problems involving trigonometric functions.

How Do You See It? Exercise

Page 159, Exercise 34 Using the graph of f, (a) determine whether dy dt is positive or negative given that dx dt is negative, and (b) determine whether dx dt is positive or negative given that dy dt is positive. Explain.

(i) (a) dx dt negative ⇒ dy dt positive

(b) dy dt positive ⇒ dx dt negative

(ii) (a) dx dt negative ⇒ dy dt negative

(b) dy dt positive ⇒ dx dt positive

Suggested Homework Assignment

Pages 157–160: 1, 7, 11, 13, 15, 17, 21, 25, 29, and 41.

Chapter 2 Project

Timing a Handoff

You are a competitive bicyclist. During a race, you bike at a constant velocity of k meters per second. A chase car waits for you at the ten-mile mark of a course. When you cross the ten-mile mark, the car immediately accelerates to catch you. The position function of the chase car is given by the equation s(t) = 15 4 t 2 5 12 t 3 , for 0 ≤ t ≤ 6, where t is the time in seconds and s is the distance traveled in meters. When the car catches you, you and the car are traveling at the same velocity, and the driver hands you a cup of water while you continue to bike at k meters per second.

Exercises

1. Write an equation that represents your position s (in meters) at time t (in seconds).

2. Use your answer to Exercise 1 and the given information to write an equation that represents the velocity k at which the chase car catches you in terms of t.

3. Find the velocity function of the car.

4. Use your answers to Exercises 2 and 3 to find how many seconds it takes the chase car to catch you.

5. What is your velocity when the car catches you?

6. Use a graphing utility to graph the chase car’s position function and your position function in the same viewing window.

7. Find the point of intersection of the two graphs in Exercise 6. What does this point represent in the context of the problem?

8. Describe the graphs in Exercise 6 at the point of intersection. Why is this important for a successful handoff?

9. Suppose you bike at a constant velocity of 9 meters per second and the chase car’s position function is unchanged.

(a) Use a graphing utility to graph the chase car’s position function and your position function in the same viewing window.

(b) In this scenario, how many times will the chase car be in the same position as you after the 10-mile mark?

10. Suppose you bike at a constant velocity of 8 meters per second and the chase car’s position function is unchanged.

(a) Use a graphing utility to graph the chase car’s position function and your position function in the same viewing window.

(b) In this scenario, how many times will the chase car be in the same position as you after the ten-mile mark?

(c) In this scenario, why might it be difficult for the driver of the chase car to successfully handoff a cup of water to you? Explain.

Section2.3

CHAPTER2 Differentiation

CHAPTER2

Differentiation

Section2.1TheDerivativeandtheTangentLineProblem

1. Theproblemoffindingthetangentlineatapoint P is essentiallyfindingtheslopeofthetangentlineatpoint P.Todosoforafunction f,if f isdefinedonanopen intervalcontaining c,andifthelimit ()()

00 limlim Δ→Δ→ +Δ− Δ == ΔΔxx fcxfc y m xx

exists,thenthelinepassingthroughthepoint () () , Pcfc withslope m isthetangentlinetothegraphof f atthe point P

2. Somealternativenotationsfor () f x ′ are () ,,,dydyfx dxdx ′

and [ ] Dx y

3. Thelimitusedtodefinetheslopeofatangentlineisalso usedtodefinedifferentiation.Thekeyistorewritethe differencequotientsothat Δx doesnotoccurasafactor ofthedenominator.

4. Ifafunction f isdifferentiableatapoint, x c = then f iscontinuousat. x c = Theconverseisnottrue.Thatis, afunctioncouldbecontinuousatapoint,butnot differentiablethere.Forexample,thefunction yx = is continuousat0, x = butisnotdifferentiablethere.

5. At () 11,,slope0. = xy

At () 22 5 2 ,,slope.xy =

6. At () 11 2 3 ,,slope. = xy

At () 22 2 5 ,,slope.xy =−

7. (a)–(c)

8. (a) ()() ()() 4152 1 413 4354.75 0.25 431 ff ff == ≈= So, ()() ()() 4143 4143 ffff >

(b)Theslopeofthetangentlineat ()1,2equals ()1. f ′ Thisslopeissteeperthantheslopeoftheline through ()1,2and ()4,5.So, ()() () 41 1. 41 ff f ′ <

9. () 35 f xx =− isaline.Slope5 =−

10. () 3 21gxx=+ isaline.3 Slope2 =

11. () ()() ()() ()() () () 0 22 0 2 0 2 0 0 22 Slopeat2,5lim 223223 lim 24435 lim 82 lim lim828 x x x x x fxf x x x xx x xx x x Δ→ Δ→ Δ→ Δ→ Δ→ +Δ− = Δ   +Δ−−−

= Δ

+Δ+Δ−− = Δ Δ+Δ = Δ =+Δ=

12. () ()() () ()() ()() 0 2 0 2 0 2 0 0 33 Slopeat3,4lim lim53(4) 5964 lim 6 lim lim(6)6 x x x x x fxf x x x xx x xx x x Δ→ Δ→

13. () ()() ()() () 0 2 0 0 00 Slopeat0,0lim 30 lim lim33 t t t ftf t tt t t Δ→ Δ→ Δ→ +Δ− = Δ Δ−Δ− = Δ =−Δ= 6 5 4 3 2 123456 1 y x f

(d) ()() ()() () () 41 11 41 3 12 3 112 1 ff yxf x x x =−+

14. () ()() () ()() 0 2 0 2 0 2 0 0 11 Slopeat1,5lim

15. () () ()() 0 0 0 7 lim 77 lim lim00 x x x fx f xxfx fx

16. () () ()() () 0

18. () () ()() ()() () 0 0 0 0 0 73 lim 7373 lim 77373 lim 7 lim lim77 x x x x x fxx fxxfx fx x xxx x xxx x x x

19. () () ()() () 0 0 0 0 2 3 3 lim 22 33 lim33 222 33 lim333 2 lim32 3 s s s s hss hsshs hs s s ss

17. () () ()() ()() () 0 0 0 0 0 5 lim 55 lim 555 lim 5 lim lim55 x x

x x fxx f xxfx fx x x xx x xxx x x

()

20. () () ()() () ()

22. ()

23. ()

24. () ()

25. () () ()() ()() ()() ()() ()() ()

27. () () ()() ()()

28. () () ()() () () () () 0 0 0 0 0 2 lim 22 lim 2 lim 2 lim 2 lim 21 2 s s s s s hss hsshs hs s sss s sss s ss s s ss sss ssss sss ss

29. (a)2()3fxx=+ () ()() () () () () ()

At ()1,4,theslopeofthetangentlineis () 212. m =−=−

Theequationofthetangentlineis

30. (a)2()21 fxxx=+−

At ()1,2,theslopeofthetangentlineis () 2124. m =+=

Theequationofthetangentlineis

(c)Graphingutilityconfirms () 4at1,2. dy dx =

31. (a) () () ()() () ()() ()() () () 3 33 00 3223 0 223 0 2 22 0 limlim 33 lim 33

At ()2,8,theslopeofthetangentis () 2 3212. m ==

Theequationofthetangentlineis ()8122 81224 1216. yx yx yx −=− −=−

32. (a) () () ()() () () ()()() ()() 3 0 33

23 323

3311

At ()1,0,theslopeofthetangentlineis () 2 313. m =−=

Theequationofthetangentlineis () 031 33. yx yx −=+ =+

33. (a) ()

At ()1,1,theslopeofthetangentlineis 11 212 m ==

Theequationofthetangentlineis () 1 11 2 11 1 22 11 22 −=− −=− =+ yx yx yx

(b)

(c)Graphingutilityconfirmsat

(b)

(c)Graphingutilityconfirms at (b)

(c)Graphingutilityconfirmsat

34. (a) ()

Theequationofthetangentlineis

35. (a) ()

At ()4,5,theslopeofthetangentlineis () 2 43 1. 44 m =−=

Theequationofthetangentlineis () 3 54 4 3 53 4 3 2. 4 yx yx yx +=+ +=+ =−

(c)Graphingutilityconfirms at (5,2) 2 4 10 4 (b)

36. (a) ()

(b) (c)Graphingutilityconfirms at 5 4 4 5 (1,0)

At ()1,0,theslopeofthetangentlineis ()12. mf ′ == Theequationofthetangentlineis () 021 22. yx yx −=− =−

37. Usingthelimitdefinitionofaderivative, () 1 2 f xx ′ =−

Becausetheslopeofthegivenlineis1,youhave 1 1 2 2. x x −=− =

Atthepoint ()2,1,thetangentlineisparallelto 0. xy+= Theequationofthislineis

()() 112 1. yx yx −−=−− =−+

38. Usingthelimitdefinitionofderivative, () 4. f xx ′ = Becausetheslopeofthegivenlineis–4,youhave 44 1. x x =− =−

Atthepoint ()1,2thetangentlineisparallelto 430. xy++= Theequationofthislineis

() 241 42. yx yx −=−+ =−−

39. FromExercise31weknowthat () 2 3. f xx ′ = Becausetheslopeofthegivenlineis3,youhave 2 33 1. x x = =±

Therefore,atthepoints ()1,1and ()1,1thetangent linesareparallelto310. xy−+= Theselineshaveequations ()() 131and131 3232. yxyx yxyx −=−+=+ =−=+

40. Usingthelimitdefinitionofderivative, () 2 3. f xx ′ = Becausetheslopeofthegivenlineis3,youhave 2 2 33 11. x xx = =  =±

Therefore,atthepoints ()1,3and ()1,1thetangent linesareparallelto340. xy−−= Theselineshave equations ()() 331and131 334. yxyx yxyx −=−−=+ ==+

41. Usingthelimitdefinitionofderivative, () 1 2 fx x x ′ = Becausetheslopeofthegivenlineis1,2youhave 11

22 1. xx x −=− =

Therefore,atthepoint ()1,1thetangentlineisparallelto 260.xy+−= Theequationofthislineis

42. Usingthelimitdefinitionofderivative, () ()32 1 . 21 fx x ′ = Becausetheslopeofthegivenlineis1,2youhave () () 32 32 11 212 11 112. x x

Atthepoint ()2,1,thetangentlineisparallelto 270.xy++= Theequationofthetangentlineis

43. Theslopeofthegraphof f is1forall x-values.

45. Theslopeofthegraphof f isnegativefor4, x < positivefor4, x > and0at4. x =

46. Theslopeofthegraphof f is–1for4, x < 1for 4, x > andundefinedat4. x =

44. Theslopeofthegraphof f is0forall x-values.

47. Theslopeofthegraphof f isnegativefor0 x < and positivefor0. x > Theslopeisundefinedat0. x =

48. Theslopeispositivefor20 x −<< andnegativefor 02. x << Theslopeisundefinedat2, x =± and0at 0. x =

49. Answerswillvary. Sample answer: yx =−

Thederivativeof =− yx is1. y ′ =− So,thederivative isalwaysnegative.

50. Answerswillvary. Sample answer:32 3 yxx =−

Notethat () 2 3632.yxxxx ′ =−=− So,0 y ′ = at0 x = and2. x =

51. No.Forexample,thedomainof () f xx = is0, x ≥ whereasthedomainof () 1 2 fx x ′ = is0. x >

52. No.Forexample, () 3 f xx = issymmetricwithrespectto theorigin,butitsderivative, () 2 3, f xx ′ = issymmetric withrespecttothe-axis. y

53. ()45 g = becausethetangentlinepassesthrough ()4,5. () 505 4 473 g ′ ==−

54. ()14 h −= becausethetangentlinepassesthrough ()1,4. () () 6421 1 3142 h′ −===

55. () 53 f xx =− and1 c =

56. () 3 f xx = and2 c =−

57. () 2 f xx =− and6 c =

58. () 2 f xx = and9 c =

59. ()02 f = and () 3, fxx ′ =−−∞<<∞ () 32fxx=−+

60. ()()() 04,00;0 fffx ′′ ==< for () 0,0xfx ′ <> for0 x >

Answerswillvary: Sample answer: () 24 fxx=+

61. Let () 00 , x y beapointoftangencyonthegraphof f Bythelimitdefinitionforthederivative, () 42. f xx ′ =− Theslopeofthelinethrough ()2,5and () 00 , x y equalsthederivativeof f at0: x ()() () ()() 0 0 0 000 22 0000 2 00 000 5 42 2 5242 54882 043 0131,3 y x x yxx xxxx xx xxx =− −=−− −−=−+ =−+ =−−  =

Therefore,thepointsoftangencyare ()1,3and ()3,3, andthecorrespondingslopesare2and–2.Theequations ofthetangentlinesare: ()() 522522 2129 yxyx yxyx −=−−=−− =+=−+

62. Let () 00 , x y beapointoftangencyonthegraphof f.

Bythelimitdefinitionforthederivative, () 2. f xx ′ =

Theslopeofthelinethrough () 1,3and () 00 , x y equals thederivativeof f at0: x

Therefore,thepointsoftangencyare ()3,9and ()1,1, andthecorrespondingslopesare6and–2.Theequations ofthetangentlinesare: ()() 361321 6921 yxyx yxyx +=−+=−− =−=−−

At () 1,13xg ′ =−−= andthetangentlineis

() 131or32.yxyx +=+=+

63. (a) ()

+Δ− ′ = Δ +Δ− = Δ +Δ+Δ− = Δ Δ+Δ = Δ =+Δ= x x x x x fxx fxxfx fx x xxx x x xxxx x xxx x xxx

() ()() () () 2 0 22 0 2 22 0 0 0 lim lim 2 lim 2 lim lim22

At () 1,12xf ′ =−−=− andthetangentlineis

() 121or21.yxyx −=−+=−−

At () 0,00xf ′ == andthetangentlineis0. y =

At () 1,12xf ′ == andthetangentlineis

21.yx=−

At () 0,00xg ′ == andthetangentlineis0. y =

At () 1,13xg ′ == andthetangentlineis

() 131or32.yxyx −=−=−

Forthisfunction,theslopesofthetangentlinesare sometimesthesame.

64. (a) ()03 g ′ =− (b) ()30 g ′ =

(c)Because () 8 3 1, g g ′ =− isdecreasing(falling)at 1. x =

(d)Because () 7 3 4, g g ′ −= isincreasing(rising)at 4. x =−

(e)Because () 4 g ′ and () 6 g ′ arebothpositive, () 6 g is greaterthan ()4, g and ()() 640.gg−>

(f)No,itisnotpossible.Allyoucansayisthat g is decreasing(falling)at2. x =

Forthisfunction,theslopesofthetangentlinesare alwaysdistinctfordifferentvaluesof x. (b) () ()() () ()()() ()() () ()() () 0 33 0 23 323 0 22 0 2 22 0 lim lim 33 lim 33 lim lim333 x x x x x gxxgx gx x xxx x x xxxxxx x xxxxx x xxxxx Δ→ Δ→ Δ→ Δ→ Δ→ +Δ− ′ = Δ +Δ− = Δ +Δ+Δ+Δ− = Δ Δ+Δ+Δ = Δ =+Δ+Δ=

65. () 12 2 f xx =

(b)Bysymmetry:

(d)

67. ()()()() () () 22424,2.12.142.13.99 3.994 20.1Exact:20 2.12 ff ff =−==−= ′′ ≈=−

68. () () () () () 13 222,2.12.31525 4 2.315252 23.1525Exact:23 2.12 ff ff === ′′ ≈=  =  

69. () 3221,2fxxxc=++=− () ()() () () 2 32 2 2 2 22 2 2lim 2 211 lim 2 2 limlim4 2 x x xx fxf f x xx x xx x x →− →− →−→− ′ −= + ++− = + + === +

70. () 2,1gxxxc=−= () ()() () 1 2 1 1 1 1 1lim 1 0 lim 1 1 lim 1 lim1 x x x x g xg g x xx x xx x x → → → → ′ = = = ==

71. () ,0gxxc== () ()() 00 0 0limlim. 0 xx x gxg g x x →→ ′ == Doesnotexist.

As 1 0,. x x x x →=→−∞

As 1 0,. x x x x + →=→∞ Therefore () g x isnotdifferentiableat0. x =

72. () 3 ,4fxc x == () ()() () () () 4 4 4 4 4 33 4 4 4lim 4 lim 4 123 lim 44 34 lim 44 33 lim 416 x x x x x x fxf f x x x xx x xx x → → → → → ′ = = = = =−=−

73. ()() 23 6,6fxxc=−=

() ()() () () 6 23 6613 6 6lim 6 601 limlim. 66 x xx fxf f x x x x → →→ ′ = ==

Doesnotexist.

Therefore () f x isnotdifferentiableat6. x =

74. ()()13 3,3gxxc=+=−

() ()() () () () 3 13 3323 3 3lim 3 301 limlim. 33 x xx gxg g x x x x →− →−→− ′ −= +− == + + Doesnotexist.

Therefore () g x isnotdifferentiableat3. x =−

75. () 7,7hxxc=+=− () ()() () 7 77 7 7lim 7 707 limlim. 77 x xx hxh h x xx xx →− →−→− ′ −= +−+ == ++

Doesnotexist. Therefore ()hx isnotdifferentiableat7. x =−

76. () 6,6fxxc=−= () ()() 6 66 6 6lim 6 606 limlim. 66 x xx fxf f x xx xx → →→ ′ = == Doesnotexist. Therefore () f x isnotdifferentiableat6. x =

77. () f x isdifferentiableeverywhereexceptat4. x =− (Sharpturninthegraph)

78. () f x isdifferentiableeverywhereexceptat2. x =± (Discontinuities)

79. () f x isdifferentiableontheinterval () 1,.−∞ (At 1=− x thetangentlineisvertical.)

80. () f x isdifferentiableeverywhereexceptat0. x = (Discontinuity)

81. () 5 fxx=− isdifferentiable everywhereexceptat5. = x Thereisasharpcornerat 5. x =

82. () 4 3 x fx x = isdifferentiable everywhereexceptat3. x = f isnotdefinedat3. x = (Verticalasymptote)

83. () 25 f xx = isdifferentiable forall0. x ≠ Thereisasharp cornerat0. x =

84. f isdifferentiableforall1. x ≠ f isnotcontinuousat1. x =

85. () 1 fxx=−

Thederivativefromtheleftis ()() 11 110 limlim1. 11xx x fxf xx→→ ==−

Thederivativefromtherightis ()() 11 110 limlim1. 11xx x fxf xx++ →→ ==

Theone-sidedlimitsarenotequal.Therefore, f isnot differentiableat1. x =

86. () 12 f xx =−

Thederivativefromtheleftdoesnotexistbecause ()() 2 11 22 12 12 110limlim 11 11 lim 11 1 lim. 1 xx x x fxf x xx x x x x x x →→ → → = =⋅ + =−=−∞ (Verticaltangent)

Thelimitfromtherightdoesnotexistsince f is undefinedfor1. x > Therefore, f isnotdifferentiableat 1. x =

87. () () () 3 2 1,1 1,1 xx fx xx  −≤  =  −>  

Thederivativefromtheleftis ()() () () 3 11 2 1 110 limlim 11 lim10. xx x fxfx xx x →→ → = =−=

Thederivativefromtherightis ()() () () 2 11 1 110 limlim 11 lim10. xx x fxfx xx x ++ →→ + → = =−=

Theone-sidedlimitsareequal.Therefore, f is differentiableat1. x = () () 10 f ′ =

88. ()()123 f xx =−

Thederivativefromtheleftdoesnotexist. ()() () () 23 11 13 1 110 limlim 11 1 lim 1 →→ → = ==−∞ xx x fxfx xx x

Similarly,thederivativefromtherightdoesnotexist becausethelimitis ∞

Therefore, f isnotdifferentiableat1. x =

89. Notethat f iscontinuousat2. x = () 21,2 43,2 xx fx xx  +≤

=  −>

Thederivativefromtheleftis ()() () () 2 22 2 215 limlim 22 lim24. xx x x fxf xx x →→ → +− = =+=

Thederivativefromtherightis ()() () 222 2435 limlimlim44. 22 xxx fxfx xx +++ →→→ ===

Theone-sidedlimitsareequal.Therefore, f is differentiableat2. x = () () 24 f ′ =

90. () 1 22,2 2,2 xx fx xx  +<  =  ≥

 f isnotdifferentiableat2 = x becauseitisnot continuousat2. x = () () ()() 2 2 1 lim223 2 lim222 x x fx fx → + → =+= ==

91. (a)Thedistancefrom ()3,1totheline40 mxy−+= is ()() 11 2222 311433 . 11 m AxByCm d ABmm −+ +++ === +++

(b) Thefunction d isnotdifferentiableat1. m =− Thiscorrespondstotheline4, yx=−+ whichpassesthrough thepoint ()3,1.

92. (a) () 2 f xx = and () 2 f xx ′ = (b) () 3 g xx = and () 32 g xx ′ =

(c)Thederivativeisapolynomialofdegree1lessthantheoriginalfunction.If () , hxxn = then () 1 hxnxn ′ =

(d)If () 4 , f xx = then () ()() ()

So,if () 4 , f xx = then () 43 f xx ′ = whichisconsistentwiththeconjecture.However,thisisnotaproofbecauseyou mustverifytheconjectureforallintegervaluesof,2. nn ≥

93. False.Theslopeis ()() 0 22 lim. x fxf x Δ→ +Δ− Δ

94. False.2 yx=− iscontinuousat2, x = butisnot differentiableat2. x = (Sharpturninthegraph)

95. False.Ifthederivativefromtheleftofapointdoesnot equalthederivativefromtherightofapoint,thenthe derivativedoesnotexistatthatpoint.Forexample,if () , f xx = thenthederivativefromtheleftat0 x = is –1andthederivativefromtherightat0 x = is1.At 0, x = thederivativedoesnotexist.

96. True.SeeTheorem2.1.

97. () () sin1,0 0,0 xxx fx x  ≠  =  =

UsingtheSqueezeTheorem,youhave () sin1,0.xxxxx −≤≤≠ So, ()() 0 limsin100 x x xf → == and f iscontinuousat0. x = Usingthealternativeformofthederivative,youhave ()() () 000 0sin101 limlimlimsin. 00 xxx fxfxx x xx→→→  ==

Becausethislimitdoesnotexist( ()sin1 x oscillatesbetween–1and1),thefunctionisnotdifferentiableat0. x = () () 2sin1,0 0,0 xxx gx x  ≠  =

UsingtheSqueezeTheoremagain,youhave () 222sin1,0.xxxxx −≤≤≠ So, ()() 2 0 limsin100 x x xg → == and g iscontinuousat0. x = Usingthealternativeformofthederivativeagain,youhave ()() () 2 000 0sin101 limlimlimsin0. 00 xxx gxgxx x xxx→→→ ===

Therefore, g isdifferentiableat () 0,00.xg ′ == 98.

3 3 1 3

Asyouzoomin,thegraphof211yx=+ appearstobelocallythegraphofahorizontalline,whereasthegraphof 21yx=+ alwayshasasharpcornerat ()0,1.2 y isnotdifferentiableat ()0,1.

Section2.2BasicDifferentiat ionRulesandRatesofChange

1. Thederivativeofaconstantfunctionis0. [] 0 d c dx =

2. Tofindthederivativeof () , n f xcx = multiply n by, c andreducethepowerof x by1. () 1 ′ = n f xncx

3. Thederivativeofthesinefunction, () sin, f xx = isthe cosinefunction, () cos. f xx ′ = Thederivativeofthecosinefunction, () cos, g xx = is thenegativeofthesinefunction, () sin. g xx ′ =−

4. Theaveragevelocityofanobjectisthechangein distancedividedbythechangeintime.Thevelocityis theinstantaneouschangeinvelocity.Itisthederivative ofthepositionfunction.

7. 12 0 = ′ = y y

8. () () 9 0 fx fx =− ′ =

9. 7 76 yx yx = ′ = 10. 12 1211 yx yx = ′ =

11. 5 5 6 6 1 5 5 yx x yx x == ′ =−=−

12. () 7 7 8 8 3 3 21 37 yx x yx x == ′ =−=−

13. () () 919 89 89 11 99 fxxx fxx x == ′ ==

14. 414 34 34 11 44 yxx yx x == ′ ==

15. () () 11 1 fxx fx =+ ′ =

16. () () 63 6 gxx gx =+ ′ =

17. () () 2 326 62 fttt ftt =−+− ′ =−+

18. 231 23 ytt yt =−+ ′ =−

19. () () 23 2 4 212 g xxx g xxx =+ ′ =+

20. 3 2 43 49 yxx yx =− ′ =−

21. () () 32 2 538 3103 stttt sttt =+−+ ′ =+−

22. 32 2 261 612 yxx yxx =+− ′ =+

23. sin 2 cos 2 y y π θ π θ = ′ =

24. () () cos sin g tt g tt π π = ′ =−

25. 21 2 1 2 cos 2sin yxx yxx =− ′ =+

26. () 4 33 72sin 742cos282cos yxx y xxxx =+ ′ =+=+

Function Rewrite Differentiate Simplify

27. 45 45 2288 7777 ′′ ===−=− yyxyxy x x

28. () 544 5 888 58 555 ′′ ==== yyxyxyx x

29. () 34 34 661818 5125125125 yyxyxy x x ′′ ===−=−

30. () () 2 2 3 1212224 2 ′′ ==== yyxyxyx x

31. () () () () 2 2 3 3 8 8,2,2 16 16 22 fxx x fxx x f == ′ =−=− ′ =−

32. () () () () 1 2 2 4 224,4,1 4 4 1 4 4 ftt t ftt t f =−=− ′ == ′ =

33. () () () () 3 2 7 11 252 21 5 ,0, 00 fxx fxx f =−+− ′ = ′ =

34. () () 4 3 23,1,1 8 18 yx yx y =−− ′ = ′ =

35. ()() ()() 2 2 41,0,1 1681 328 032088 =+ =++ ′ =+ ′ =+= yx xx yx y

36. ()()() () () 2 2 24,2,8 21632 416 28168 fxx xx fxx f =− =−+ ′ =− ′ =−=−

37. ()() () ()() 4sin,0,0 4cos1 04113 f f f θθθ θθ =− ′ =− ′ =−=

38. ()() () () 2cos5,,7 2sin 0 gtt gtt g π π =−+ ′ = ′ =

39. () () 22 3 3 53 6 262 fxxx fxxxx x =+−

′ =+=+

40. () () 33 242 4 23 9 32932 fxxxx fxxxx x =−+

′ =−−=−−

41. () () 223 3 4 4 4 4 12 2122 gtttt t gtttt t =−=− ′ =+=+

42. () () 2 2 3 3 3 883 6 868 f xxxx x fxx x =+=+

′ =−=−

43. () () 32 2 2 3 33 34 34 88 1 xx f xxx x x fx xx −+ ==−+

′ =−=

44. () () 3 21 2 2 425 425 5 858 xx hxxx x hxxxx x ++ ==++

′ =−=−

45. () () 2 121232 32 123252 2 52 348 348 3 212 2 3424 2 tt g tttt t gtttt tt t +− ==+−

′ =−+ −+ =

46. () () 5 1432313 13 1131343 5 43 26 26 144 2 33 1446 3 ss hssss s hssss ss s ++ ==++ ′ =+− +− =

47. ()23 2 1 31 yxxxx yx =+=+ ′ =+

48. () () 2243 322 2323 8989 yxxxxx yxxxx =−=−

′ =−=−

49. () () 31213 1223 23 66 112 2 22 fxxxxx fxxx x x =−=− ′ =−=−

50. () () 2313 1323 1323 4 2121 3333 fttt fttt tt =−+ ′ =−=−

51. () () 12 12 65cos65cos 3 35sin5sin f xxxxx f xxxx x =+=+

′ =−=−

52. () () 13 3 43 43 2 3cos23cos 22 3sin3sin 33 fxxxx x f xxxx x =+=+ ′ =−−=−−

53. () () 22 2 1 5cos35cos95cos 3 185sin =−=−=− ′ =+ yxxxxx x yxx

54. () 3 3 4 4 33 2sin2sin 28 9 2cos 8 9 2cos 8 yxxx x yxx x x =+=+ ′ =+ =−+

55. (a) () () 42 3 253 810 fxxx fxxx =−+− ′ =−+

At ()()()() 3 1,0:1811012 f ′ =−+=

Tangentline: () 021 22 yx yx −=− =− (b)and(c)

56. (a)3 2 3 33 yxx yx =− ′ =−

At ()() 2 2,2:3239 y ′ =−=

Tangentline: ()292 916 9160 yx yx xy −=− =− −−= (b)and(c)

57. (a) () () 34 43 74 74 2 2 33 22 fxx x fxx x == ′ =−=−

At ()() 3 1,2:1 2 f ′ =−

Tangentline: () 3 21 2 37 22 3270 yx yx xy −=−− =−+ +−=

(b)and(c)

58. (a) ()()232 2 236 326 yxxxxxx yxx =−+=+− ′ =+−

At ()()() 2 1,4:312161 y ′ −=+−=−

(b)and(c)

(1,2) 10 55 (1,4) 10

59. () ()() 42 3 2 23 44 41 411 00,1 yxx yxx xx xxx yx =−+ ′ =− =− =−+ ′ =  =±

Horizontaltangents: ()()() 0,3,1,2,1,2

60. 3 2 310forall. yxx yxx =+ ′ =+>

Tangentline: ()() 411 3 30 yx yx xy −−=−− =−− ++=

Therefore,therearenohorizontaltangents.

61. 2 2 3 3 1 2 2cannotequalzero. yx x yx x == ′ =−=−

Therefore,therearenohorizontaltangents. 3 33 1 (1,0) 4 3 3 5 (2,2) 7 1 2

62. 29 200 yx yxx =+ ′ ==  =

At0,1. xy==

Horizontaltangent: () 0,9

63. sin,02 1cos0 cos1 yxxx yx xx

At:

Horizontaltangent: () , ππ

64. 32cos,02 32sin0 32 sinor 233 yxxx yx xx π

66. () 2,23fxkxyx==−+

() 2 f xkx ′ = andslopeoftangentlineis2. m =− () 2 22 1 fx kx x

Horizontaltangents:

65. () 2,61fxkxyx=−=−+ () 2 f xx ′ =− andslopeoftangentlineis6. m =− () 6 26 3 fx x

67. () 3 ,3 4 k fxyx x ==−+ () 2 k fx x ′ =− andslopeoftangentlineis3 4 m =−

68. () ,4fxkxyx==+ () 2 k fx x ′ = andslopeoftangentlineis1. m = () 2 1 1 2 2 4 ′ = = = = fx

74. If f isquadratic,thenitsderivativeisalinearfunction. () () 2 2 f xaxbxc fxaxb =++ ′ =+

75. Thegraphofafunction f suchthat0 f ′ > forall x and therateofchangeofthefunctionisdecreasing (i.e.,as x increases, ′ f decreases)would,ingeneral,look likethegraphbelow.

69. ()()()() 6 g xfxgxfx ′′ =+  =

70. ()()()() 22 g xfxgxfx ′′ =  =

71. ()()()() 55 g xfxgxfx ′′ =−  =−

72. ()()()() 313 g xfxgxfx ′′ =−  =

73.

If f islinearthenitsderivativeisaconstantfunction. () () f xaxb fxa =+ ′ =

76. (a)Theslopeappearstobesteepestbetween A and B (b)Theaveragerateofchangebetween A and B is greater thantheinstantaneousrateofchangeat B (c)

77. Let () 11 , x y and () 22 , x y bethepointsoftangencyon2 yx = and265, yxx=−+− respectively.

Thederivativesofthesefunctionsare:

Because2 11 yx = and222265:yxx=−+− ()() 22 21221 2 2121 65 26 xxx yy mx xxxx −+−− ===−+

22110,2xyx =  == and14 y = So,thetangentlinethrough ()1,0and ()2,4isSo,thetangentlinethrough ()2,3and ()1,1is

22123,1xyx =  == and11 y =

78. 1m istheslopeofthelinetangentto2 yxm = istheslopeofthelinetangentto1. yx = Because 1222 111 11and.yxymyym x xx ′′ =

Thepointsofintersectionof yx = and1yx = are 1211.xxx x =

At21,1.xm=±=− Because211,mm =− thesetangentlinesareperpendicularatthepointsofintersection.

79. () () 3sin2 3cos fxxx fxx =++ ′ =+

Because () cos1,0 xfx ′ ≤≠ forall x and f doesnot haveahorizontaltangentline.

80. () () 53 42 35 595 f xxxx fxxx =++ ′ =++

Because () 42 590,5. xxfx ′ +≥≥ So, f doesnot haveatangentlinewithaslopeof3.

81. () () () 12 ,4,0 11 22 10 24 42 42 42 4,2 fxx fxx x y x x xxy xxx xx xy =−

Thepoint ()4,2isonthegraphof f.

Tangentline: () 02 24 44 484 044 yx yx xy −=− −=− =−+

83. (a)Onepossiblesecantisbetween () 3.9,7.7019and ()4,8:

82. () () () 2 2 ,5,0 2 fx x fx x = ′ =− 2 2 2 20 5 102 2 102 1022 410 54 , 25 y xx xxy xx x xx x xy −=

Thepoint 54 , 25

isonthegraphof f.Theslopeofthe tangentlineis 58 225 f

Tangentline:485 5252 2520820 825400 yx yx xy

(b) () () () ()() 12 33 423 22 34834 fxxf

Theslope(andequation)ofthesecantlineapproachesthatofthetangentlineat ()4,8asyouchoosepointscloserandcloserto ()4,8.

(c)Asyoumovefurtherawayfrom ()4,8,theaccuracyoftheapproximation T getsworse.

(d)–3–2–1–0.5–0.100.10.5123 12.8285.1966.5487.70288.

84. (a)Nearbypoint: () 1.0073138,1.0221024 () () 1.02210241

Secantline:11 1.00731381 3.02211 yx yx −=− =−+ (Answerswillvary.)

(b) () ()() 32 31132 fxx Txxx ′ = =−+=−

(c)Theaccuracyworsensasyoumoveawayfrom ()1,1.

(d) TheaccuracydecreasesmorerapidlythaninExercise85because3 yx = isless“linear”than32 yx =

–3–2–1–0.5–0.100.10.5123 –8–100.1250.72911.3313.37582764

–8–5–2–0.50.711.32.54710

85. False.Let () f xx = and () 1. gxx=+ Then ()() , f xgxx ′′ == but ()() f xgx ≠

86. True.If2, a yxbx + =+ then () () () 211 22. a a dy axbaxb dx +− + =++=++

87. False.If2, y π = then0. dydx = (2 π isaconstant.)

88. True.If ()() , f xgxb =−+ then ()()() 0. f xgxgx ′′′ =−+=−

89. False.If () 0, fx = then () 0 fx ′ = bytheConstantRule.

90. False.If () 1 , n n f xx x == then () 1 1 n n n fxnx x + ′ =−=

91. () [ ]35,1,2=+ftt () 3. ′ = ft So, ()() 123.′′ ==ff

Instantaneousrateofchangeistheconstant3.

Averagerateofchange: ()() 21118 3 211 == ff

92. () [ ] () 27,3,3.1 2 ftt ftt =− ′ = Instantaneousrateofchange: At ()() 3,2:36 f ′ = At ()() 3.1,2.61:3.16.2 f ′ =

Averagerateofchange: ()() 3.132.612 6.1 3.130.1 ff ==

93. () [] () 2 1 ,1,2 1 fx x fx x =− ′ =

Instantaneousrateofchange: ()() () 1,111 11 2,2 24 f f ′  =

′  =

Averagerateofchange: ()() ()() 211211 21212 ff ==

94. () () sin,0, 6 cos fxx fxx π  = 

′ =

Instantaneousrateofchange: ()() 0,001 13 ,0.866 6262 f f

Averagerateofchange: ()() () () () 601203 0.955 6060 ff π πππ ==≈

95. (a) () () 2 161362 32 stt vtt =−+ =−

(b) ()() 21 1298134648ft/sec 21 ss =−=−

When () 1:132ft/sectv==− When () 2:264ft/sectv==−

(d)2 2 1613620 13621362 9.226sec 164 t tt −+= =  =≈ (e) 13621362 32 44 81362295.242ft/sec v

96. () () () () () ()() ()() 2 2 2 1622220 3222 3118ft/sec 1622220 112heightafterfalling108ft 16221080 228270 2 232222 86ft/sec sttt vtt v sttt tt tt t v =−−+ =−− =− =−−+ = −−+= −−+=

97. () () ()() ()() 2 00 2 4.9 4.9120 9.8120 59.8512071m/sec 109.81012022m/sec sttvts tt vtt v v =−++ =−+ =−+ =−+= =−+=

98. (a) () ()() 22 00 4.94.9214 9.8 sttvtst stvtt =−++=−+ ′ ==−

(b) ()() 52 Averagevelocity52 91.5194.4 3 34.3m/sec = = =− ss

(c) ()() ()() 29.8219.6m/sec 59.8549.0m/sec s s ′ =−=− ′ =−= (d) () 2 2 2 4.92140 4.9214 214 4.9 6.61sec =−+= = = ≈ stt t t t (e) ()() 6.619.86.6164.8m/sec v =−≈−

99. From ()0,0to ()() () 11 22 4,2,mi/min. sttvt =  = () () 1 26030mi/h vt == for04 t <<

Similarly, () 0 vt = for46. t << Finally,from () 6,2 to () 10,6, ()()41mi/min.60mi/h.sttvt =−  ==

(Thevelocityhasbeenconvertedtomilesperhour.)

100. ThisgraphcorrespondswithExercise101.

Time(inminutes) s

101. 32 ,3 dV Vss ds ==

When3 6cm,108cm dV s ds == percmchangein s t Time(inminutes) 246810 10 20 30 40 50 60 Velocity (in mi/h) v t 246810 2 6 10 4 8 (0,0) (4,2)(6,2) (10,6) Distance (in miles)

102. 2,2 dA A ss ds ==

When6m, s = 12m2 dA ds = permchangein s

103. (a)Usingagraphingutility, () 0.4170.02.Rvv=−

(b)Usingagraphingutility, () 2 0.00560.0010.04.Bvvv=++

104. ()() () 2 gallonsoffuelusedcostpergallon 15,00052,200 3.48 52,200 C xx dC dxx =

(d)

(e)0.01120.418 dT v dv =+

For () 40,400.866vT ′ =≈

For () 80,801.314vT ′ =≈

For () 100,1001.538vT ′ =≈

(f)Forincreasingspeeds,thetotalstoppingdistance increases.

Thedriverwhogets15milespergallonwouldbenefitmore.Therateofchangeat15 x = islargerinabsolutevaluethanthat at35. x =

105. () 12 2 s tatc =−+ and () s tat ′ =−

107. 2 yaxbxc =++

Becausetheparabolapassesthrough ()0,1and ()1,0,youhave: ()()() ()()() 2 2 0,1:1001 1,0:01111 abcc abba =++  = =++  =−−

So, () 211.yaxax=+−−+ Fromthetangentline1, yx=− youknowthatthederivativeis1atthepoint ()1,0. () ()() 21 1211 11 2 13 yaxa aa a a ba ′ =+−− =+−− =− = =−−=−

Therefore,2231.yxx=−+

108. 2 1 ,0 1 yx x y x => ′ =−

At (),,ab theequationofthetangentlineis () 22 112 or. x yxay aaaa −=−−=−+

The x-interceptis () 2,0. a The y-interceptis 2 0,. a 

Theareaofthetriangleis () 112 22. 22 Abha a  ===

109. 3 2 9 39 yxx yx =− ′ =−

Tangentlinesthrough () 1,9: ()() () () 2 332 322 3 2 9391 993399 02323 0or yxx xxxxx xxxx xx +=−− −+=−−+ =−=− ==

Thepointsoftangencyare ()0,0and () 381 28,.At ()0,0,theslopeis ()09. y ′ =− At () 381 28,,theslopeis ()39 24 y ′ =−

TangentLines: () ()8193 842 927 44 090and 9 9094270 −=−−+=−− =−=−− +=++= yxyx yxyx xyxy x 123 1 2 ) a,(( ,)=a b 1 a y

110. 2 2 yx yx = ′ =

(a)Tangentlinesthrough ()0,: a () 22 2 20 2 yaxx xax ax ax −=− −= −= ±−=

Thepointsoftangencyare () ,.aa±−− At (),,aa theslopeis () 2.yaa ′ −=− At (),,aa theslopeis () 2.yaa ′ −−=−− () ()

Tangentlines:2and2 22 yaaxayaaxa yaxayaxa +=−−−+=−−+− =−+=−−+

Restriction: a mustbenegative.

(b)Tangentlinesthrough (),0:a () () 22 2 02 22 022 yxxa xxax x axxxa −=− =− =−=−

Thepointsoftangencyare ()0,0and () 2 2,4. aa At ()0,0,theslopeis ()00. y ′ = At () 2 2,4, aa theslopeis ()24.

Restriction: None, a canbeanyrealnumber. 111. () 3 2 ,2 ,2 axx fx xbx  ≤

f mustbecontinuousat2 x = tobedifferentiableat2. x = () () () 3 22 2 22 limlim8 84 limlim484 xx xx fxaxa ab ab fxxbb

() 2 3,2 2,2 axx fx xx

For f tobedifferentiableat2, x = theleftderivativemustequaltherightderivative. ()() 2 1 3 4 3 3222 124 84 a a a ba = = = =−=−

112. () cos,0 ,0 xx fx axbx <  =  +≥ 

()() 0cos011 fbb ===  = () sin,0 ,0 xx fx ax −< ′ =  > 

So,0. a =

Answer:0,1 ab==

113. () 1sinf xx = isdifferentiableforall, x nn π ≠ an integer.

() 2sin f xx = isdifferentiableforall0. x ≠

Youcanverifythisbygraphing1 f and2 f andobserving thelocationsofthesharpturns.

114. Let () cos. f xx = () ()() () () 0 0 00 lim limcoscossinsincos coscos1sinlimlimsin 0sin1sin x x xx fxxfx fx x

=−=−

115. Youaregiven: f RR → satisfying

()() ()() * f xnfx fx n +− ′ = forallrealnumbers x and allpositiveintegers n.Youclaimthat () ,,. f xmxbmbR =+∈ Forthiscase, () () [ ]mxnbmxb f xmm n

++  −+  ′ ===

Furthermore,thesearetheonlysolutions:

Notefirstthat () ()() 21 1, 1 fxfx fx +−+ ′ += and ()()() 1. f xfxfx ′ =+− From ()*youhave ()()() ()()()() ()() 22 211 1. fxfxfx f xfxfxfx fxfx ′ =+− =  +−+

Thus, ()()1. fxfx ′′ =+

Let ()()() 1. g xfxfx =+−

Let ()()() 010.mgff ==−

Let ()0. bf = Then ()()() ()() ()()()() () 10 constant0 1 gxfxfx gxgm f xfxfxgxm fxmxb ′′′=+−=

=+−==

Section2.3ProductandQuotientRulesandHigher-OrderDerivatives

1. Tofindthederivativeoftheproductoftwodifferentiable functions f and g,multiplythefirstfunction f bythe derivativeofthesecondfunction g,andthenaddthe secondfunction g timesthederivativeofthefirst function f.

2. Tofindthederivativeofthequotientoftwodifferentiable functions f and, g where()0, gx ≠ multiplythe denominatorbythederivativeofthenumeratorminus thenumeratortimesthederivativeofthedenominator, allofwhichisdividedbythesquareofthedenominator.

3. 2 2 tansec cotcsc secsectan csccsccot d xx dx d xx dx d x xx dx d x xx dx = =− = =−

4. Higher-orderderivativesaresuccessivederivativesofa function.

5. ()()() ()()()()() 2315 235152 1015210 2017 =−− ′ =−−+− =−++− =−+ gxxx gxxx xx x 6. ()() ()()()() 3 23 323 32 345 34353 912315 121215 yxx yxxx xxx xx =−+ ′ =−++ =−++ =−+ 7. () () () () () () 2122 12212 3232 12 32 12 22 12 11 1 21 2 11 2 22 51 22 1515 22 =−=− ′ =−+− =−+− =−+ == httttt httttt tt t t t tt t t

8. () () () () () () 2122 12212 323212 32 12 2 88 1 28 2 1 24 2 54 2 58 2 gsssss gsssss sss s s s s =+=+ ′ =++ =++ =+ + =

9. () ()() () () 3 32 23 2 cos sincos3 3cossin 3cossin fxxx f xxxxx xxxx xxxx = ′ =−+ =− =−

10. () () sin 1 cossin 2 1 cossin 2 gxxx gxxxx x x xx x =  ′ =+

11. () () ()()() () () () 222 5 51155 555 x fx x xx xx fx xxx = ′ ===−

12. () () ()() ()() () () () 2 2 2 22 2 2 2 31 25 256312 25 123062 25 6302 25 t gt t ttt gt t ttt t tt t = + +−− ′ = + +−+ = + ++ = +

13. () () () () () () () 12 33 312122 32 33 1232 3 32 11 1 13 2 1 16 21 15 21 == ++ +− ′ = + +− = + = + xx hx xx x xxx hx x xx xx x xx

14. () 2 21 x fx x = + () ()() () () () () () () 212 2 3232 2 32 2 2 212 21 42 21 32 21 32 21 +− ′ = + +− = + + = + + = + xxxx fx x xxx x xx x xx x

15. () () ()() () 2 2 223 sin cossin2cos2sin x gx x xxxx x xx gx x x = ′ == 16. () () () () () 3 32 324 cos sincos3sin3cos t ft t tttt ttt ft t t = + ′ ==−

17. () ()() () ()() ()() () 32 322 4234322 432 4325 46232534 62428961512820 158211620 020 =++− ′ =++++−+ =+++++−++− =+++− ′ =− fxxxxx fxxxxxxx xxxxxxxxx xxxx f

18. () ()()

19.

() () ()()()() ()() () () 2 2 2 2 2 sin cossin1 cossin 63212 636 3318 336 x

20.

29.

=−=− ++ +− ′ =− + ++− = + +− = + x fxxx xx xx fx x xx x xx x

x fxxx xx xx x f xxx x x x xx xx xx x x

+−−

() () ()() () () () () () () 44

() () 1212 1232 32 31 3 3131 222 ==− + ′ =+= x fxxx x x fxxx x Alternatesolution: () () ()() () () 12 1212 12 32 3131 1 331 2 1 31 2 31 2 ==

= + = + = xx fx x x xxx fx x xx x x x 34. () () () () () 31312 13121223 1623 1623 33 11 3 23 5 6 51 6 =+=+

Alternatesolution: () () () 35613 1623 1623 33 551 66 =+=+ ′ =+=+ fxxxxx fxxx x x

35.

36. () ()

38. () () ()() () () () () 2 2 2222 222 21 2 11 121221222

39. () ()()() () ()()() ()()() ()()() ()()()() ()()

Note: Youcouldsimplifyfirst: () ()() 32 256fxxxxx=+−−

40. () ()()() () ()()()()()()()()() ()()()()()() ()

41. () ()() 2 2 sin cos2sincos2sin fttt f ttttttttt = ′ =+=+

42. ()() ()()()()() () 1cos 1sincos1 cos1sin f f θθθ θθθθ θθθ =+ ′ =+−+ =−+

43. () () 22 cos sincossincos t ft t tttttt ft tt = −−+ ′ ==−

44. () () () () 3 32 324 sin cossin3cos3sin x fx x xxxx x xx fx x x = ′ ==

45. () () 22 tan 1sectan fxxx f xxx =−+ ′ =−+=

46. 22 cot 1csccot yxx yxx =+ ′ =−=−

47. () () 414 34 34 6csc6csc 1 6csccot 4 1 6csccot 4 g ttttt gtttt tt t =+=+ ′ =− =−

48. () () 1 2 2 1 12sec12sec 12sectan 1 12sectan hxxxx x hxxxx xx x =−=− ′ =−− =−

49. () ()()()() () () () 2 22 2 2 31sin33sin 2cos2cos 3cos2cos33sin2sin 2cos 6cos6sin6sin 4cos 3 1tansectan 2 3 sectansec 2 x x y xx x xxx y x xxx x xxx xxx == ′ = −+− = =−+− =−

50. () 2 2 sec sectansec sectan1 x y x x xxx y x xxx x = ′ = =

51. () 2 2 2 cscsin csccotcos cos cos sin coscsc1 coscot =−− ′ =− =− =− = yxx yxxx x x x xx xx

52. sincos cossinsincos yxxx yxxxxxx =+ ′ =+−=

53. () () () 2 222 tan sec2tansec2tan fxxx f xxxxxxxxx = ′ =+=+

54. () ()()() sincos sinsincoscoscos2 fxxx f xxxxxx = ′ =−+=

55. () () 2 2 2 2sincos 2cos2sinsin2cos 4cos2sin yxxxx yxxxxxxx xxxx =+ ′ =++−+ =+−

56. () () 2 5sectan 5sectan5secsectan h h θθθθθ θθθθθθθθ =+ ′ =+++

57. () () () ()() ()()()() () () 2 2 2 1 25 2 12111 225 22 281 2 +  =−  +   +−+ +  ′  =+−  +   +  +− = + x gxx x xx x gxx x x xx x () Formofanswermayvary.

or duplicated, or posted to a publicly accessible website, in whole or in part.

58. () () ()()()() () () () 2 22 2 2 cos 1sin 1sinsincoscos 1sin sinsincos 1sin 1sin 1sin 1 1sin x fx x x xxx fx x xxx x x x x = ′ = −++ = = = (Formofanswermayvary.)

59. ()()()() () () ()() () 22 2 1csc 1csc 1csccsccot1csccsccot2csccot 1csc1csc 223 43 612 x y x xxxxxx x x y xx y π + = −−−+ ′ ==

60. () () () tancot1 0 10 fxxx fx f == ′ = ′ =

61. () () ()()() () () () 22 22 sec sectansec1sectan1 sectan11 t ht t ttttttt ht tt h πππ π ππ = ′ == ′ ==

62. ()() ()()() 22 sinsincos sincossinsincoscos sincossinsincoscos sin2cos2 sincos1 422 fxxxx f xxxxxxx x xxxxx xx f πππ =+ ′ =−++ =−++ =+

′ =+=

64. (a) ()()() () ()()() ()() ()() 2 2 22 2 24,1,5 2241 244 344 13;Slopeat1,5 fxxx fxxxx xxx xx f =−+− ′ =−++ =−++ =−+ ′ =−− Tangentline: ()()53138yxyx −−=−  =− (b)

(c)Graphingutilityconfirms3 dy dx = at () 1,5. 6 13 3 (1,4) 15 36 (1,5) 3

63. (a) () ()()() () ()()()() ()() 3 3 332 32 412,1,4 411234 413648 4689 13;Slopeat1,4 fxxxx fxxxxx xxxxx xxx f 2 =+−−− ′ =+−+−+ =+−+−+− =−+− ′ =−− Tangentline: () 43131yxyx +=−−  =−− (b) (c)Graphingutilityconfirms3 dy dx =− at () 1,4.

65. (a) () () () ()()() () () () () () 22 2 ,5,5 4 4114 44 4 54;Slopeat5,5 54 x fx x xx fx xx f =−

Tangentline: () 545425yxyx −=+  =+

(b)

Tangentline:

(3( ,2

(c)Graphingutilityconfirms4 dy dx = at ()5,5.

66. (a) () () () ()()()() () () () () 22 3 ,4,7 3 31316 33 6 46;Slopeat4,7 1 x fx x xx fx xx f + = −−+ ′ ==− ′ ==−

Tangentline: () 764631yxyx −=−−  =−+

(b)

(c)Graphingutilityconfirms6 dy dx =− at ()4,7.

67. (a) () () 2 tan,,1 4 sec 2;Slopeat,144 fxx fxx f

69. () () () ()()() () () () () () 2 2 2222 2 8 ;2,1 4 408216 44 1621 2 442 fx x xx x fx xx f = + +−

+ () 1 12 2 1 2 2 240 yx yx yx

70. () () ()()() () () () () () 2 2 2222 2 273 ;3, 92 9027254 99 5431 3 992 fx x xx x fx xx

71. () () ()()() () () () () 2 22 2222 2 168 ;2, 165 161616225616 1616 25616412 2 2025 x fx x xxx x fx xx f  =−− +

+− ′ == ++ ′ −== () 812 2 525 1216 2525 2512160 yx yx yx +=+ =− −+=

72. () () ()()() () () () 2 22 2222 2 44 ;2, 65 6442244 66 24162 2 1025 x fx x xxx x fx xx f

=  +  +− ′ == ++ ′ == () 42 2 525 216 2525 252160 yx yx yx −=− =+ −−=

73. () () () 12 2 23 3 21 2 21 22 x fxxx x x fxxx x ==− −+ ′ =−+= () 0 fx ′ = when1, x = and ()11. f = Horizontaltangentat ()1,1.

74. () () ()() ()() () () 2 2 22 22 22 1 122 1 2 1 x fx x x xxx fx x x x = + +− ′ = + = + () 0 fx ′ = when0. x =

Horizontaltangentisat ()0,0.

75. () () ()()() () () () () 2 2 2 2 22 1 121 1 22 11 x fx x xxx fx x xx xx xx = ′ = == () 0 fx ′ = when0 x = or2. x =

Horizontaltangentsareat ()0,0and ()2,4.

76. () () ()()()() () () () ()() () 2 2 22 22 22 2 2222 4 7 7142 7 728 7 8771 77 = ′ = −−+ = −+ =−=− x fx x xxx fx x xxx x xx xx xx () 0 fx ′ = for () () 11 1,7;1,7 214xff=== f hashorizontaltangentsat 1 1, 2    and 1 7,. 14  

77. () () ()() () () 22 1 1 112 11 + = −−+ ′ == x fx x xx fx xx 1 263; 2 yxyx +=  =−+ Slope: 1 2 () () ()() 2 2 21 12 14 12 1,3;10,32 =− −= −=± =−−== x x x xff () () 111 01 222 117 23 222 yxyx yxyx −=−+  =−− −=−−  =−+

2y + x =7 y x f (x)= x +1 x 1 2 64

4 6 6 (3,2) (1,0)

2246

2y + x =1

78. () () () () () 22 1 11 11 x fx x xx fx xx = ′ == Let ()() (),,1 xyxxx=− beapointoftangencyonthe graphof f () () () ()() () ()() ()() 2 2 2 511 11 451 111 4511 41040 1 2210,2 2 xx x

() () 11 1,22;4,21 22 ffff

Twotangentlines: () 1 1441 2 2124 yxyx yxyx

79. () ()() () () () ()()() () () () () () () () 22 22 23316 22 255416 22 54324 2 222 xx fx xx xx gx xx xxx gxfx xxx +− ′ == ++ +−+ ′ == ++ ++ ==+=+ +++ f and g differbyaconstant.

80. () ()()() () ()()() () () 22 22 cos3sin31cossin cos2sin21cossin sin2sin35 5 xxxx x xx fx xx xxxx x xx gx xx xxxxx gxfx xx ′ == +−+ ′ == +−+ ===+ f and g differbyaconstant.

81. (a) ()()()()() ()()()()()() 1 111111461 2 pxfxgxfxgx pfgfg ′′′ =+  ′′′ =+=+−=  

(b) () ()()()() () () ()() 2 2 31701 4 33 g xfxfxgx qx gx q ′′ ′ = ′ ==−

82. (a) ()()()()() () ()() 1 48104 2 p xfxgxfxgx p ′′′ =+ ′ =+=

(b) () ()()()() () () ()() 2 2 4241123 7 4164 g xfxfxgx qx gx q ′′ ′ = ′ ===

83. ()() () 3212 12122 Area6565 5185 9cm/sec 22 A ttttt t Attt t ==+=+ + ′ =+=

84. () () 23212 11 22 22 Vrhttttπππ  ==+=+   () 12123 12 1332 in./sec 224 t Vttt t ππ +  ′ =+=   85. () 2 32 200 100,1 30 40030 100 30 x Cx xx dC dxx x

x y (2,2) (1,5) 1,1 (2) 2 424 2 6 y =4x +1 y = x +4 f (x)= x x 1 (a)When (b)When (c)When Astheordersizeincreases,thecostperitemdecreases.

86. ()

87. (a) [] ()()()() () 2 1 sec cos cos01sin 1sin1sin secsectan coscoscoscoscos cos x x xx ddxx x xx dxdxxxxxx x =

(b) [] ()()()() () 2 1 csc sin sin01cos 1cos1cos csccsccot sinsinsinsinsin sin x x xx ddxx x xx dxdxxxxxx x =

===−=−⋅=−

+ ===−=−=−

88. () () [ ) ()() sec csc,0,2 fxx gxx fxgx π = = ′′ = 3 3 3 1sin sectansin coscos sectancsccot111tan1tan1 1cos csccotcos sinsin x xxx xx xxxxxx x xxx xx

37 , 44 x ππ =

89. (a) () () 101.71593 2.1287 =+ =+ htt ptt (b)

(c) 101.71593 2.1287 + = + t A t A representstheaveragehealthcareexpendituresper person(inthousandsofdollars). (d) () 2 25,842.6 4.411205.482,369 ′ ≈ ++ At tt () A t ′ representstherateofchangeoftheaverage healthcareexpendituresperpersonforthegiven year t.

90. (a) () sin csc csccsc1 r rh rhr hrrr θ θ θθ = + += =−=− (b) ()() () () csccot 30 6 40002380003mi/rad

91. () () () 274 27 2 fxxx fxx fx =+−

92. () ()

93. ()

94. ()

95. () ()

97. () () ()() sin cossin sincoscos sin2cos fxxx fxxxx f xxxxx xxx = ′ =+ ′′ =−++ =−+

98. () () ()() cos cossin sinsincoscos2sin = ′ =− ′′ =−−+=−− fxxx fxxxx f xxxxxxxx

99. () () () () () 2 32 csc csccot csccsccotcsccot csccotcsc fxx fxxx f xxxxxx xxx = ′ =−+ ′′ =−−−− =+

100. () () () () () () 2 22 sec sectan secsectansectan secsectan fxx fxxx f xxxxxx xxx = ′ = ′′ =+ =+

101. () () () 325 235 85 85 2 3 5 66 66 2525 fxxx fxxx fxxxx x ′ =− ′′ =− ′′′ =+=+

xx fx x xxxx fx x xxxxxx xx x xxxxx fx x xxxxx x xxxx x xxxx x x + = −+−+ ′ = == −+ ′′ =

96. () () ()() ()() () () () ()() ()()

102. () () () () 35445 415 15 44 55 fxxx fxx x == ==

103. () () () () () () () () () () () () () 3 4 5 6 7 8 sin cos sin cos sin cos sin f xx f xx f xx f xx f xx f xx f xx ′′ =− =− = = =− =− =

104. () () () () 4 5 cos cossin fttt f tttt = =−

105. ()()() ()()() ()()() () 2 2 2222 224 0 f xgxhx f xgxhx fgh =+ ′′′ =+ ′′′ =+ =−+ =

106. ()() ()() ()() 4 224 fxhx fxhx fh =− ′′ =− ′′ =−=−

107. () () () () ()()()() () () ()()()() () ()()()() () 2 2 2 2222 2 2 1234 1 10 gx fx hx hxgxgxhx fx hx hggh f h = ′′ ′ =

108. ()()() ()()()()() ()()()()() ()()()() 22222 3412 14 fxgxhx f xgxhxhxgx fghhg = ′′′ =+ ′′′ =+ =+−− =

109. Polynomialsofdegree1 n (orlower)satisfy () () 0. n fx = Thederivativeofapolynomialofthe0th degree(aconstant)is0.

110. Todifferentiateapiecewisefunction,separatethefunctionintoitspieces,anddifferentiateeachpiece. If

() 2, f xx ′ = and () 2. fx ′′ = Noticethat ()() 00,00,ff ′ == but () 0 f ′′ doesnotexist(theleft-handlimitis2,whereastheright-handlimitis2).

111. Itappearsthat f iscubic,so f ′ wouldbequadraticand f ′′ wouldbelinear.

112.

Itappearsthat f isquadratic so f ′ wouldbelinearand f ′′ wouldbeconstant. 113. 114.

115. Thegraphofadifferentiablefunction f suchthat ()20,0ff ′ =< for2, x −∞<< and0 f ′ > for 2 x <<∞ would,ingeneral,looklikethegraph below.

Onesuchfunctionis ()() 2 2. fxx=−

116. Thegraphofadifferentiablefunction f suchthat0 f > and0 f ′ < forallrealnumbers x would,ingeneral, looklikethegraphbelow.

117. () ()() () () 2 2 36,06 2 327m/sec 36m/sec vttt atvtt v a =−≤≤ ′ ==− = =− Thespeedoftheobjectisdecreasing.

119. () ()() ()() 2 8.2566 16.5066 16.50 sttt vtstt atvt =−+ ′ ==+ ′ ==− Averagevelocityon:

57.750 0,1is57.75 10 9957.75 1,2is41.25 21 123.7599 2,3is24.75 32 132123.75 3,4is8.25 43 = = = =

120. (a) s positionfunction v velocityfunction a accelerationfunction

(b)Thespeedoftheparticleistheabsolutevalueofitsvelocity.So,theparticle’sspeedisslowing downontheintervals ()0,43and ()83,4anditspeedsupontheintervals () 43,83and ()4,6. x f y t(sec)01234 s(t)(ft)057.7599123.75132 6649.53316.50 –16.5–16.5–16.5–16.5–16.5

118. () ()() ()()()() () () 22 100 215 21510010021500 215215 t vt t tt atvt tt = + +− ′ === ++

(a) () () 2 2 1500 52.4ft/sec 2515 a ==  +  

(b) () () 2 2 1500 101.2ft/sec 21015 a =≈

 (c) () () 2 2 1500 200.5ft/sec 22015 a =≈  +  

121. () () ()()()()()1221! n n fxx f xnnnn = =−−=

Note: ()() !1321read“factorial” nnnn =−⋅⋅

123. ()()() f xgxhx =

fxgxhxhxgx

fxgxhxgxhxhxgxhxgx

gxhxgxhxhxgx

122. () () () ()()()()()() () 11 1 112211! nn n nn fx x nnnn fx xx ++ = == 

(a) ()()()()() ()()()()()()()()() ()()()()()() ()()()()()()()()()()()()() ()()()()()()()() () ()() () ()()()()()()() 44 2 22 33 333

fxgxhxgxhxgxhxgxhxhxgxhxgx

gxhxgxhxgxhxgxhx

fxgxhxgxhxgxhxgxhx

=++++ ()()()() ()() () ()() () () ()()()()()()() () ()() 4 44 3 464 g xhxgxhx

Note: () !1321nnn=−⋅⋅ (read“n factorial”)

124. ()()() ()()()()()() ()()()()()() 2 23 xfxxfxfx xfxxfxfxfxxfxfx xfxxfxfxfxxfxfx

=++=+

Ingeneral, () () () () () () 1 . n nn xfxxfxnfx

125. () () 1 sin cossin n nn fxxx f xxxnxx = ′ =+

When () 1:cossin nfxxxx ′ ==+

When () 2 2:cos2sin nfxxxx ′ ==+

When () 32 3:cos3sin nfxxxxx ′ ==+

When () 43 4:cos4sin nfxxxxx ′ ==+

Forgeneral () 1 ,cossin. nn nfxxxnxx ′ =+

126. ()

() () 1 1 1 cos cos sincos sincos sincos n n nn n n x fxxx x f xxxnxx x xxnx xxnx x + == ′ =−− =−+ + =−

When () 2 sincos 1: x xx nfx x + ′ ==−

When () 3 sin2cos 2: x xx nfx x + ′ ==−

When () 4 sin3cos 3: x xx nfx x + ′ ==−

When () 5 sin4cos 4: x xx nfx x + ′ ==−

Forgeneral () 1 sincos ,. n x xnx nfx x + + ′ =−

127. 23 3232 32 112 ,, 21 22220 yyy xxx xyxyxx xx ′′′ ==−=

128. () () 3 2 22 2610 66 12 12 2–121226624 yxx yx yx y yxyyxxxx =−+

137. ()()() ()() () ()

Section2.4TheChainRule

1. Tofindthederivativeofthecompositionoftwo differentiablefunctions,takethederivativeoftheouter functionandkeeptheinnerfunctionthesame.Then multiplythisbythederivativeoftheinnerfunction.

() () () () f gxfgxgx

129. () 2sin3 2cos 2sin 2sin2sin33 yx yx yx yyxx =+ ′ = ′′ =− ′′ +=−++=

130. ()() 3cossin 3sincos 3cossin 3cossin3cossin0 yxx yxx yxx yyxxxx =+ ′ =−+ ′′ =−−

131. False.If ()(), yfxgx = then ()()()() dy f xgxgxfx dx ′′ =+

132. True. y isafourth-degreepolynomial. 0 n n dy dx = when4. n >

133. True ()()()()() ()()()() 00 0 hcfcgcgcfc fcgc ′′′ =+ =+ =

134. True

135. True

136. True

2. The(Simple)PowerRuleis () 1 nn d xnx dx = The GeneralPowerRuleusestheChainRule:

3. () 654yx=− 65ux=− 4 yu =

9. () ()() () 3 2 2 27 3272 627 yx yx x =− ′ =− =−

12. ()() () () () 23 13 3 92 26 929 392 ftt ftt t =+ ′ =+= +

15. () () () () () 313 22 223 2232 32 6161 144 6112 36161 yxx xx yxx x x =+=+ ′ =+== + +

16. () () () () () 414 22 234 2343 42 2929 1 292 4 99 yxx yxx xx x x =−=−

17. () ()() () 1 2 2 2 1 121 2 yx yx x =− ′ =−−=

18. () () () () () () () 21 2 22 2222 1 45 45 4552 5225 4554 sttt tt stttt tt tttt ==−− ′ =−−−−− ++ == −−+−

′ =−+

==− + + hsss hsss ss s s

13. () () () () () () 2212 212 2122 253253 1 25310 2 1010 5353 =−+=−+

14. () () () () () 2212 212 2 4343 13 436 243 gxxx x gxxx x =−=− ′ =−−=−

19. () () () ()()()() () 33 33 4 32 2 34 6 62 2 6323 54 2 gss s gsss s s ==− ′ =−− =−

20. () () () () 4 4 5 5 3 32 2 12 122 2 yt t yt t =−=−− ′ =−=

21. () () () () () 12 32 32 3 1 35 35 1 353 2 3 235 3 235 yx x yx x x ==+ + ′ =−+ = + =− +

22. () () () () () () () 212 2 232 232 23 1 2 2 1 22 2 2 2 gtt t gttt t t t t ==−

23. ()() ()()() ()() ()() 27 762 6 6 2 2272 2227 294 fxxx f xxxxx x xxx xxx =− ′ =−+− =−

24. ()() ()()()()()() ()() ()() 3 23 2 2 25 3252251 25625 2585 fxxx fxxxx xxx xx =− ′ =−+− =−

25. () () () () () () () () () 2212 121222 1212222 12 222 2 2 11 1 1211 2 11 11 12 1 yxxxx yxxxx xxx xxx x x =−=−

′ =−−+−   =−−+−

26. () () () () () () () 222212 1212 222 22 212 2 2 1616 1 216162 2 216 16 323 16 yxxxx yxxxxx x xx x xx x =−=− ′ =−+−−

27. () () () () () () () () () () () 212 2 121222 2122 1212222 2 12 222 2 2323 2 11 1 1112 2 1 11 1 11 1 11 11 xx y x

30.

34. () () () () () () () () () ()()() 243 22 43342222 21 32141224121 gxx gxxxxxxx =++ ′ =+++=+++

35. cos4 4sin4 yx dy x dx = =−

36. sin cos yx dy x dx π ππ = =

37. () () 2 5tan3 15sec3 g xx g xx = ′ =

38. () ()() sec6 sec6tan66 6sec6tan6 hxx hxxx x x = ′ = =

39. () () ()() () 222 222222 22 sinsin cos22cos 2cos yxx yxxxx xx ππ

40. () ()()()() ()()() 2 22 22 csc12 csc12cot122122 412csc12cot12 yx yxxx xxx =− ′ =−−−    =−−−

41. () ()()() 22 sin2cos2 sin22sin2cos22cos2 2cos22sin2 2cos4 hxxx hxxxxx xx x = ′ =−+ =− = Alternatesolution: () () () 1 2 1 2 sin4 cos442cos4 hxx hxxx = ′ ==

= ′ =+

42. () () () () () () () () () () 2 22 11 22 1111111 2222222 1111 2222 sectan secsectansectan secsectan g g θθθ

43. () () ()() 2 2 4 222 33 cotcos sinsin sinsincos2sincos sin sin2cos1cos sinsin xx fx xx x xxxx fx x x xx xx == ′ = ==

44. () ()()() 22 coscossin csc coscossinsin cossincos2 v gvvv v g vvvvv vvv ==⋅ ′ =+− =−=

45. 2 2 4sec 8secsectan8sectan yx y xxxxx = ′ =⋅=

46. ()() ()()() ()() 225cos5cos 10cossin 10sincos 5sin2 g ttt gttt tt t ππ πππ πππ ππ == ′ =− =− =−

47. () () () ()()()() 2211 44 1 4 1 2 sin2sin2 2sin2cos22 sin2cos2sin4 f f θθθ θθθ θθθ == ′ = ==

48. ()() ()()()() ()() 2 2 2 2cot2 4cot2csc2 4cot2csc2 π πππ πππ =+ ′   =+−+   =−++ htt httt tt

49. ()() ()()()()()() ()()() 2 22 22 3sec1 3sec1tan121 61sec1tan1 ftt ftttt ttt π ππππ ππππ =− ′′ =−−− =−−−

50. () ()()()() () 2 2 22 5cos 5sin2 10sin yx yxx xx π πππ ππ = ′ =− =−

51. () ()() 2 2 sin3cos cos3cos6sin yxx yxxxx =+ ′ =+−

52. () ()() cos5csc sin5csc5csccot yxx yxxxx =+ ′ =−+−

53. () () () ()() () () 12 12122 2 sincot3sincot3 1 coscot3cot3csc33 2 3coscot3csc3 2cot3 ππ ππππ πππ π ==

=− yxx yxxx xx x

54. () () () () ()() () () () 2 1/22 cossintan 1sinsintancostansec sinsintansintancostansec 22sintan yx x xx yxxxx x

55. () 2 232 22 1 1 134 21 x y x x x y xx + = + ′ = +

Thezeroof y ′ correspondstothepointonthegraphof y wherethetangentlineishorizontal.

56. ()32 2 1 1 21 = + ′ = + x y x y xx y ′ hasnozeros.

57. () () 1 1 21 + = + ′ =− + x y x x x y xx y ′ hasnozeros. 2 15 y y ′ 2 66

58. () () 11 11 2121 =−++ ′ =+ −+ gxxx gx xx g ′ hasnozeros.

59. 2 2 cos1 sincos1 sincos1 π πππ πππ + = = ++ =− x y x dyxxx dxx xxx x

Thezerosof y ′ correspondtothepointsonthegraphof y wherethetangentlinesarehorizontal.

60. 2 2 1 tan 11 2tansec yx x dy x dxxx = =−

Thezerosof y ′ correspondtothepointsonthegraphof y wherethetangentlinesarehorizontal.

61. () sin3 3cos3 03 yx yx y = ′ = ′ = 3cyclesin [] 0,2π 62. () sin 2 1 cos 22 1 0 2 =

63. () () () () () () () () 2212 21/2 2122 2 88,1,3 24 14 828 28 28 1455 1 18193 yxxxx x x yxxx x x

64. () () ()() () () 31/5 4/5 32 2 34/5 34,2,2 1 3494 5 94 534 1 2 2 yxx yxxx x xx y =+

65. () () () ()() () () 31 2232 32 1 52,2, 2 15 523 2 603 2 1005

66. () () () () () () () () () 22 22 23 23 11 3,4, 316 223 2323 3 5 4 32 fxxx xx x fxxxx x x f

67. () ()() () () () 2 2 3 3 4 42,0,1 2 8 82 2 8 01 8 yx x yx x y ==+ +

68. () () () () () ()()() 23 23 24 4 4 42,1,4 2 12222 112100 yxx xx yxxx y ==−− ′ =−−− ′ =−−=

69. () () 3 2 3 26sec4,0,25 3sec4sec4tan44 12sec4tan4 00 yx yxxx xx y =−

70. () () () 112 212 2 12 coscos,, 2 11sin cossin 22cos yxxx x x yxxx x x π π

=+=+

′ =−+−=−− () /2 y π ′ isundefined.

71. (a) () () () () () () () 21/2 21/2 2 27,4,5 12 274 227 8 4 5 fxx x fxxx x f =−

= Tangentline: () 8 548570 5 yxxy −=−  −−= (b)

73. (a) () () ()()() () 32 3223 43,1,1 243122443 124 yx yxxxx y =+− ′ =+=+ ′ −=−

Tangentline: () 124124230yxxy −=−+  ++= (b)

72. (a) () () () () () () () () () () 2212 1/21/222 2 2 2 11 55,2,2 33 111 525 323 1 5 253 4113 23 3339 fxxxxx fxxxxx x x x f =+=+  ′ =+++ 

Tangentline: () 13 2213980 9 yxxy −=−  −−= (b)

6 2 21 14 (1,1) 6 5 1 2 (1,4) 0 2 2 2 (,0) π π

74. (a) () () () () () () () () () 22/3 21/3 21/3 1/3 9,1,4 24 92 339 42 1 383 fxx x fxxx x f =− ′ =−−= ′ ==−

Tangentline: () 2 4123140 3 yxxy −=−−  +−= (b)

Tangentline: () 888yxx=−=−ππ (b) 2 66 6 (4,5)

75. (a) ()() () () sin8,,0 8cos8 8 fxx fxx f π π = ′ = ′ =

76.

Horizontaltangentat () 1,4

83. ()() ()()()() ()()()() 4 33 22 527 2027714027 420277294027

fxx fxxxxx fxxxxxx xxxx xxx =+

=++ 85. () () ()()() ()()() () () 1 2 3 3 3 1 116 116 11611 2211611 242116 242 116 ==−

=−−

=−− =− = fxx x fxx fxx x x

86. () () () ()() ()() () 2 2 3 4 4 8 82 2 162 48 482 2 fxx x fxx fxx x ==− ′ =−− ′′ =−=

87. () () () () () 2 2 22 222 sin 2cos 22sin2cos 2cos2sin fxx fxxx f xxxxx xxx = ′ =  ′′ =−+  =−

88. () ()() () ()() () () () 2 2 222 24222 2222 222 sec 2secsectan 2sectan 2secsec2tan2sectan 2sec4sectan 2secsec2tan 2sec3sec2 π

= ′ = = ′′ =+ =+ =+ =− fxx fxxxx xx f xxxxxx xxx xxx xx

89. () () () () ()()() ()()() () 3 22 164 99 1 9 31,1, 331331 2313186 124 hxx hxxx hxxx h =+ ′ =+=+ ′′ =+=+ ′′ =

90. () () () () () () () () 12 32 52 52 11 4,0, 42 1 4 2 33 4 444 3 0 128

==+  +  ′ =−+ ′′ =+= + ′′ = fxx x fxx fxx x f

91. ()() () ()() () () ()() () ()() () 2 22 22 222 cos,0,1 sin22sin 2cos22sin 4cos2sin 00 = ′ =−=− ′′ =−− =−− ′′ = fxx f xxxxx f xxxxx xxx f

92. () ()() ()()()() ()() 2 2 tan2,,3 6 2sec2 4sec2sec2tan22 8sec2tan2 323 6 π π  =   ′ = ′′ =⋅ =

93.

′′ =   gtt gtt g tttt tt g

Thezerosof f ′ correspondtothepointswherethegraph of f hashorizontaltangents.

94.

f ′

f isdecreasingon () ,1−∞− so f ′ mustbenegativethere. f isincreasingon () 1, ∞ so f ′ mustbepositivethere.

95. (a) ()() ()()()()() 3 3333 gxfx g xfxgxfx = ′′′′ =  =

Therateofchangeof g isthreetimesasfastasthe rateofchangeof. f

(b) () () () ()()() () 2 22 22 gxfx g xfxxgxxfx = ′′′′ =  =

Therateofchangeof g is2 x timesasfastastherate ofchangeof f

96. () () 2 25 31 x rx x = +

(a)If () () () , f x hx g x = thenwrite ()()()() 1 hxfxgx = andusetheProductRule.

(b) ()()() ()()()()()()() ()() () () 2 32 3 3 2531 252313312 625231 31 632 31 rxxx rxxxx xx x x x =−+ ′ =−−+++ −−++ = + −+ = + (c) () ()()()()()() () ()()() () () 2 4 3 3 312252313 31 312625 31 632 31 xxx rx x xx x x x +−−+ ′ = + +−− = + −+ = +

(d)Answerswillvary.

97. (a) ()()()() 2 g xfxgxfx ′′ =−  =

(b) ()()()() 22 hxfxhxfx ′′ =  =

So,youneedtoknow ()3. f x ′ ()()()() ()()()() 1 3 03031 1333412 rf rf ′′ =−=−−= ′′ −=−=−−=

(d) ()()()() 22sxfxsxfx ′′ =+  =+

So,youneedtoknow ()2. fx ′ + ()() 1 3 20,etc.sf ′′ −==−

98. (a) ()()() ()()()()() ()()()()() 5326324 fxgxhx f xgxhxgxhx f = ′′′ =+

(b) ()()() ()()() () ()()()() 53223 fxghx fxghxhx fgg = ′′′ = ′′′ =−=−

Notpossible,youneed () 3 g ′ tofind ()5. f ′

(d) ()()

()()() ()()() 3 2 2 3 5336162 fxgx fxgxgx f =   ′′ =   ′ =−=

99. (a) ()()() ()() () ()()() ()

()()() ()()()()() 1 2 11 22 ,14,1,41 111411 hxfgxggf hxfgxgx hfggfg ′′ ===−=− ′′′ = ′′′′′ ===−−=

(b) ()()() ()()() ()()() ()

()()() ()()() ,56,51,6doesnotexist. 55561 sxgfxffg sxgfxfx sgffg ′′ ===− ′′′ = ′′′′ ==−

() 5 s ′ doesnotexistbecause g isnotdifferentiableat6. x 210123 4124

100. (a) ()()() ()()() () ()()() ()()() 1 333512 hxfgx hxfgxgx hfggf = ′′′ = ′′′′ ===

(b) ()()() ()()() () ()()() ()()()()() 99982122 sxgfx sxgfxfx sgffg = ′′′ = ′′′′ ===−=−

101. (a) ()

()()()() () 1 2 2 132,400331 1132,4003311132,400 331 Fv Fv v =− ′ =−−−=

When30,1.461. vF ′ =≈ (b) ()

()()()() () 1 2 2 132,400331 1132,4003311132,400 331 Fv Fv v =+ ′ =−+−= +

When30,1.016. vF ′ =≈−

102. [] [] 11 34 11 34 cos12sin12 12sin1212cos12 4sin123cos12 ytt vytt tt =− ′ ==−− =−−

When8,0.25ftand4ft/sec. tyv π ===

103. 0.2cos8t θ =

Themaximumangulardisplacementis0.2 θ = (because 1cos81). t −≤≤

[] 0.28sin81.6sin8 d tt dt θ =−=−

When3,1.6sin241.4489 tddt θ ==−≈ rad/sec.

104. cos yAt ω =

(a)Amplitude: 3.5 1.75 2 1.75cos A yt ω == = Period: 2 10 105 1.75cos 5 t y ππ ω π  == = (b)1.75sin0.35sin555 tt vy πππ π  ′ ==−=−

105. (a) ()()27.3sin0.491.9057.1Ttt=−+

(b)

Themodelisagoodfit. (c) ()() 13.377cos0.491.90Ttt ′ =−

0 20 20 13 0 013

(d)Thetemperaturechangesmostrapidlyaroundspring (March–May)andfall(Oct.–Nov.).Thetemperature changesmostslowlyaroundwinter(Dec.–Feb.)and summer(Jun.–Aug.).Yes.Explanationswillvary.

106. (a)Accordingtothegraph ()() 41.CC ′′ > (b)Answerswillvary.

107. () () () () () () 22 22 23 23 3 400140012002 2 4800 240022 2 Nt t t Nttt t   =−=−+  +  ′ =+= +

(a) ()00 N ′ = bacteria/day

(b) () () ()3 480014800 1177.8 1227 N ′ ==≈ + bacteria/day

(d) () () ()3 4800314,400 310.8 921331 N ′ ==≈ + bacteria/day

(e) () () ()3 4800419,200 43.3 1625832 N ′ ==≈ + bacteria/day

(f)Therateofchangeofthepopulationisdecreasingas . t →∞ 0 013

108. (a) () () 12 1 010,000 01 10,00010,0001 1 k V t k Vk Vt t = + === + ==+ +

(b) () () () 32 32 32 15000 10,0001 21 5000 11767.77dollars/year 2 dV t dt t V  =−+=

+ ′ =≈−

109. () sin f xx β =

(a) () () () () 2 3 44 cos sin cos sin f xx f xx f xx f x ββ ββ ββ ββ ′ = ′′ =− ′′′ =− =

(b) ()()() 222sinsin0fxfxxx βββββ ′′ +=−+=

110. (a)Yes,if ()() f xpfx += forall x,then ()(), f xpfx′′ += whichshowsthat f is periodicaswell.

(b)Yes,if ()()2, g xfx = then ()() 22. g xfx′′ = Because f ′ isperiodic,sois g ′

111. (a) ()()() () ()()() ()111 rxfgxgx rfgg ′′′ = ′′′ =

Notethat ()14 g = and () 505 4. 624 f ′ ==

Also, ()10. g ′ = So, ()10. r ′ =

(b) ()()() () ()()() ()444 s xgfxfx sgff ′′′ = ′′′ =

′ ===

Notethat () 55641 4, 22622fg

and () 5 4. 4 f ′ = So, () 155 4. 248 s

112. (a) ()() ()() 22 sincos10 2sincos2cossin0 gxxxgx gxxxxx ′ =+=  = ′ =+−=

(b) ()() 22 tan1sec 1 x x g xfx += += Takingderivativesofbothsides, ()() g xfx′′ = Equivalently, () 2secsec2tan2sectan f xxxxxx ′ =⋅= and () 22 2tansec2sectan, g xxxxx ′ =⋅= which arethesame.

113. (a)If ()(), f xfx−=− then () () ()()() ()() 1

 ′′ −−=− ′′ −= dd f dxdxxfx fxfx fxfx

So, () f x ′ iseven.

(b)If ()(), f xfx−= then () () ()()() ()() 1 .   =    ′′ −−= ′′ −=− dd f dxdxxfx fxfx fxfx

So, f ′ isodd.

114. () () 2 2212 2 1 2 2 ,0 uu dd uuuuu dxdx uuu uu u u =  ′  ==   ′ ′ ==≠

115. () () 35 355 3, 353 gxx x gxx x =−  ′ =≠   

116. () () 2 2 2 9 9 2,3 9 fxx x fxxx x =−   ′ =≠±  

117. () () cos sincos,0 hxxx x hxxxxx x = ′ =−+≠

118. () () sin sin cos, sin fxx x f xxxk x π =  ′ =≠   

(b)

(d)Theaccuracyworsensasyoumoveawayfrom6. x π = 121. True

(d)Theaccuracyworsensasyoumoveawayfrom

Section2.5ImplicitDifferentiation

1. Answerswillvary. Sample answer: Intheexplicitform ofafunction,thevariableisexplicitlywrittenasa functionof x.Inanimplicitequation,thefunctionisonly impliedbyanequation.Anexampleofanimplicit functionis25. xxy+= Inexplicitformitwouldbe () 2 5. yxx =−

2. Answerswillvary. Sample answer: Givenanimplicit equation,firstdifferentiatebothsideswithrespectto x Collectalltermsinvolving y ′ ontheleft,andallother termstotheright.Factorout y ′ ontheleftside.Finally, dividebothsidesbytheleft-handfactorthatdoesnot contain y ′

3. Youuseimplicitdifferentiationtofindthederivative y ′ whenitisdifficulttoexpress y explicitlyasafunction of x

4. If y isanimplicitfunctionof, x thentocompute, y ′ you differentiatetheequationwithrespectto. x Forexample, if21, xy = then220. yxyy ′ += Here,thederivativeof 2 y is2. yy ′

12. () () 2 122 2 2 2 2 1 1 2 2 2 22 2 22 2 2 2 4 2 xyxy xyxyyxyxy xy yxyxy xyxy x y xyxy xyxy y xy xy y x x xy xyxyy y x xxy =+

= ′ = 13. () 322 222 222 22 2 3212 336420 43632 632 43 xxyxy xxyxyxyyy xyxyxyxy xyxy y xyx −+=

14. () 42 432 432 32 4 839 488630 86843 843 86 xyxyxy xyxyxyyxyyy x xxyyyxyy yxyy y x

15. () sin2cos21 cos4sin20 cos 4sin2 xy xyy x y y +=

16. () ()() () 2 sincos2 2sincoscossin0 cossin0 cos sin xy xyxyy xyy x y y ππ ππππππ ππππ

17. () () () 2 2 csc1tan csccot1tansec csccot1tan sec xxy x xyxyy x xy y xy =+

18. () 2 2 22 cot csc1 11 tan 1csccot yxy yyy yy yy =− ′′ −=− ′ ===−

19. [] () ()() () () sin cos coscos cos 1cos yxy yxyyxy yxxyyyxy yxy y xxy = ′′ =+ ′′ −= ′ =

′ ==−

20. () () 2 2 2 1 sec 11 1sectan 11 coscot sec1tan1 x y y yyy y yy yyyy = ′ =−

21. (a)22 22 2 64 64 64 xy yx yx += =− =±− (b)

(d)Implicitly:220 xyy x y y ′ += ′ =− y1=64 x 2 124412 12 4 12 y2=64 x 2 y x

(c)Explicitly: () () 212 2 2 1 642 264 64 =±−−= ==− ±−  dyx xx dx x xx y x

22. (a) () () 22 222 22 2 2536300 36300252512 25 12 36 5 12 6 xy yxx yx yx += =−=− =− =±−

(b)

(c)Explicitly:

(d)Implicitly:50720 5025 7236 xyy x x y yy ′ +⋅= ′ ==−

23. (a)22 22 22 2 2 1616 1616 16 1 1616 16 4 yx yx xx y x y −= =+ + =+= ±+ =

(c)Explicitly:

(d)Implicitly:221616 3220 322 2 3216 yx yyx yyx x x y yy −= ′ −= ′ = ′ ==

(c)Explicitly:(d)Implicitly:

25. () 6 10 = ′ += ′ =− ′ =− xy xyy xyy y y x

At () 1 6,1: 6 y ′ −−=−

26. 3 3 23 32 2 3 36 2 30 3 33 = = ′ += ′ =− ′ ==− xy xy xyxy xyxy x yy y x x

At () ()32 1,2:6 1 y ′ ==−

27. ()() ()() () () () 2 2 2 22 22 22 22 49 49 492492 2 49 196 2 49 98 49 = + +−− ′ = + ′ = + ′ = + x y x x xxx yy x x yy x x y yx

At ()7,0: y ′ isundefined.

28. ()() ()() () () 2 3 3 322 2 32 24 232 36 4 36 362363 12 36 72108 1236 = + +−− ′ = + +− ′ = + x y x x xxx yy x xxx y yx

At ()6,0: y ′ isundefined(divisionby0).

29. () ()() () () 333 322333 22 22 22 22 33 330 0 220 22 2 2 +=+ +++=+ += += ′′ +++= ′ +=−+ + ′ =− + xyxy xxyxyyxy xyxy xyxy xyxyxyyy x xyyyxy yyx y x xy

At ()1,1:1 y ′ −=−

30. () 33 22 22 2 2 61 3366 3663 63 36 xyxy x yyxyy yxyyx yx y yx +=− ′′ +=+ ′ −=− ′ = At () 181262 2,3: 2712155 y ′ ===

31. () ()() () () () () () 2 2 2 2 2 2 2 2 tan 1sec1 1sec sec tan tan1 sin 1 xyx yxy x y y x y x y xy xy x x += ′ ++= −+ ′ = + −+ = ++ =−+ =− + At ()0,0:0 y ′ =

32. [] cos1 sincos0 cos sin 1 cot cot xy xyyy y y x y y x y x = ′ −+= ′ = = = At 1 2,: 323 y π  ′ =  

33. () () () () () 2 2 2 2 2 22 48 420 2 4 284 4 16 4 xy xyyx xy y x xx x x x += ′ ++= ′ = + −+ = + = + At () 321 2,1: 642 y ′ ==− 2 8 Or,youcouldjustsolvefor:4 yy x  =  + 

34. () ()()() () 23 22 22 4 4213 3 24 xyx xyyyx x y y yx −= ′ −+−= + ′ =

At ()2,2:2 y ′ =

35. () ()()() ()() 2 222 222 32232 23232 23232 32 232 4 22248 444448 444844 442 2 xyxy xyxyyxyyx xxyyxyyyxyxy x yyyyxyxyxxy yxyyxxyxxy xyxxy y xyyx += ′′ ++=+ ′′′ +++=+ ′′′ +−=−− ′ +−=−− ′ = +−

At ()1,1:0 y ′ =

36. () 33 22 22 22 22 60 33660 3663 632 362 xyxy xyyxyy yyxyx yxyx y yxyx +−=

−=− ′ == At () () () () 163169 48324 ,: 3364983405 y  ′ ===

37. ()()() () 2 345,6,1 234 2 3 yx yy y y −=− ′ −= ′ =

At () 2 6,1:1 13 y ′ ==−

Tangentline: ()116 7 yx yx −=−− =−+

38. ()()() ()() ()() () 22 2337,4,4 22230 32 2 3 xy xyy yyx x y y ++−= ′ ++−= ′ −=−+ + ′ =−

At () 6 4,4:6 1 y ′ =−=−

Tangentline: ()464 628 yx yx −=−− =−+

39. () 2222 22 2 2 940,4,23 221880 182 28 xyxy xyyxyxyy xxy y x yy −−=− ′′ +−−= ′ =

At () ()()() ()() 1842412 4,23: 21623163 2413 483236 y ′ −= ===

Tangentline: () 3 234 6 38 3 63 yx yx −=+ =+

40. () 2323 1313 1313 13 5,8,1 22 0 33 xy xyy xy y yx += ′ +=  ′ ==−

At () 1 8,1: 2 y ′ =−

Tangentline: () 1 18 2 1 5 2 yx yx −=−− =−+

41. ()() () ()()() 2 2222 22 3100,4,2 62210022 xyxy xyxyyxyy +=− ′′ ++=− () ()()() 2 11

At4,2: 61648410084 960480800400 880160 yy yy y y ′′ ++=− +=−′′ ′ =− ′ =−

Tangentline: () 2 11 230 1111 24 112300 yx yx yx −=−− +−= =−+

42. () () 2222 2242 223 2,1,1 2 2244 yxyx yxyx yyxxyyyx += += ′′ ++= () 1 3

At1,1: 2244 62 yy y y ′′ ++= ′ = ′ =

Tangentline: () 1 3 12 33 11yx yx −=− =+

43. Answerswillvary. Sample answers: 2 2 22 2 2 2 2 2 4 2 xyy x x yxxy x xy xyy =  = +=  = += +=

44. Theequation2221 xy++= implies221, xy+=− whichhasnorealsolutions.

45. (a) () 22 1,1,2 28 0 4 4 xy yy x x y y += ′ += ′ =−

At ()1,2:2 y ′ =−

Tangentline: () 221 24 yx yx −=−− =−+ (b) ()() 222 22222 2 0 0000 2 0 22 0000 2222 22 10 ,Tangentlineat, ′ ′ +=  +=  = −=− −=+ xyxyybx y ababay bx yyxxxy ay yyyxxx bbaa

Because 22 00 221,xy ab += youhave00221.yyxx ba +=

Note: Frompart(a), () () 1211 1124, 2842 xy yxyx +=  =−+  =−+

Tangentline.

46. (a) () 22 1,3,2 68 0 34 43 4 3 xy xy y yx y x y y −=− ′ −= ′ = ′ = At () () () 43 3,2:2 32 y ′ −==−

Tangentline: () 223 24 yx yx +=−− =−+ (b) ()() 222 22222 2 0 0000 2 0 22 0000 2222 22 10 ,Tangentlineat, ′ ′ −=  −=  = −=− −=− xyxyyxb y ababya xb yyxxxy ya yyyxxx bbaa

Because 22 00 221,xy ab −= youhave00221.xxyy ab −=

Note: Frompart(a), () 2 31 1124, 6824 y xy xyx −=  +=  =−+

Tangentline.

47. 2 2 2 222 2 tan sec1 1 cos, sec22 sec1tan1 1 1 yx yy yyy y yyx y x ππ = ′ = ′ ==−<< =+=+ ′ = +

48. 22 22 22 2 cos sin1 1 ,0 sin sincos1 sin1cos sin1cos1 1 ,11 1 yx yy yy y yy yy yyx yx x π = ′ −⋅= ′ =<< += =− =−=− ′ =−<<

xyxy xyxyy xyxy xy y x

53.

7sin2

1212 5 11 0 22 xy xyy y y x += ′ +=

= At () 2 9,4: 3 y ′ =− Tangentline: () 2 49 3 23300 yx xy −=−− +−= 56. ()()()() () () () 2 2 222 2222 2 22 1 1 1112122 2 11 12 21 x y x xxx x xx yy xx xx y yx = + +−− +−+ ′ == ++ +−

+ At () () 2 51441 2,: 5105 25 41 5 y

Tangentline: () 51 2 5105 105102 10580 yx yx xy −=− −=− −+= 1 114 9 (9,4) 1 1 5 52,5 () 1

57. 2225 220 xy xyy x y y += ′ += ′ =

At ()4,3:

Tangentline: () 4 3443250 3 yxxy −=−  +−=

Normalline: () 3 34340 4 yxxy −=−  −=

58. 2236 220 xy xyy x y y += ′ += ′ =−

At ()6,0;slopeisundefined.

Tangentline:6 x = Normalline:0 y =

At ()3,4: Tangentline: () 3 4334250 4 yxxy −=+  −+=

Normalline: () 4 43430 3 yxxy −=+  +=

59. 222 220

slopeofnormalline xyr xyy x y y y x += ′ += ′ == =

slopeoftangentline

At () 5,11,slopeis 5 11

Tangentline: () 5 115 11 1111525 511360 yx yx xy −=− −=−+ +−=

Normalline: () 11 115 5 551111511 5110 yx yx yx −=− −=− −=

(5,11)

Let () 00 , x y beapointonthecircle.If00, x = thenthetangentlineishorizontal,thenormallineisverticaland,hence,passes throughtheorigin.If00, x ≠ thentheequationofthenormallineis () 0 00 0 0 0 y yyxx x y yx x −=− = whichpassesthroughtheorigin.

60. () 24 24 2 1at1,2 yx yy y y = ′ = ′ ==

Equationofnormallineat ()1,2is

() 211,3. yxyx −=−−=−

Thecentersofthe circlesmustbeonthenormallineandatadistanceof 4unitsfrom ()1,2.Therefore, ()() () 22 2 13216 2116 122. xx x x −+   =  −= =±

Centersofthecircles: () 122,222 +− and () 122,222 −+

Equations: () () () () 22 22 12222216 12222216 xy xy −−+−+= −++−−=

62. 22 48440 82840 8844 242 xyxy xyyy x x y yy +−++= ′′ +−+= ′ == ++

Horizontaltangentsoccurwhen1: x = ()() () 22 2 4181440 4400,4 yy yyyyy +−++= +=+=  =−

Horizontaltangents: ()() 1,0,1,4

Verticaltangentsoccurwhen2: y =− ()() () 22 2 4284240 484200,2 xx xxxxx +−−+−+= −=−=  =

61. 22 25162001604000 50322001600 20050 16032 xyxy xyyy x y y ++−+= ′′ ++−= + ′ =

Horizontaltangentsoccurwhen4: x =− ()() () 2 25161620041604000 1000,10 yy yyy ++−−+= −=  =

Horizontaltangents: ()() 4,0,4,10

Verticaltangentsoccurwhen5: y = () 2 254002008004000 25800,8 xx xxx ++−+= +=  =−

Verticaltangents: ()() 0,5,8,5

Verticaltangents: ()() 0,2,2,2 x 842106 2 10 6 4 (0,5) (4,10) (4,0) (8,5) y x 11234

63. Findthepointsofintersectionbyletting24yx = intheequation2226. xy+= ()() 2 246and310 xxxx+=+−=

Thecurvesintersectat () 1,2. ± :: 42024 22 EllipseParabola xyyyy x yy yy ′′ +== ′′

At ()1,2,theslopesare: 11yy ′′ =−=

At () 1,2,theslopesare: 11yy ′′ ==−

Tangentsareperpendicular.

64. Findthepointsofintersectionbyletting23 yx = intheequation22235. xy+= 2332 235and3250 xxxx+=+−=

Intersectwhen1. x =

Pointsofintersection: () 1,1

At ()1,1,theslopesare: 32 23yy ′′ ==−

At ()1,1,theslopesare: 32 23yy ′′ =−=

Tangentsareperpendicular.

65. andsin yxxy =−=

Pointofintersection: () 0,0 :sin: 11cos sec yxxy yyy yy =−= ′′ =−=

At ()0,0,theslopesare: 11yy ′′

Tangentsareperpendicular.

66. Rewritingeachequationanddifferentiating:

Foreachvalueof x,thederivativesarenegative reciprocalsofeachother. So,thetangentlinesareorthogonalatbothpointsof intersection. 67. 22

Atthepointofintersection (),,x y theproductofthe slopesis ()() ()() 1. xyKxKxK −=−=− Thecurves areorthogonal. 69.

Atanypointofintersection () ,x y theproductofthe slopesis ()() 1. yxxy −=− Thecurvesareorthogonal.

70. (a)Theslopeisgreaterat3. x =−

(b)Thegraphhasverticaltangentlinesatabout ()() 2,3and2,3.

(c)Thegraphhasahorizontaltangentlineatabout ()0,6.

71. (a) ()

Usestartingpoint B.

(b)24 1 394 4 yxx =  =− 24 42 2 3616 16360 16256144 828 2 xx xx x =− −+= ±− ==±

Notethat () 2 282882717. x =±=±=± So,therearefourvaluesof x: 17,17,17,17 −−−−++

Tofindtheslope, () () 2 3 8 28. 23 x x yyxxy ′′ =−  =

For () 1 17,77, 3 xy ′ =−−=+ andthelineis

()() () 1 11 77173778723. 33 yxx

=++++=+++

For () 1 17,77, 3 xy ′ =−=− andthelineis ()() () 2 11 77173772387. 33 yxx

=−−++=−+−

For () 1 17,77, 3 xy ′ =−+=−− andthelineis

()() () () 3 11 77173772387. 33 yxx 

=−−+−+=−−−−

For () 1 17,77, 3 xy ′ =+=−+ andthelineis

()() () () 4 11 77173778723. 33 yxx 

=−+−−+=−+−+

(c)Equating3 y and4 : y ()() ()() ()() ()() 11 7717377173 33 77177717 77777777777777 16714 87 7 xx xx xxxx x x −−+−+=−+−−+ −+−=+−− +−−−+=−−+−− = =

If87, 7 x = then5 y = andthelinesintersectat 87 ,5. 7

72. 11 0 22 x yc dy dx xy y dy dx x += += =−

Tangentlineat () () 0 0000 0 ,: y x yyyxx x −=−−

x-intercept: () 000,0 xxy +

y-intercept: () 0,000 yxy +

Sumofintercepts:

73. 11 11 1 1 1 ;,integersand0 pq qp qp pp qq p pqpq p yxpqq yx qyypx pxpxy y qyqy pxp xx qxq => = ′ = ′ =⋅=⋅ =⋅=

So,if1 ,,then. nn yxnpqynx ′ ===

75. () () 22 2 1,4,0 49 22 0 49 9 4 90 44 944 += ′ += ′ = = −−= xy xyy x y y xy yx xxy

Points:6,8and6,8

But,2222 94364369. x yyx+=  =− So,22293643691. xxyxx −+==−  =

Pointsonellipse: 3 1,3 2  ±

() 3993

′ ===−

At1,3: 242 4323 33

′ −=

x y y y

At1,3: 22

Tangentlines: () () 33 423 22 33 423 22 yxx yxx =−−=−+ =−=−

2 12 1 2,slopeoftangentline xy yy y y = ′ = ′ =

Considertheslopeofthenormallinejoining () 0,0x and () () 2 ,,x yyy = ontheparabola.

2 0 2 0 2 0 0 2 1 2 1 2 y y yx yx yx −= −=− =−

(a)If014, x = then2111 424, y =−=− whichis impossible.So,theonlynormallineisthe x-axis ()0. y =

(b)If012, x = then200. yy =  = Sameaspart(a).

(c)If01, x = then21 2 yx == andtherearethree normallines.

The x-axis,thelinejoining () 0,0x and 11 2,, 2 

andthelinejoining () 0,0x and 11 , 22 

(d)Iftwonormalsareperpendicular,thentheirslopes are–1and1.So, () 2 0 00 0 01 21 2 and 12113 1. 14424 y yy yx xx x −=−=  = =−  −=− 

Theperpendicularnormallinesare 3 4 yx=−+ and 3 4 yx=−

Section2.6RelatedRates

1. Arelated-rateequationisanequationthatrelatesthe ratesofchangeofvariousquantities.

77. (a) 22 1 328 22 0 3284 xy x yyx y y += ′ ′ +=  = (b)

At () () 41 4,2: 422 y ′ ==−

Slopeofnormallineis2. ()224 26 yx yx −=− =−

Secondpoint: 2846 , 1717   

2. Answerswillvary.Seepage153. 3. 1 2 2 yx dydx dtdt x dxdy x dtdt =  =   =

(a)When4 x = and3: dxdt = () 13 3 244 dy dt ==

(b)When25 x = and2: dydt = () 225220 dx dt == 4 6 4 6

4. () 2 35 65 1 65 yxx dydx x dtdt dxdy dtxdt =− =− =

(a)When3 x = and2: dx dt =

() 635226 dy dt =   = 

(b)When2 x = and4: dy dt = () () 14 4 6257 dx dt ==

5. 4 0 xy dydx xydtdt dyydx dtxdt dxxdy dtydt = +=

=−

(a)When8,1/2, xy== and10: dxdt = () 1/25 10 88 dy dt =−=−

(b)When1,4, xy== and6: dydt =− () 13 6 42 dx dt =−−=

6. 2225 220 xy dxdy xydtdt dyxdx dtydt dxydy dtxdt += +=

(a)When3,4, xy== and8: dxdt = () 3 86 4 dy dt =−=−

(b)When4,3, xy== and2: dydt =−

() 33 2 42 dx dt =−−=

7. 2 21 2 4 yx dx dt dydx x dtdt =+ = =

(a)When1: x =−

()() 4128cm/sec dy dt =−=−

(b)When0: x =

()() 4020cm/sec dy dt ==

(c)When1: x =

()() 4128cm/sec dy dt == 8. () () () () 2 22 2222 1 ,6 1 2 1 212 6 11 dx y xdt dyxdx dtdt x x x xx == + =⋅ + == ++

(a)When2: x =−

()() () 22 12224 1225 dy dt ==  +−  in./sec

(b)When0: x =

() () 2 120 0 10 dy dt == + in./sec

(c)When2: x =

()() ()22 12224 in./sec 1225 dy dt ==− +

(a)When:

(b)When: 4 x

(c)When0:

(a)When: 6 x

(c)When: 3 x π =

4sin423cm/sec 32

11. 2 4 2 Ar dr dt dAdr r dtdt π π = = =

When ()() 2 37,2374296cmmin. dA r dt ===ππ

12. () 23 4 13 33 2 42 = = == s A ds dt dAdssds s dtdtdt

When ()() 353332 41,4113fthr. 22 dA s dt ===

13. 3 2 4 3 3 4 Vr dr dt dVdr r dtdt π π = = =

(a)When9, r = ()()23 493972in./min. dV dt ==ππ

When36, r = ()()23 436315,552in./min. dV dt ==ππ

(b)If drdt isconstant, dVdt isproportionalto2 r

14. 432 ,4 3 800 dVdr Vrr dtdt dV dt ==ππ = (a)2 4 drdVdt dtr π = At () 2 30,8002cmmin. 4309 dr r dt π π === At () 2 8008 85,cmmin. 485289 dr r dt π π ===

(b) dr dt dependson2, r not r

15. 3 2 6 3 Vx dx dt dVdx x dtdt = = =

(a)When2, x = ()()23 32672cm/sec. dV dt ==

(b)When10, x = ()()23 31061800cm/sec. dV dt ==

16. 62 6 12 sx dx dt dsdx x dtdt = = =

(a)When2, x = ()() 12261442cm/sec. ds dt ==

(b)When10, x = ()() 121067202cm/sec. ds dt ==

17. [] () 22 3 2 2 119because23 334 3 4 10 94 49 Vrhhhrh h dV

When15, h = () () 2 4108 ft/min. 915405 dh dt π π ==

19. (a)Totalvolumeofpool ()()()()()() 13 21261612144m 2 =+=

Volumeof1mofwater ()()() 13 16618m 2 == (seesimilartrianglediagram)

%poolfilled () 18 100%12.5% 144 ==

(b)Becausefor02,6, hbh≤≤= youhave ()() 12 633618 2 Vbhbhhhh ==== () 1111 36m/min. 41441441144 dVdhdh h dtdtdth ==  ===

20. ()()121266since 2 Vbhbhhbh ====

(a)1 12 12 dVdhdhdV h dtdtdthdt =  = When () () 11 1and2,2ft/min. 1216 dVdh h dtdt ====

(b)If ()() 31133 in./minft/minand2ft,then122ft/min. 832324 dhdV h dtdt

21. 222 25 220 2 because2. xy dxdy xydtdt dyxdxxdx dtydtydt += += =⋅==

(a)When () 277 7,57624,ft/sec. 2412 dy xy dt =====−

When () 2153 15,40020,ft/sec. 202 dy xy dt =====−

When () 22448 24,7,ft/sec. 77 dy xy dt ====−

(b) 1 2 1 2 Axy dAdydx xy dtdtdt =  =+

Frompart(a)youhave7 7,24,2,and. 12 dxdy xy dtdt ====− So, () 175272 7242ft/sec. 21224 dA dt

Using7 7,24,2, 12 dxdy xy dtdt ====− and24 cos, 25 θ = youhave () () 2 2 2417712rad/sec. 25241212 24 d dt

22. 2225 220

23. When226,12663,yx==−= and () 22121083612.sxy=+−=+=

() ()() () 2 22 12 221212 12 +−= +−−= +−= xys dxdyds x ys dtdtdt dxdyds xys dtdtdt

Also,22212. xy+= 220dxdydyxdx xydtdtdtydt +=  =

So, ()12. dxxdxds xys dtydtdt  +−= 

dxxdsdxsyds xxs dtydtdtxdt

m/sec(horizontal) () 3 631 6155m/sec(vertical) dyxdx dtydt ==⋅=

24. Let L bethelengthoftherope. (a)22 144 22 4 since4ft/sec Lx dLdx Lx dtdt dxLdLLdL dtxdtxdt =+ =

(b)If

When13: L = () 21441691445 41352 10.4ft/sec 55 =−=−= =−=−=− xL dx dt

Speedoftheboatincreasesasitapproachesthedock.

25. (a)

() () 222 450 600 222 sxy dx dt dy dt dsdxdy sxydtdtdt x dxdtydydt ds dts =+ =− =− =+ + =

When225 x = and300, y = 375 s = and ()() 225450300600 750mi/h. 375 ds dt −+− ==−

(b)3751h30min 7502 t ===

26. 222 202because0 xys dxdsdy xs dtdtdt dxsds dtxdt +=

+==

When10,100257553, sx==−== () 10480 2401603277.13mi/h. 533 dx dt ===≈

29. (a) () 15 15156 6 5 3 5 5525 5ft/sec 333 y yxy yx yx dx dt dydx dtdt =

=⋅== (b) () 25105ft/sec 33 dyx dydx dtdtdt =−=−=

Section 2.6 Related Rates 191

27. 222 90 20 25 22 sx x dx dt dsdxdsxdx sx dtdtdtsdt =+ = =− =  =⋅

When2220,90201085,xs==+= () 2050 255.42ft/sec. 108585 ds dt =−=≈−

28. 222 90 902070 25 sx x dx dt dsxdx dtsdt =+ =−= = =⋅

When2270,907010130,xs==+= () 70175 2515.35ft/sec. 10130130 ds dt ==≈

30. (a)

20206

31. () 122sin,1 26 t xtxy π =+=

(a)Period: 2 12seconds 6 π π =

Lowestpoint: 3 0, 2

(c)When1,

32. () 322sin,1 5 xttxy π =+=

(a)Period:22seconds π π =

(c)When 2 3115 ,1and 1044

==−=

xy 3311 sinsin 10526 ππ =  =  = ttt 22 3 cos 5 1 220 dx t dt xy dxdydyxdx xydtdtdtydt ππ = += +=  =

So, 3103 cos 15456 995 255125 dy dt

Speed 95 0.5058m/sec 125 π =≈

33. Becausetheevaporationrateisproportionaltothesurfacearea, () 2 4.dVdtkr π = However,because () 3 43,Vr π = you have24. dVdr r dtdt π = Therefore, ()22 44. drdr krrk dtdt ππ =  =

34.(i) (a)negativepositive dxdy dtdt 

(b)positivenegative dydx dtdt  (ii) (a)negativenegative dxdy dtdt 

(b)positivepositive dydx dtdt 

35. (a) ()3 dydtdxdt = meansthat y changesthreetimes asfastas x changes. (b) y changesslowlywhen0or. x xL≈≈ y changes morerapidlywhen x isnearthemiddleofthe interval.

36. No.32 ,3 dVds Vss dtdt == If ds dt isconstant,then dV dt is3s 2 timesthatconstant.

37. 12 1 2 12 222 12 111 1 1.5 111 RRR dR dt dR dt dRdRdR R dtRdtRdt =+ = = ⋅=⋅+⋅

When150 R = and275: R = 30 R = () () () () () 2 22 11 3011.50.6ohm/sec 5075 dR dt

38. 1 VIR dVdRdI IR dtdtdt dIdVIdR dtRdtRdt = =+ =−

When12,4,3, dV VR dt === and 12 2,3 4 dRV I dtR ==== and () () 133 32amperessec.444 dI dt =−=−

39. sin18 x y °= ()() 2 1 0 sin1827584.9797mi/hr xdydx ydtydt dxxdy dtydt

40. 2 2 tan 50 4m/sec 1 sec 50 1 cos 50 y dy dt ddy dtdt ddy dtdt θ θ θ θ θ = = ⋅= =⋅

When50,, 4 y π θ == and 2 cos. 2 θ = So, () 2 121 4rad/sec. 50225

41. () () () 2 2 222 sin10 1ft/sec 10 cos 10 sec 1025 1 252510 1012 255212521 221 0.017rad/sec 525 x dx dt ddx dtxdt ddx dtxdt θ

42.

44. ()() 10rev/sec2rad/rev20rad/sec d dt θ ==ππ

(a) () cos 30 1 sin 30 30sin 30sin20 600sin x ddx dtdt dxd dtdt θ θ θ θ θ θπ πθ = −= =− =− =−

(b)

(a)When30, θ =° 120130rad/hrad/min. 42 d

(b)When60, θ

(c)When75, θ =° 2 120sin75111.96rad/h1.87rad/min. d dt θ =°≈≈

43. () 2 2 tan 50 30260rad/minrad/sec 1 sec 50 50sec x d

() sin1or90180. 2 nn π θθπ =  =+°+⋅° () isleastwhenor180.dxdtnnθπ=⋅°

(d)For30, θ =° () 1 600sin30600300cm/sec. 2 dx dt =−°=−=−πππ

For60, θ =° ()600sin60 3 6003003cm/sec. 2 dx dt π ππ =−° =−=−

45. (a) () 22 12 sin2sin 22 coscos 22 11 2sincos 2222 2sincossin 2222 b bs s h hs s Abhss ss θθ θθ θθ θθ θ =  = = 

(b) 21 coswhererad/min. 22 dAsdd dtdtdt θθ θ ==

When 22 313 ,. 62228 dAss dt π θ

When 22 11 ,. 32228 dAss dt π θ

(a)When20030,ft/sec. 3 dx dt π θ =°=

(b)When60,200ft/sec. dx dt θπ =°=

(c)When70,427.43ft/sec. dx dt θπ =°≈

(c)If s and d dt θ isconstant, dA dt isproportionaltocos. θ

46. tan50tan 50 x x θθ =  = 2 2 2 50sec 250sec 1 cos, 2544 dxd dtdt d dt d dt θ θ θ θ θππ θθ = = =−≤≤

47. (a)Usingagraphingutility, () 32 0.00960.55910.5461.5.rffff =−+−

(b) () 2 0.02881.11810.54 drdrdfdf ff dtdfdtdt ==−+

For9,16.3 tf== fromthetableundertheyear 2009.

Also,1.25, df dt = soyouhave ()() () () 2 0.028816.31.11816.310.541.25 0.03941millionparticipantsperyear. dr dt =−+ =−

48. () () () 2 4.920 9.8 14.92015.1 19.8 ytt dy t dt y y =−+ =− =−+= ′ =− Bysimilartriangles:2012

20240 y xx xxy = −=

When15.1: y = () () 2024015.1 2015.1240 240 4.9 xx x x −= −= = 20240 20 20 xxy dxdydx x y dtdtdt dxxdy dtydt −= =+ = At () 2404.9 1,9.897.96m/sec. 2015.1 dx t dt ==−≈−

49. 2225;xy+= accelerationofthetopoftheladder 2 2 dy dt =

Firstderivative:220 0 dxdy xydtdt dxdy xydtdt += +=

Secondderivative: 22 22 2222 22 0 1 dxdxdxdydydy xy dtdtdtdtdtdt dydxdxdy x dtydtdtdt +⋅++⋅=

Firstderivative:22

Secondderivative:

4. () () ()() () () () () 0 00002 6 lim 66 666666limlimlimlim x xxxx fx x fxxfx fx x xxx x xxx x xxxxxxxxxxxx

5. () () ()() () ()() ()() 2 2 2 2 2 2 23,2 2 2lim 2 232 lim 2 221 lim 2 lim212215 x x x x gxxxc gxg g x xx x xx x x → → → → =−= ′ = = −+ = =+=+=

6. () 1 ,3 4 fxc x == + () ()() ()() () 3 3 3 3 3 3lim 3 11 lim47 3 74 lim 347 11 lim 4749 x x x x fxf f x x x x xx x → →

7. f isdifferentiableforall3. x ≠

8. f isdifferentiableforall1. x ≠−

9. 25 0 y y = ′ =

10. () () 6 0 ft ft π = ′ = 11. () () 32 2 11 322 f xxx f xxx =− ′ =−

12. () () 54 43 32 '158 g sss g sss =− =−

13. () () 31213 1223 32 6363 31 3 hxxxxx hxxx x x =+=+ ′ =+=+

14. () () 1256 1211615 26 fxxx fxxx =− ′ =+

15. () () 2 3 3 2 3 44 33 gtt gtt t = ′ ==−

16. () () 4 4 5 5 88 55 3232 55 hxx x hxx x == ′ =−=−

17. () () 45sin 45cos f f θθθ θθ =− ′ =−

18. () () 4cos6 4sin g g αα αα =+ ′ =−

19. () () sin 3cos 4 3sincos 4 f f θ θθ θ θθ =− ′ =−−

20. () () 5sin 2 3 5cos 2 3 g g α αα α α =− ′ =−

21. () () ()() () 3 3 4 4 4 27 27,3,1 81 273 81 31 3 fxx x fxx x f == ′ =−=− ′ =−=−

22. ()() () () 2 34,1,1 64 1642 fxxx fxx f =−− ′ =− ′ =−=

23. ()() () () 5 4 43sin,0,0 203cos 0312 fxxxx fxxx f =+− ′ =+− ′ =−=

24. ()() () () 5cos9,0,5 5sin9 05sin099 fxxx fxx f =− ′ =−− ′ =−−=−

25. () 200 100 FT Ft T = ′ =

(a)When () 4,450TF ′ == vibrations/sec/lb.

(b)When () 1 3 9,933TF ′ == vibrations/sec/lb.

26. 62 12 = = Sx dS x dx When () 4,124482in.in. dS x dx ===

27. () 2 0000 16;600,30sttvtssv=−++==−

(a) () ()() 2 1630600 3230 sttt stvtt =−−+ ′ ==−−

29. () ()() () ()() ()() () 22 22 3232 32 32 5846 58244610 10162032104060 20604432 4515118 fxxxx fxxxxxx x xxxxx xxx xxx =+−− ′ =+−+−− =+−−+−− =−−− =−−−

30. () ()() () ()()()() 3 32 332 32 2534 2533465 61518241520 24243020 gxxxx gxxxxx xxxxx xxx =+− ′ =++−+ =++−+− =−+−

31. ()() ()() 91sin 91cos9sin 9coscos9sin fxxx fxxxx x xxx =− ′ =−+ =−+

32. () ()() () 5 54 54 2cos 2sincos10 2sin10cos fttt f ttttt tttt = ′ =−+ =−+

(b) ()() 31 Averagevelocity31

366554 2 94ft/sec s s = = =−

(d) () 2 01630600sttt==−−+ UsingagraphingutilityortheQuadraticFormula, 5.258 t ≈ seconds.

(e)When ()() 5.258,325.25830198.3ft/sec.tvt≈≈−−≈−

28. () ()() ()() ()() 2 00 2 1 2 16450 32 232264ftsec 5325160ftsec =−++ =−+ ′ ==− =−=− =−=− stgtvts t vtstt v v

33. () () ()() ()() () () () 2 2 22 22 2 22 1 1 12112 1 1 1 xx fx x x xxxx fx x x x +− = −+−+− ′ = −+ =

34. () () ()()()() () () () () () 2 2 22 22 22 22 2222 27 4 42272 4 28414 4 2148274 44 x fx x xxx fx x xxx x xx xx xx + = + +−+ ′ = + +−− = + −+− −−+ == ++

35. ()() 4 34 2 34 2 cos cos4sin cos 4cossin cos x y x x xxx y x xxxx x = ′ = + =

36. () ()() () 4 43 425 sin cossin4cos4sin x y x xxxx x xx y x x = ′ ==

37. 2 2 3sec 3sectan6sec yxx y xxxxx = ′ =+

38. 2 22 tan sec2tan yxx yxxxx =− ′ =−−

39. cossin sincoscossin yxxx yxxxxxx =− ′ =−+−=−

40. () () () 4 342 342 cot3cos 4cotcsc3cos3sin 4cotcsc3cos3sin gxxxxx g xxxxxxxx x xxxxxx =+ ′ =+−+− =−+−

41. ()()() () ()()() ()() () 2 2 222 25,1,6 2251 245345 13454 fxxx fxxxx xxxxx f =++−

=+++ =+++=++ ′ −=−+=

Tangentline: () 641 410 yx yx −=+ =+

42. ()()() () ()()() ()() 2 2 22 2 461,0,4 426611 222461 3425 fxxxx fxxxxx xxxx xx =−+− ′ =−+++− =−−++− =+− ()0002525 f ′ =+−=−

Tangentline: ()4250 254 yx yx −=−− =−+

43. () () ()() () () () 22 11 ,,3 12 112 11 12 8 214 x fx x xx fx xx f +  =−  −−+ ′ ==

′ ==−

Tangentline: 1 38 2 81 yx yx

+=−−

=−+

44. () () ()()()() () () 2 2 cos,,1 cos2 cossincossin cos 2sin cos 2 2 21 x fx x x xxx fx x x x f π π 1+  =  1−

1−−−1+ ′ = 1− = 1−

Tangentline:122 21 yx yx π π  −=−−

45. () () () 3 2 8512 245 48 gttt gtt gtt =−−+ ′ =−− ′′ =−

46. () () () 22 3 4 4 67 1214 36 361414 hxxx hxxx hxx x =+ ′ =−+ ′′ =+=+

47. () () () 52 32 12 75 2 225225 44 15 fxx fxx f xxx = ′ = ′′ ==

48. () () () 515 45 95 95 2020 4 1616 55 fxxx fxx fxx x == ′ = ′′ ==−

49. () () ()() 2 2 3tan 3sec 6secsectan 6sectan f f f θθ θθ θθθθ θθ = ′ = ′′ = =

50. () () () 10cos15sin 10sin15cos 10cos15sin httt httt httt =− ′ =−− ′′ =−+

51. () () ()() 2 2 4cot 4csc 8csccsccot 8csccot gxx gxx g xxxx xx = ′ =− ′′ =−− =

52. () ()() () () () () 2 32 12csc 12csccot12csccot 12csccsc12cotcsccot 12csccsccot =− ′ =−−= ′′ =−+− =−+ htt httttt htttttt ttt

53. () ()() () ()() 2 2 2 20,06 2 320311m/sec 3236m/sec =−≤≤ ′ ==− =−= =−=− vttt atvtt v a

54. () () ()() () () () 2 22 90 410 41090904 410 900225 41025 = + +− = + == ++ t vt t tt at t tt

(a) () () 2 90 16.43ft/sec 14 225 14.59ft/sec 49 v a =≈ =≈

(b) () () () 2 2 905 515ft/sec 30 225 51ft/sec 15 v a == == (c) () () () 2 2 9010 1018ft/sec 50 225 100.36ft/sec 25 v a == ==

55. () ()()() 4 33 73 47372873 yx yxx =+ ′ =+=+

56. () () () () 23 2222 6 36266 yx yxxxx =− ′ =−=−

57. () () () () () 23 23 24 24 1 5 5 352 6 5 ==+ + ′ =−+ =− + yx x yxx x x

58. () () () ()()() () 2 2 3 3 1 51 51 10 2515 51 ==+ + ′ =−+=− + fxx x fxx x

59. () ()()() 5cos91 5sin91945sin91 yx yxx =+ ′ =−+=−+

60. ()() 4 4334 6sin3 6cos31272cos3 yx y xxxx =− ′ =−=−

61. () () 2 sin2 24 111 cos221cos2sin 242 xx y yxxx =− ′ =−=−=

62. ()() () 75 64 52 53 secsec 75 secsectansecsectan sectansec1 sectan xx y yxxxxxx xxx xx =− ′ =− =− =

63. () ()()()() ()() ()() ()() 5 45 45 4 4 61 5616611 306161 613061 61361 yxx yxxx xxx xxx xx =+ ′ =+++ =+++ =+++ =++

64. () () () () () ()()()() () () ()() () () 52 23 52322232 32 223 32 23 5 2 15 13512 13155 18325 fsss f ssssss sssss ssss =−+ ′ =−++− =−−++ =−−+

65. () () () () ()() () () () () 3 21212 2 32 2 5/2 1 2 5 515 3 55 325 525 310 25 x fx x xxx x fx x x xx x x x xx x  =  +    +−+  

′ =  

+ +  

 +−  = +  +  + = +

66. () () ()()()() () ()() () 2 2 2 22 2 2 23 5 3 53152 2 33 25103 3 x hx x x xx x hx x x xxx x +  =

67. () () () () () () () 3 12232 3 1,2,3 13 13 221 12 22 23 fxx x fxxx x f =−− ′ =−−= ′ −==−

68. () () () () () () () () () 32 223 223 1,3,2 12 12 331 231 3 342 fxx x fxxx x f =− ′ =−= ′ ==

69. () () () () () ()()()() () () () 12 1212 1 2 8 ,0,8 31 3118313 31 143 011 1 x fx x xxx fx x f + = + +−++ ′ = + ′ ==−

70. () () () () ()()()()()() () () ()() 3 32 6 31 ,1,4 43 433313434 43 3434 145 1 x fx x xxx fx x f + = −−+− ′ = ′ ==−

76. () 2 2 22 22 sin sin2sincos 2sincos2sincos2cos2sin 4sincos2cossin yxx yxxxx yxxxxxxxx xxxxx = ′ =+ ′′ =++− =+−

71. 11 csc2,, 242 csc2cot2 0 4 yx yxx y π π  =   ′ =−

72. 2 csc3cot3,,1 6 3csc3cot33csc3 033 6 yxx yxxx y π π  =+ 

=−=−

73. () ()()() ()()()() 3 22 85 38582485 24285838485 yx yxx yxx =+ ′ =+=+ ′′ =+=+

74. () ()()()() ()()()() () 1 22 3 3 1 51 51 1515551 50 52515 51 yx x yxx yx x =>+ + ′ =−+=−+ ′′ =−−+= +

75. () () ()() 2 2 cot csc 2csccsccot 2csccot fxx fxx f xxxx xx = ′ =− ′′ =−−⋅ =

77. () () () 2 21 22 700 410 700410 14002 410 T tt Ttt t T tt = ++ =++ −+ ′ = ++

(a) () () 2 When1, 140012 18.667deg/h. 1410 t T = −+ ′ =≈− ++ (b) () () 2 When3, 140032 7.284deg/h. 91210 t T = −+ ′ =≈− ++ (c) () () 2 When5, 140052 3.240deg/h. 252010 t T = −+ ′ =≈− ++ (d) () () 2 When10, 1400102 0.747deg/h. 1004010 t T = −+ ′ =≈− ++

78. () () () () ()() 11 cos8sin8 44 11 sin88cos88 44 2sin82cos8 Attime, 4 11 cos8sin8 44444 11 1ft. 44 2sin82cos8 444 20212ft/sec ytt ytt

2264 220 22 xy

80. () 23 2 2 2 46 24430 4324 24 34 xxyy xxyyyy x yyxy x y y yx +−=

81. () () () 33 3223 3232 3232 32 32 22 22 4 330 33 33 3 3 3 3 xyxy xyxyxyyy x yxyyyxy yxxyyxy yxy y xxy yyx y x xy −=

82. () () 4 14 22 28 82 2 8 24 84 29 932 xyxy y x yy yx xyyxyxyy xxyyxyy xyy y xxy x yy xxy xy xy =− ′′ +=− ′′ +=− ′ +=− ′ = + = +− =

83. () () sincos cossinsincos coscossinsin sinsin coscos xyyx x yyyyxyx yxyxyxy yxy y xxy = ′′ +=−+ ′ −=−− + ′ =

84. () ()() ()() () () () cos 1sin1 sin1sin 1sin csc11 sin += ′ −++= ′ −+=++ ++ ′ =−=−+− + xyx yxy yxyxy xy yx xy

85. () () () 2210 220 At3,1,3 Tangentline:1333100 1 Normalline:1330 3 xy xyy x y y y yxxy yxxy

86. () 2220 220

87. 2units/sec 1 24 2 = = =  == yx dy dt dydxdxdy x x dtdtdtdt x (a)1When,22units/sec. 2 dx x dt ==

(b)When1,4units/sec. dx x dt ==

(c)When4,8units/sec. dx x dt ==

88. Surface2area6,length Axx === ofedge 8 dx dt = ()() 2 12126.58624cm/sec dAdx x dtdt === 89. () ()() () 2 22 tan 32rad/min sec tan1661 x d dt ddx dtdt dx x dt

1 When, 2 115 61km/min450km/h. 42 x dx dt π ππ =

=+==

90. ()

ProblemSolvingforChapter2

1. (a) () 2 22 2 ,Circle ,Parabola xyrr xy +−= =

Substituting: () () 22 222 2 2 20 210 yrry yryrry yryy yyr

(b)Let () ,x y beapointoftangency: ()() 221220, x xybxybyy by ′′ +−=  +−=  = Circle

22, yxyx ′ =  = Parabola

Equating: () 2 21 11 22 x x by by byby =

Also, () 221 xyb+−= and2 yx = imply:

Center: 5 0, 4

Graph2 yx = and 2 251. 4 xy

2. Let () 2 , aa and () 2 ,25bbb−+− bethepointsoftangency.For2,2 y xyx ′ == andfor225, yxx=−+−

22.yx ′ =−+ So,2221, abab =−+  += or1. ab =− Furthermore,theslopeofthecommontangentlineis

For2,11 bab==−=− andthepointsoftangencyare ()1,1and ()2,5.Thetangentlinehasslope () 2:12121 yxyx−−=−=  =−−

For1,12 bab=−=−= andthepointsoftangencyare ()2,4and ()1,8.Thetangentlinehasslope () 4:44244 yxyx −=−  =−

3. Let () () 32 2 32. pxAxBxCxD pxAxBxC =+++ ′ =++

() At1,1:

() At1,3: =1Equation1

32=14Equation2

ABCD ABC +++ ++ =3Equation3 32=2Equation4

AddingEquations1and3:222 B D +=−

ABCD ABC +−+− ++−

SubtractingEquations1and3:224 AC+= () 1 2225.DB=−−=−

AddingEquations2and4:6212 AC+=

SubtractingEquations2and4:416 B =

So,4 B = and () 1 2225.DB=−−=− Subtracting224 AC+= and6212, AC+= youobtain482. AA =  = Finally, () 1 2420.CA=−= So, () 32 245.pxxx=+−

4. () cos f xabcx =+

() sin f xbccx ′ =−

() At0,1:1Equation1 33 At,:cosEquation2 4242

FromEquation1,1. ab =− Equation2becomes () 31 1coscos. 4242 bbbbcc ππ

5. (a) () () 2,2,Slope4at2,4 Tangentline:442

(b)Slopeofnormalline:1 4

Normalline:

FromEquation3, () 1 .So: sin4 b ccπ = () () 111 cos sin4sin442 1 1cossin

Graphingtheequation () 1 sincos1, 244 cc gcc

youseethatmanyvaluesof c willwork.Oneanswer: () 1331 2,,cos2 2222 cbafxx ==−=

(d)Letbeapointontheparabola

Tangentlineatis

Normallineatis

Tofindpointsofintersection,solve:

Secondintersectionpoint:

(c)Tangentline:0 y =

Normalline:0

Thenormallineintersectsasecondtimeat

7. (a)42222

(c)Differentiatingimplicitly:

Fourpoints:

8. (a)

30 90100 (0,30) (90,6) y x (100,3) Not drawn to scale

9. (a)Linedeterminedby ()0,30and ()90,6: () 3062444 30030 090901515 yxxxyx −=−=−=−  =−+

When100: x = () 410 100303 153 y =−+=>

Asyoucanseefromthefigure,theshadowdeterminedbythemanextendsbeyondtheshadowdeterminedbythechild.

(b)Linedeterminedby ()0,30and ()60,6: () 30622 30030 06055 yxxyx −=−=−  =−+

30 6070 (0,30)

(60,6) y x (70,3) Not drawn to scale

When70: x = () 2 703023 5 y =−+=<

Asyoucanseefromthefigure,theshadowdeterminedbythechildextendsbeyondtheshadowdeterminedbytheman.

(c)Need ()()() 0,30,,6,10,3 dd + collinear. () 3066324380feet 01010 d dddd =  =  = −−+

(d)Let y bethedistancefromthebaseofthestreetlighttothetipoftheshadow.Youknowthat/5. dxdt =−

For80, x > theshadowisdeterminedbytheman.

5 3064 yyx yx =  = and525 44 dydx dtdt ==

For80, x < theshadowisdeterminedbythechild.

1010100

30399 yyx yx =  =+ and1050 99 dydx dtdt ==−

Therefore: 25 ,80 4 50 ,080 9 x dy dt x

−>

−<<

/ dydt isnotcontinuousat80. x =

ALTERNATESOLUTIONforparts(a)and(b):

(a)Asbefore,thelinedeterminedbytheman’sshadowis 4 30 15 m yx=−+

Thelinedeterminedbythechild’sshadowisobtainedbyfindingthelinethrough ()0,30and () 100,3: () 30327 30030 0100100 c yxyx −=−  =−+

Bysetting0, mc yy== youcandeterminehowfartheshadowsextend:

41 Man:030112.5112 152 271 Child:030111.11111 1009 m c yxx yxx =  =  == =  =  ==

Theman’sshadowis117 1121111 2918 −= ftbeyondthechild’sshadow.

(b)Asbefore,thelinedeterminedbytheman’sshadowis

2 30 5 m yx=−+

Forthechild’sshadow, () 30327 30030 07070 c yxyx −=−  =−+ 2 Man:03075 5 277007

Child:03077 7099 m c yxx yxx =  =  = =  =  ==

Sothechild’sshadowis77 77752 99 −= ftbeyondtheman’sshadow.

10. (a) () 1323 23 1 3 1 18 3 12cm/sec dydx yxxdtdt dx dt dx dt =  = = = (b) () ()() ()() 2222 22 1// 22 2 812219849 cm/sec 6446817 x dxdtydydt dDdxdy Dxyxyxy dtdtdt xy +

+ === + (c) ()() 2 2 // tansec x dydtydxdt yd xdtx θ θθ =  ⋅=

Fromthetriangle,sec688. θ = So ()() () 81212164 rad/sec. 6468646817 d dt θ ===−

11. (a) () () 2 27 5 27 5 27ft/sec ft/sec vtt at =−+ =− (b) () () ()() 2 2727 55 27 10 270275seconds 55275673.5feet =−+=  =  = =−++= vtttt s (c)TheaccelerationduetogravityonEarthisgreaterinmagnitudethanthatonthemoon.

12. () ()() ()()() () () () () 0000 11 limlimlimlim xxxx ExxExExExExExEx ExExEx xxxx Δ→Δ→Δ→Δ→ +Δ−Δ−

Forexample: () x E xe = θ 8 2 68

Δ−  Δ− ′ ====  ΔΔΔΔ 

But, () ()() () 00 01 0limlim1. xx ExEEx E xxΔ→Δ→ Δ−Δ− ′ === ΔΔ So, ()()()() 0 ExExEEx ′′ == existsforall x

13. () ()() ()()() () 000 limlimlim xxx LxxLxLxLxLxLx Lx x xxΔ→Δ→Δ→ +Δ−+Δ−Δ ′ === ΔΔΔ

Also, () ()() 0 0 0lim. x LxL L x Δ→ Δ− ′ = Δ But, ()00 L = because ()()()()() 0000000. LLLLL =+=+  =

So, ()() 0 LxL ′′ = forall x.Thegraphof L isalinethroughtheoriginofslope ()0. L′

14. (a) (b) 0 0 sin lim0.0174533 sin Infact,lim. 180 z z z z

0.01745240.01745330.0174533

(d) () () () ()() () 90sin90sin1 1802 180cos1801 180 sincos 180 S C dd SzczcczCz dzdz

(e)Theformulasforthederivativesaremorecomplicatedindegrees.

15. ()() j tat ′ =

(a) () j t istherateofchangeofacceleration.

(b) () () () ()() 2 8.2566 16.566 16.5 0 s ttt vtt at atjt =−+ =−+ =− ′ ==

Theaccelerationisconstant,so () 0. jt =

Chapter 2, Section 1, page 100

Theline5 yx=− appearstobetangenttothe graphof f atthepoint ()0,5becauseitseemsto intersectthegraphatonlythatpoint.

Chapter 2, Section 2, page 110 a.

Theexponentof f becomesthecoefficientof f ′ andthepowerof x decreasesby1. ()() 1 nn xnx

Chapter 2, Section 4, page 134

a. UsingtheQuotientRule:

UsingtheChainRule:

b. Usingalgebrabeforedifferentiating:322

UsingtheChainRule: ()()22 32131212 ′ =+=++ yxxx

c. UsingatrigonometricidentityandtheProductRule:

UsingtheChainRule:

Ingeneral,theChainRuleissimpler.

Chapter 2, Section 6, page 152

Therateofchangeinthevolumedoesdependonthevaluesof

CHAPTER11

VectorsandtheGeometryofSpace

Section11.1V ectorsinthePlane

1. Answerswillvary. Sample answer:Ascalarisareal numbersuchas2.Avectorisrepresentedbyadirected linesegment.Avectorhasbothmagnitudeanddirection. Forexample3,1hasdirection6π andamagnitudeof 2.

2. Noticethat () 6,724,16. QP =−=−−−−= v

Hence, Q istheinitialpointand P istheterminalpoint.

3. (a)51,424,2 v =−−= (b)

4. (a) () 42,336,0 v =−−−−−=− (b)

5. 53,622,4 31,842,4 u v

6. () () 14,805,8 72,715,8 u v uv

9. (b)52,503,5 v =−−= (c)35 vij =+ (a),(d)

10. (b) () 34,661,12 v =−−−=− (c)12 vij =−+ (a),(d)

11. (b)68,132,4 v =−−−=−−

12. (b) () 50,145,3 v =−−−−−=− (c)53 vij =−+ (a)and(d). 2345

7. 60,236,5 93,5106,5 u v uv

8. ()() 114,4115,3 250,101315,3 u v uv

5(3,5) (2,0) (5,5) v y 642 826810 4 6 2 4 6 8 x v y (4,6) (3,6) (1,12) x (8,3) (6,1) (2,4) v 2 4248 6 2 4 6 y x v 6422 4 2 2 (5,3) (0,4) (5,1) y

13. (b)66,620,4 v =−−= (c)4vj = (a)and(d).

14. (b) () 37,1110,0 v =−−−−−=−

15. (b)35 14 2233 ,31, v =−−=−

16. (b)0.840.12,1.250.600.72,0.65 v =−−= (c)0.720.65 vij =+ (a)and(d).

17. () 1 2 41 23 3,5 u u Q

22 8,15 81517 = =+= v v 22. () 22 24,7 24725 =− =−+= v v 23. ()() 22 5 1526 =−− =−+−= vij v 24. 22 33 331832 =+ =+== vij v 25. (a)223,56,10 v == (b)39,15 v −=−− (c)735 21 222 , v = 46 6 4 2 2 x v (6,6) (0,4) (6,2) y x v 86422468 3 2 1 2 3 (10,0) ((7,1) 3,1) y 12 3 2 21 x v 1 (2,3( 3 2 4 (,3 ( 5 (1,3 ( y x v 0.751.00 0.251.25 0.50 1.25 1.00 0.75 0.50 0.25 (0.72,0.65) (0.84,1.25) (0.12,0.60) y x (6,10) (3,5) v 2v y 2246810 2 2 4 6 8 10 x (9,15) (3,5) v 3v y 12963 1536 6 9 12 15 3 6 x (3,5) v v y 35 2 21 2 7 2 (,( 3369121518 3 3 6 9 12 15 18

(d)10 2 33 2, v =

26. (a)442,38,12 v =−=−

(b)3 1 22 1, v −=−

Section 11.1 Vectors in the Plane 1093

27. 4,9,2,5uv==−

(a)8 22 3334,9,6 u ==

(b)332,56,15 =−=− v

(d)2524,952,5 8,1810,25 18,7 uv+=+− =+− =−

28. 3,8,8,7=−−=uv

(a)16 22 333 3,82, =−−=−− u

(b)338,724,21 == v

(d)2523,858,7 6,1640,35 34,19 uv+=−−+ =−−+ = 29.

Twiceaslongasgivenvector u.

(d)612,18 u −=−

35. 3,12 v = 22 312153 v =+= v u v == 3,12312 , 153153153 17417,unitvector 1717 = =

36. 5,15 25225250510 5,1510310,unitvector 5101010 v v v u v =− =+== ===−

37. 35 , 22 v = 22 3534 222 v

35 , 2235 , 343434 2 334534,unitvector 3434 v u v

(a)011 u =+= (b)9932 v =+= (c)3,2 9413 uv uv +=− +=+= (d) (e)13,3 32 1 v v v v =− = (f)13,2 13 1 uv uv uv uv + =− + + = + 0,1 1 u u u u = = x y

38. ()() 22 6.2,3.4 6.23.45052 6.2,3.4312172,unitvector 525050 v v v u v =− =−+== ===−

39. 1,1,1,2uv=−=−

(a)112 u =+= (b)145 v =+=

40. 0,1,3,3uv==−

41. 1 1,,2,3 2 uv==

(a)15 1 42 u =+=

(b)4913 v =+=

42. 2,4,5,5uv=−= (a)41625 u =+=

(b)252552 v =+=

3,2 133.606 1,2 52.236 2,0 2 2135 u u v v uv uv u+vuv =− =≈ =− =≈ +=− += ≤+ ≤+ 45. 1 0,30,1 3 660,10,6 0,6 u u u u v ==

1 1,1 2 4221,1 22,22 u u u u v =

= 

= 47. 112 1,2, 555 12 55,5,25 55 5,25 u u u u v =−=−

=−=−

1 3,3 23 1 23,3 3 1,3 u u u u v =

50. ()() 5cos120sin120 553553 , 2222 vij ij =  °+°   =−+=−

51. ()() 2cos150sin150 33,1 vij ij =  °+°  

52. ()() 4cos3.5sin3.5 3.99250.2442 3.9925,0.2442 vij ij =  °+°   ≈+ =

53. ()() ()() cos0sin0 3232 3cos453sin45 22 2323223232 , 2222 uiji vijij uvij =°+°= =°+°=+

54. ()() ()() 4cos04sin04 2cos302sin303 535,3 uiji vijij uvij =°+°= =°+°=+ +=+=

55. ()() ()() ()() 2cos42sin4 cos2sin2 2cos4cos22sin4sin2 2cos4cos2,2sin4sin2 uij vij uvij =+ =+ +=+++ =++

56. ()() ()() ()() () 5cos0.55sin0.5

5cos0.55sin0.5

5cos0.55sin0.5 10cos0.510cos0.5,0 uij ij vij uvi

57. Theforcesactalongthesamedirection. θ =0°

58. Theforcescancelouteachother.180. θ =°

59. 0. XYYZZX++= Vectorsthatstartandendatthesamepointhavea magnitudeof0.

60. (a)True. d hasthesamemagnitudeas a butisintheoppositedirection.

(b)True. c and s havethesamelengthanddirection.

(c)True. a and u aretheadjacentsidesofaparallelogram.So,theresultantvector,, + au isthediagonalofthe parallelogram, c

(d)False.Thenegativeofavectorhastheoppositedirectionoftheoriginalvector.

(e)True. () 0 +=+−=adaa

(f)False. () ()()()() 2 2222224 uvuuu btbbbuu −=−−= −+=−+=−=−

61. 4,51,21,1 ab ==+− v 4 52 ab ab =+ =−

Addingtheequations,933. aa =  = Thenyouhave4431. ba=−=−= 3,1ab==

62. 7,21,21,1 ab =−−=+− v 7 22 ab ab −=+ −=−

Addingtheequations,933. aa −=  =−

Thenyouhave () 7734.ba=−−=−−−=− 3,4ab=−=−

63. 6,01,261,1 a =−=+− v 6 02 ab ab −=+ =−

Addingtheequations,632. aa −=  =− Thenyouhave () 6624.ba=−−=−−−=− 2,4ab=−=−

64. 0,61,21,1 v ab ==+−

0 62 ab ab =+ =−

Addingtheequations,632. aa =  = Thenyouhave2. ba=−=− 2,2ab==− X Y Z

65. 1,31,21,1 ab =−=+− v 1 32 ab ab =+ −=−

Addingtheequations,2 23. 3 aa −=  =−

Thenyouhave 25 11. 33 ba  =−=−−=   25 , 33 ab=−=

67. ()()()2,2,36fxxfxxf ′′ ===

Section 11.1 Vectors in the Plane 1097

66. 1,81,21,1 ab =−=+− v 1 82 ab ab −=+ =−

Addingtheequations,7 73. 3 aa =  =

Thenyouhave710 11. 33 ba=−−=−−=− 710 , 33 ab==−

(a)6. m = Let1,6,37, ww== then 1 1,6. 37 w w ±=±

(b)1 6 m =− Let6,1,37, ww=−= then 1 6,1. 37 w w ±=±−

68. (a)2. m =− Let1,2, w =− 5, w = then 1 1,2. 5 w w ±=±−

(b)1 2 m = Let2,1,5, ww== then 1 2,1. 5 w w ±=±

69. ()()32 ,33fxxfxx ′ === at1. x =

(a)3. m = Let1,3, w = 10, w = then 1 1,3. 10 w w =±

(b)1 3 m =− Let3,1,10, ww=−= then 1 3,1. 10 w w =±−

70. ()()32 ,312fxxfxx ′ === at2. x =−

(a)12. m = Let1,12, w = 145, w = then 1 1,12. 145 w w =±

(b)112 m =− Let12,1,145, ww=−= then 1 12,1. 145 w w =±−

71. () 252 f xx =−

() 2 3 254 x fx x ′ == at3. x =

(a)3 4 m =− Let4,3, w =− 5, w = then 1 4,3. 5 w w =±−

(b)4 3 m = Let3,4,5, ww== then 1 3,4. 5 w w =±

()()()25,2,12fxxfxxf ′′ =−+=−=− x y 2246810 2 4 6 8 10 (3,9) (a) (b) 1 3123 1 1 2 3 4 x y (1,4) (a) (b) x y 12 1 2 (a) (1,(b) 1) 42 624 4 6 10 x y (a) (b) x y 112345 1 2 3 4 (3,4) (a)(b)

Magnitude:22430.8890440.18 +≈ newtons (c) (d)

M decreasesbecausetheforceschangefromacting inthesamedirectiontoactingintheopposite directionasincreasesfromto

77. ()()() () () 123 123 123 7512575125 2222 75cos3075sin30100cos45100sin45125cos120125sin120

79. () () cos30sin30 cos130sin130 CB CA uuij vvij  ==°+° ==°+°

Verticalcomponents:

Horizontalcomponents:

sin30sin1303000uv°+°= cos30cos1300uv°+°=

Solvingthissystem,youobtain 1958.1 u ≈ pounds, 2638.2 v ≈ pounds.

Section 11.1 Vectors in the Plane 1099

80. 1 24 arctan0.8761or50.2 20 θ

=≈°

2 24 arctan1.9656or112.6 10 θπ  =+≈°

() () 11 22 cossin cossin uuij vvij θθ θθ =+ =+

Verticalcomponents: 12 sinsin5000uvθθ+=

Horizontalcomponents: 12 coscos0uvθθ+=

Solvingthissystem,youobtain 2169.4 u ≈ and3611.2. v ≈

81. Horizontalcomponentcos 1200cos61193.43ftsec v θ = =°≈

Verticalcomponentsin 1200sin6125.43ftsec v θ = =°≈

82. Tolifttheweightvertically,thesumoftheverticalcomponentsof u and v mustbe100andthesumofthehorizontal componentsmustbe0.

() () cos60sin60 cos110sin110 uuij vvij =°+° =°+°

So,sin60sin110100, uv°+°= or 3 sin110100. 2 uv  +°=

Andcos60cos1100 uv°+°= or 1 cos1100. 2 uv

+°=

Multiplyingthelastequationby ()3andaddingtothefirstequationgives () sin1103cos11010065.27pounds uv °−°=  ≈

+°=

Then, 1 65.27cos1100 2 u

gives44.65pounds u ≈

(a)Thetensionineachrope:44.65lb, 65.27lb u v = =

(b)Verticalcomponents:sin6038.67lb, sin11061.33lb u v °≈ °≈

83. () () 900cos148sin148 100cos45sin45 uij vij =°+° =°+° ()() ()() 22 900cos148100cos45900sin148100sin45

692.53547.64

547.64 arctan38.34;38.34NorthofWest 692.53 692.53547.64882.9kmh

DirectionNorthofEast:N84.46E

Speed:336.35mih ≈

85. False.Weighthasdirection.

86. True

87. True

88. True

89. True

90. True

91. True

92. False 0 ab==

93. False 2 aba ij+=

94. True

95. 22 22 cossin1, sincos1 u v θθ θθ =+= =+=

96. Letthetrianglehaveverticesat ()() 0,0,,0, a and (),.bc

Let u bethevectorjoining ()0,0and (),,bc asindicated inthefigure.Then v,thevectorjoiningthemidpoints,is () 222 22 11 22 abac bc bc vij i+j iju. +

So,1 xy+= and0. xy−= Solvingyouhave 1 2 xy== x a + bc a 22 2 (,( (( ,0 (b, c) (a (0,,0) 0) u v y u s r v

97. Let u and v bethevectorsthatdeterminethe parallelogram,asindicatedinthefigure.Thetwo diagonalsare uv + and vu So, ()() ,4. x ruvsvu =+=− But, ()()()()x yxyxy urs uvvuuv =− =+−−=++−

99. Thesetisacircleofradius5,centeredattheorigin.

100. Let0cos xvt α = and2

(x, y) x y α

If 22 0 2 0 22, vgx y g v ≤− then α canbechosentohitthepoint (),.x y Tohit ()0,: y Let90. α =° Then 222 200 0 0 1 1, 222 vvg yvtgtt ggv  =−=−−

andyouneed

ThesetHisgivenby0, x ≤ 0 y < and 22 0 2220 vgx y g v ≤−

Note: Theparabola 22 0 2220 vgx y g v =− iscalledthe“parabolaofsafety.”

Section11.2SpaceCoor dinatesandVectorsinSpace

1. 0x isdirecteddistanceto yz-plane.

0y isdirecteddistanceto xz-plane.

0z isdirecteddistanceto xy-plane.

2. The y-coordinateofanypointinthe xz-planeis0.

4. Thenonzerovectors u and v areparallelifthereexistsa scalarsuchthat c = uv sincos 22 tantan 2 coscos 22

3. (a)4 x = isapointonthenumberline. (b)4 x = isaverticallineintheplane. (c)4 x = isaplaneinspace.

(0,4,5) 8 z 6 2 2 4 6 6 6

9. 3, x =− 4, y = 5: z = () 3,4,5

10. 7, x = 2, y =− 1: z =− () 7,2,1

11. () 0,12:12,0,0yzx===

12. ()0,3,2:0,3,2xyz===

13. Thepointis1unitabovethe xy-plane.

14. Thepointis6unitsinfrontofthe xz-plane.

15. Thepointisontheplaneparalleltothe-plane yz that passesthrough3. x =−

16. Thepointis5unitsbelowthe xy-plane.

17. Thepointistotheleftofthe xz-plane.

18. Thepointmorethan4unitsawayfromthe yz-plane.

19. Thepointisonorbetweentheplanes3 y = and 3. y =−

20. Thepointisinfrontoftheplane4. x =

21. Thepoint () ,,x yz is3unitsbelowthe xy-plane,and beloweitherquadrantIorIII.

22. Thepoint () ,,x yz is4unitsabovethe xy-plane,and aboveeitherquadrantIIorIV.

23. Thepointcouldbeabovethe xy-planeandsoabove quadrantsIIorIV,orbelowthe xy-plane,andsobelow quadrantsIorIII.

24. Thepointcouldbeabovethe xy-plane,andsoabove quadrantsIandIII,orbelowthe xy-plane,andsobelow quadrantsIIorIV.

25. ()()() 222 842165 1611 1832 d =−+−+− =++ ==

26. () () ()() 222 315131 41616 366 d =−−−+−+−− =++ ==

27. ()()() 222 302284 9016 255 d =−+−+− =++ ==

28. () () ()() 222 538741 4125 30 d =−−−+−+−− =++ =

29. ()()() 0,0,4,2,6,7,6,4,8ABC () ()() 222 222 222 222 263497 641219614 421524575 24549196 AB AC BC BCABAC =++== =++−== =+−+−== ==+=+ Righttriangle

30. ()()() 3,4,1,0,6,2,3,5,6 94114 012526 911626 ABC AB AC BC =++= =++= =++= Because, ACBC = thetriangleisisosceles.

31. ()()() 1,0,2,1,5,2,3,1,1ABC 0251641 AB =++= 41914 AC =++= 436141 BC =++=

Because, ABBC = thetriangleisisosceles.

32. ()()() 4,1,1,2,0,4,3,5,1 41914 136037 125935 ABC AB AC BC =++= =++= =++=

Neither

33. () 4808620 ,,6,4,7 222 ++−+  =  

39. Centerismidpointofdiameter: 2113313 ,,,2,1 2222 ++−  =

Radiusisdistancefromcentertoendpoint: ()() 2 22 3121 1231114 242 d  =−+−++=++=

−+−+−=

()() 2 22 321 21 24xyz

40. Centerismidpointofdiameter: () 244053 ,,3,2,1 222 −−+−+  =−−  

34. 7522231 ,,1,0, 2222  =−  

35. () 314860 ,,2,6,3 222 +++  =  

36. () 5293733 ,,,3,5 2222  +−

37. Center: () 7,1,2

Radius:1 ()()() 222 7121xyz −+−++=

38. Center: () 1,5,8

Radius:5 ()()() 222 15825xyz ++++−=

Radiusisdistancefromcentertoendpoint: () () ()() () 22 2 430231141621 d =−−−+−+−−=++= ()()() 222 32121xyz ++−++=

41. Center: () 7,7,6

Tangentto xy-plane

Radiusis z-coordinate,6.

()()() 222 77636xyz ++−+−=

42. Center: () 4,0,0

Tangentto yz-plane

Radiusisdistanceto yz-plane,4. () 222 416xyz+++=

43. ()()() ()()() 222 222 222 26810 216981611916 13425 xyzxyz xxyyzz xyz ++−+++= −+++++++=−+++ −++++=

Center: () 1,3,4

Radius:5

44.

45.

Center: 9 ,1,5 2

Radius:109 2

46.

Center: () 1 3 ,1,0

Radius:1

Center: () 1 2 3,,1

Radius:4

47. (a)24,42,312,2,2 v =−−−=− (b)222 vijk =−++ (c)

48. (a)40,05,314,5,2 v =−−−=− (b)452 vijh =−+ (c)

49. (b) () 31,32,434,1,1 v =−−−−= (c)4vijk =++ (a),(d)

(3,3,4) (0,0,0) (4,1,1) v xy 12 (4,3,7) (6,4,9) (2,1,2) z 9

50. (b) ()() 42,31,726,4,9 v =−−−−−−=− (c)649 vijk =++ (a),(d) x

51. 43,12,601,1,6 113638 v v =−−−=− =++=

Unitvector:1,1,6116 ,, 38383838 =

52. () 21,42,241,6,6 v =−−−−−=− 1363673 v =++=

Unitvector:1,6,6166 ,, 73737373 =

53. 04,52,204,3,2=−−−=− v () 222 432169429=−++=++= v

Unitvector: 1432 4,3,2,, 29292929 −=−

54. () () 222 11,22,300,0,3 0033 =−−−−−−=− =++−= v v

Unitvector: 1 0,0,30,0,1 3 −=−

55. ()()() 123,,0,6,23,5,6qqq −=− () 3,1,8 Q =

56. () () () () 123 521 232 4 3 ,,0,2,1,, 1,,3 qqq Q −=− =−

57. (a)22,4,4 v =

58. (a)2,2,1 v −=−−

(b)24,4,2 v =−

(b)1,2,2 v −=−−−

59. 1,2,32,2,14,0,4 3,0,0 z =−+ =−−+− = uvw

60. 1 532 5,10,156,6,32,0,2 3,4,20 zuvw =−− =−−−− =−

61. 1 3 3 1 3 3 93 91,2,334,0,4 9,18,2712,0,12 21,18,15 −= =+ =+ =+− =+− = zuw zuw zuw

62. 123 123 11 22 33 2321,2,32,2,14,0,43,,0,0,0 0,6,93,3,30,0,0 0300 6302 9303 0,2,3 zzz zzz zz zz zz uvwz z +−+=+−−−+=

63. (a)and(b)areparallelbecause 6,4,1023,2,5 −−=−− and 10 42 333 2,,3,2,5. −=−

64. (b)and(d)areparallelbecause ()33412 32234 2 ijkijk −+−=−−+ and ()3933 12 482234 ijkijk −+=−+

65. 342 zijk =−++

(a)isparallelbecause6842. ijkz−++=

66. 7,8,3 z =−−

(b)isparallelbecause () 14,16,6. z z −=−

67. ()()() 0,2,5,3,4,4,2,2,1 PQR 3 2 3,6,9 2,4,6 3,6,92,4,6 PQ PR = = =

So, PQ and PR areparallel,thepointsarecollinear.

68. ()()() 1 2 4,2,7,2,0,3,7,3,9 6,2,4 3,1,2 3,1,26,2,4 PQR PQ PR =−− =− −=−−−

So, PQ and PR areparallel.Thepointsarecollinear.

69. () 1,2,4, P ()() 2,5,0,0,1,5QR 1,3,4 1,1,1 PQ PR =− =−−

Because PQ and PR arenotparallel,thepointsarenot collinear.

70. ()()() 0,0,0,1,3,2,2,6,4 1,3,2 2,6,4 PQR PQ PR =− =−

Because PQ and PR arenotparallel,thepointsarenot collinear.

71. ()()()() () 2,9,1,3,11,4,0,10,2,1,12,5 1,2,3 1,2,3 2,1,1 2,1,1 ABCD AB CD AC BD = = =− =−

Because ABCD = and, ACBD = thegivenpoints formtheverticesofaparallelogram.

72. ()()()() 1,1,39,1,2,11,2,9,3,4,4 8,2,5 8,2,5 2,3,7 2,3,7 ABCD AB DC AD BC =−− =−− =− =−

Because ABDC = and, ADBC = thegivenpoints formtheverticesofaparallelogram.

73. () 222 1,0,1 101 112 =− =−++ =+= v

74. ()()() 222 5,3,4 534 25916 50 52 =−−− =−+−+− =++ = = v

75. 350,3,5 092534 vjk v =−=− =++=

76. 252,5,1 425130 vijk v =+−=− =++=

Section 11.2 Space Coordinates and Vectors in Space 1107

77. 231,2,3 14914 vijk v =−−=−− =++=

78. 4374,3,7vijk=−++=− 1694974 v =++=

79. 2,1,2 4143 v v =− =++=

(a) 1 2,1,2 3 v v =− (b) 1 2,1,2 3 v v −=−−

80. 6,0,8 3606410 v v = =++= (a) 1 6,0,8 10 v v = (b) 1 6,0,8 10 v v −=−

81. 453 162595052 =−+ =++== vijk v

(a) () 122232 453 525210 =−+=−+ vijkijk v

(b) () 122232 453 525210 −=−−+=−+− vijkijk v

82. 53 259135 =+− =++= vijk v

(a) () 13533535 53 3573535 =+−=+− vijkijk v

(b) () 13533535 53 3573535 −=−+−=−−+ vijkijk v

83. 10100,3,3 32 111010 100,,0,, 2222 u v u == ==

84. 331,1,1 3 111333 3,,,, 333333 u v u == ==

85. () 2,2,1 3332211 ,,1,1, 22323332 u v u ====−

86. 4,6,214217 77,, 214141414 u v u ===

87. ()() 2cos30sin30 30,3,1 vjk jk =  ±°+±°   =±=±

88. () () () () 52 5cos45sin45or 2 5cos135sin13552 2 vikik vikik =°+°=+ =°+°=−+

89. ()()() 2 3 3,6,3 2,4,2 4,3,02,4,22,1,2 v v =−− =−− +−−=−

90. () () () 210 33 1013 33 5,6,3 ,4,2 1,2,5,4,2,6,3 v v =− =− +−=

91. Asphereofradius4centeredat () 111,,. x yz ()()() 211 222 111 ,, 4 v xxyyzz xxyyzz =−−− =−+−+−= ()()() 222 11116xxyyzz −+−+−=

92. ()()() ()()() 222 0 222 1112 1114 rr xyz xyz −=−+−+−= −+−+−= Thisisasphereofradius2andcenter () 1,1,1.

93. Thesetofallpoints () ,,x yz suchthat1 > r represent outsidethesphereofradius1centeredattheorigin.

94. (a) ()() ,,3,3,3xyz = 3,3,33,0,0 33,30,300,3,3 =− =−−−= v

(b) ()() ,,4,4,8xyz = 4,4,84,0,0 44,40,800,4,8 =− =−−−= v

95. Theterminalpointsofthevectors,tt uuv + and s t uv + arecollinear.

96. (a)

(b) () 0 abaabb wuvijk =+=+++= 0,0,0aabb=+==

So, a and b arebothzero.

(d) () 23 1,2,3 aabb aabb ijkijk +++=++ =+== Notpossible

97. Let α betheanglebetween v andthecoordinateaxes. ()()() () coscoscos 3cos1 13 cos 33 33 1,1,1 33 vijk v vijk ααα α α =++ == == =++=

302,50018,125

98. () () 2 2 5507550100

99. (a)Theheightoftherighttriangleis2218. hL=−

Thevector PQ isgivenby0,18,. PQh =−

Thetensionvector T ineachwireis0,18,ch T =− where248. 3 ch == So,80,18, h h T =− and ()22222 2222 888 181818,18. 1818 T L ThLL h

(0,0,0) (0,18,0) 18 (0,0, h) 0100 0 30 L =18 T =8 (c) isaverticalasymptoteandisa horizontalasymptote.

20253035404550

18.411.5109.39.08.78.6

(e)Fromthetable,10 T = implies30 L = inches.

100. AsinExercise99(c), x a = willbeaverticalasymptote.So, 0 lim. ra T → =∞

101. 11 22 3 123 0,70,115,0,70,115 60,0,115,60,0,115 45,65,115,45,65,115 0,0,500 AB C AC C AD C 3 F F F FFFF

So: () 23 13 123 60450 70650 115500 CC CC CCC

Solvingthissystemyields1210428 6923 ,,CC== and3112 69 C = So:

102. Let A lieonthe y-axisandthewallonthe x-axis.Then ()() 0,10,0,8,0,6,AB== ()

C =− and 8,10,6,10,10,6.ABAC=−=−−

102,259ABAC==

Thus,12420,650 ABAC ABAC FF==

12237.6,297.0,178.2423.1,423.1,253.9185.5,720.1,432.1FFF=+≈−+−−≈−− 860.0lb F ≈

103. ()() 2 dAPdBP =

Sphere;center: 41 ,3,, 33 

Section11.3TheDotProductofTwoVectors

1. Thevectorsareorthogonal(perpendicular)ifthedot productofthevectorsiszero.

2. If 2 arccos30, v =° v then 2 cos30. v °= v

So,theanglebetween v and j is30°.

3. 3,4,1,5uv==−

(a) ()() 314517 uv⋅=−+=

(b) ()() 334425 uu⋅=+=

(d) () 171,517,85 uvv⋅=−=−

(e) ()()() 3331751uvuv⋅=⋅==

4. 4,10,2,3uv==−

(a) ()() 4210322 uv⋅=−+=

(b) ()() 441010116 uu⋅=+=

(d) () 222,344,66 uvv⋅=−=−

(e) ()()() 3332266uvuv⋅=⋅==

5. 6,4,3,2u=v−=−

(a) ()()() 634226 uv⋅=−+−=−

(b) ()()() 664452 uu⋅=+−−=

(d) () 263,278,52 uvv⋅=−−=−

(e) ()()() 3332678uvuv⋅=⋅=−=−

6. 7,1,4,1 =−−=−−uv

(a) ()() 741129⋅=−−+−−=uv

(b) ()() 771150⋅=−−+−−=uu

(d) () 294,1116,29⋅=−−=−−uvv

(e) ()()() 3332987 ⋅=⋅==uuuv

7. 2,3,4,0,6,5uv=−=

(a) ()()()()() 2036452 uv⋅=+−+=

(b) ()()()() 22334429 uu⋅=+−−+=

(d) () 20,6,50,12,10 uvv⋅==

(e) ()()() 33326uvuv⋅=⋅==

8. 5,0,5,1,2,1=−=−uv

(a) ()()() 51025110⋅=−−++=uv

(b) ()()()()() 55005550⋅=−−++=uu

(d) () 101,2,110,20,10⋅=−=−uvv

(e) ()()() 3331030 ⋅=⋅==uvuv

9. 2, uijkvik =−+=−

(a) ()()()() 2110111 uv⋅=+−+−=

(b) ()()()()() 2211116 uu⋅=+−−+=

(d) ()uvvvik ⋅==−

(e) ()()() 33313uvuv⋅=⋅==

10. 22,32 uijkvijk =+−=−+

(a) ()()()() 2113225 uv⋅=+−+−=−

(b) ()()()() 2211229 uu⋅=++−−=

(d) ()()53251510 uvvijkijk ⋅=−−+=−+−

(e) ()()() 333515uvuv⋅=⋅=−=−

11. 1,1,2,2uv==− 0 cos0 28 θ === uv uv

(a) 2 π θ = (b)90 θ =°

12. 3,1,2,1uv==− 51 cos 1052 θ === uv uv

(a) 4 π θ = (b)45 θ =°

13. 3,24 uijvij =+=−+ 21 cos 102052 θ === uv uv

(a) 1 arccos1.713 52 θ  =−≈  

(b)98.1 θ ≈°

ππ ππ θ

=+=+    =+=−+

uijij vijij uv uv

=−+=−

14. () cossin31 6622 3322 cossin 4422 cos 32122 13 22224

(a) () 27 arccos13 42 π θ

=−=

(b)105 θ =°

15. 1,1,1,2,1,1uv==− 22 cos 363 θ === uv uv

(a)2arccos1.080 3 θ =≈

(b)61.9 θ ≈°

16. 32,23 uijkvij =++=− ()() 32230 cos0 θ +−+ === uv uvuv

(a)=2π θ

(b)90 θ =°

17. 34,=23=+−+ uijvjk 8813 cos 51365 θ ⋅−− === uv uv

(a)813arccos2.031 65 θ  =−≈

(b)116.3 θ ≈°

18. 23,2 uijkvijk =−+=−+ 99321 cos 14622114 θ ==== uv uv

(a) 321 arccos0.190 14 θ  =≈

(b)10.9 θ ≈°

19. cos uv uv θ ⋅ = ()()85cos20 3 uv π ⋅==

20. ()() cos 5 4025cos5003 6 uv uv uv θ π = ⋅==−

21. 12 234,3,, notparallel 0orthogonal uv uv uv c ==− ≠  ⋅= 

22. () 1 3 1 6 2,24 parallel uijvij uv =−−=− =− 

23. 6,2 notparallel 80notorthogonal Neither ujkvijk uv uv c =+=−− ≠  ⋅=−≠ 

24. 23,2 uijkvijk =−+−=+− notparallel c uv ≠  0orthogonal uv⋅= 

25. 2,3,1,1,1,1 notparallel 0orthogonal uv uv uv c =−=−−− ≠  ⋅= 

26. cos,sin,1, sin,cos,0 notparallel 0orthogonal u v uv uv c θθ θθ =− =−

≠ 

⋅= 

27. Thevector1,2,0joining () 1,2,0and ()0,0,0is perpendiculartothevector2,1,0joining () 2,1,0 and ()0,0,0:1,2,02,1,00 ⋅−=

Thetrianglehasarightangle,soitisarighttriangle.

28. Considerthevector3,0,0joining ()0,0,0and ()3,0,0,andthevector1,2,3joining ()0,0,0and () 1,2,3:3,0,01,2,330 −⋅=−<

Thetrianglehasanobtuseangle,soitisanobtusetriangle.

29. ()() () 13 22 2,0,1,0,1,2,,,0ABC 53 22 11 22 2,1,1 ,,1 ,,2 AB AC BC =− =−− =−− 53 22 11 22 2,1,1 ,,1 ,,2 BA CA CB =−− =− =− 3 2 1 2 53 44 510 120 20

ABAC BABC CACB ⋅=+−> ⋅=−−+> ⋅=++>

Thetrianglehasthreeacuteangles,soitisanacute triangle.

30. ()()() 2,7,3,1,5,8,4,6,1 ABC 3,12,5 2,13,4 5,1,9 AB AC BC =− =− =− 3,12,5 2,13,4 5,1,9 BA CA CB =−− =−− =−− 6156200 1512450 1013360

ABAC BABC CACB

⋅=−+−> ⋅=−+> ⋅=++>

Thetrianglehasthreeacuteangles,soitisanacute triangle.

31. 22,1443 =++=++=uijku 222 1 3 2 3 2 3 144 999 cos1.2310or70.5 cos0.8411or48.2 cos0.8411or48.2 coscoscos1

αβγ =  ≈° =  ≈° =  ≈° ++=++=

γγ αβγ =+−=++= =  ≈° =  ≈° =  ≈° ++=++=

32. 222 53,259135 5 cos0.5639or32.3 35 3 cos1.0390or59.5 35 1 cos1.7406or99.7 35 2591 coscoscos1 353535 uijku αα

33. 7,491151 =+−=++=uijku

7 cos11.4 51 αα =  ≈°

1 cos82.0 51 ββ =  ≈° 1 cos98.0 51 γγ =−  ≈°

34. 435,169255052uijku=−++=++== 222 4 cos2.1721or124.4 52 3 cos1.1326or64.9 52 51 cosor45 5224 16925 coscoscos1 505050

35. 222 0,6,4,0361652213 cos0or90 2 3 cos0.5880or33.7 13 2 cos2.1588or123.7 13 94 coscoscos01 1313

36. 1,5,2,125430uu=−=++=

1 cos1.7544or100.5 30

5 cos0.4205or24.1 30 2 cos1.1970or68.6 30 1254 coscoscos1 303030

37. ) 6,7,1,4uv==

(a)1 w = ()() 2 22 proj 6174 1,4 14 34 1,42,8 17 v uv uv v  ⋅ =    + = + ==

(b) ) 216,72,84,1wuw=−=−=−

38. 9,7,1,3uv==

(a)1 w = ()() 2 2 proj 9173 1,3 13 30 1,33,9 10 v uv uv v  ⋅ =    + = + ==

(b)219,73,96,2 wuw=−===−

39. 232,3,55,1uijvij =+==+=

(a)1 w = ()() 2 2 proj 2531 5,1 51 1351 5,1, 2622 v uv uv v  =    + = + == (b)215115 2,3,, 2222 wuw=−=−=−

40. 232,3,323,2uijvij =−=−=+=

(a)1 w = ()()() 2 22 proj 2332 3,2 32 03,20,0 v uv uv v  =    +− = + ==

(b)212,3 wuw=−=−

41. 0,3,3,1,1,1uv==−

(a)1 w = ()()() proj2 013131 1,1,1

(b)210,3,32,2,22,1,1 wuw

42. () 8,2,0,2,1,1uv==−

(a)1 w

(b)218,2,06,3,32,1,3 wuw=−=−−=−

44. 5,58 =−−=−++ uijkvijk

45.u isavectorand vw isascalar.Youcannotadda vectorandascalar.

46. 2 c uv vuuvu v  ⋅ =  =   

47. Yes,22 22 11 uvvu

and v areparallel.

43. 924,44 =−−−=+ uijkvjk

(b)

48. (a)Orthogonal,2π θ = (b)Acute,02π θ << (c)Obtuse,2π <<θπ

49. 3240,1450,2235 2.25,2.95,2.65 = = u v ()()() 32402.2514502.9522352.65 $17,490.25 ⋅=++ = uv

Thisrepresentsthetotalrevenuetherestaurantearnedon itsthreeproducts.

50. 3240,1450,2235 2.25,2.95,2.65 = = u v Decreasepricesby2%:0.98v Newtotalrevenue: () 0.983240,1450,22352.25,2.95,2.650.9817490.25 $17,140.45 ⋅= =

51. Answerswillvary. Sample answer: 13 42 uij =−+ Want0. uv⋅= 122 vij =+ and122 vij−=−− areorthogonalto u.

52. Answerswillvary. Sample answer: 94. uij =− Want0. uv⋅= 49 vij =+ and49 vij−=−− areorthogonalto u

53. Answerswillvary. Sample answer: 3,1,2. u =− Want0. uv⋅= 0,2,1 v = and0,2,1 v −=−− areorthogonalto u

54. Answerswillvary. Sample answer: 4,3,6. u =− Want0 uv⋅=

0,6,3 v = and0,6,3 v −=−− areorthogonalto u

55. Let s = lengthofaside. ,, 3 s ss s v v = =

3 sss s ss s v v v v θ θ = = = = == =≈°

57. (a)GravitationalForce48,000 Fj =− () ()() () 12 1 cos10sin10 48,000sin10 8335.1cos10sin10 8335.1lb vij Fv wvFvv v v ij w =°+° ==⋅

≈ (b) () 21 48,0008335.1cos10sin10 8208.546,552.6 wFw jij ij =− =−+°+° =− 247,270.8lb w ≈

58. (a)GravitationalForce5400 =− Fj () ()() () 12 1 cos18sin18 5400sin18 1668.7cos18sin18 1668.7lb =°+° ==⋅ =−° ≈−°+° = vij Fv wvFvv v v ij w (b) () 21 54001668.7cos18sin18 1587.04884.3 =− =−+°+° ≈− wFw jij ij 25135.7lb ≈ w y x v s s s z y x v1 v2 (s, s,0) (s,s,s) z

59. 13 85 22 10 425ft-lb W Fij vi Fv

= =⋅=

60. ()()() proj cos20 cos206550 3054.0ft-lb PQ WPQ PQ = =° =° ≈ F F

61. ()1600cos25sin25 2000 Fij vi =°+° = () 16002000cos25

2900.2km-N Fv W =⋅=° ≈ ≈

2,900,184.9Newtonmeters(Joules)

62. () ()() proj cos60 1 40040 2 8000Joules PQ WPQ PQ = =° = = F F

63. False.

Forexample,let1,1,2,3 uv== and1,4. w = Then235 uv⋅=+= and145. uw⋅=+=

64. True

() 000 wuvwuwv ⋅+=⋅+⋅=+= so, w and uv + areorthogonal.

65. (a)Thegraphs2 1 yx = and13 2 yx = intersectat ()0,0and ()1,1.

(b)12 y x ′ = and223 1 3 y x ′ =

At ()0,0,1,0 ± istangentto1 y and0,1 ± is tangentto2 y

At () 1 1,1,2 y ′ = and21 . 3 y ′ = 1 1,2 5 ± istangentto11,3,1 10 y ± istangent

to2 y

(c)At ()0,0,thevectorsareperpendicular ()90. °

At ()1,1, ()() 11 1,23,151510 cos 11502 45 θ θ === =°

66. (a)Thegraphs3 1 yx = and13 2 yx = intersectat ()1,1, ()0,0and ()1,1.

(b)2 13 yx ′ = and223 1 3 y x ′ =

At ()0,0,1,0 ± istangentto1 y and0,1 ± is tangentto2 y

At ()1,1,13 y ′ = and21 3 y ′ = 1 1,3 10 ± istangentto11,3,1 10 y ± istangent

to2 y

At () 1 1,1,3 y ′ −−= and21 3 y ′ = 1 1,3 10 ± istangentto11,3,1 10 y ± istangent to2 y

(c)At ()0,0,thevectorsareperpendicular ()90. °

At ()1,1, ()() 11 1,33,1631010 cos. 11105

0.9273or53.13 θ θ === ≈°

Bysymmetry,theangleisthesameat () 1,1. y x 112 1 1 2 y = x 2 (1,1) (0,0) y = x 1/3 y x 2112 1 2 1 2 (1,1) (0,0) (1,1) y = x 3 y = x 1/3

67. (a)Thegraphsof2 11 yx =− and221 yx=− intersectat ()1,0and ()1,0.

(b)12 y x ′ =− and22. y x ′ =

At () 1 1,0,2 y ′ =− and22. y ′ = 1 1,2 5 ±− istangentto11,1,2 5 y ± istangentto2 y

At () 1 1,0,2 y ′ −= and22. y ′ =− 1 1,2 5 ± istangentto11,1,2 5 y ±− istangentto2 . y

(c)At ()1,0,113 cos1,21,2. 555 θ =−⋅−=

0.9273or53.13 θ ≈°

Bysymmetry,theangleisthesameat ()1,0.

68. (a)Tofindtheintersectionpoints,rewritethesecond equationas3 1.yx += Substitutingintothefirst equation

() 26 10,1. yxxxx +=  =  =

Therearetwopointsofintersection, ()0,1and ()1,0,asindicatedinthefigure.

(b)Firstequation:

()() () 21 1211 21 yxyyy y ′′ +=  +=  = +

At () 1 1,0,. 2 y ′ =

Secondequation:32 13.yxyx ′ =−  = At ()1,0,3. y ′ = 1 2,1 5 ± unittangentvectorstofirstcurve, 1 1,3 10 ± unittangentvectorstosecondcurve

69. Inarhombus, uv = Thediagonalsare uv + and .uv

()()()() 220 uvuvuvuuvv uuvuuvvv uv +⋅−=+⋅−+⋅ =⋅+⋅−⋅−⋅ =−= So,thediagonalsareorthogonal.

⇔+⋅+=−⋅− ⇔+=− ⇔ y x 112 2 1 (1,0) (0,1) y = x 31 x =( y +1) 2 u v u + v u v u v u + v u − v

At ()0,1,theunittangentvectorstothefirstcurve are0,1, ± andtheunittangentvectorstothesecond curveare1,0. ±

(c)At ()1,0, 1151 cos2,11,3. 510502 θ =⋅== 4 π θ ≈ or45° At ()0,1thevectorsareperpendicular,90. θ =°

70. If u and v arethesidesoftheparallelogram,thenthe diagonalsare uv + and, uv asindicatedinthe figure. theparallelogramisarectangle. ()()()() 22 0 22 Thediagonalsareequalinlength. uv uvuv uvuvuvuv uvuv

⇔⋅= ⇔⋅=−⋅

71. (a)

(b)Lengthofeachedge:22202kkk ++=

(d)1

,,0,,,,

72. cos,sin,0,cos,sin,0uv==ααββ

Theanglebetween u and v is α β ( Assumingthat αβ) > Also, () ()() cos coscossinsin 11 coscossinsin. uv uv αβ α

α β −= + = =+

1. × uv isavectorthatisperpendicular(orthogonal)to both u and v

2. If u and v aretheadjacentsidesofaparallelogram,then A =×uv

73. ()() ()() () 2 22 222 2 2 uvuvuv uvuuvv =uuvuuvvv uuvv uuvvuv +=+⋅+ =+⋅++⋅ ⋅+⋅+⋅+⋅ =+⋅+ ≤++≤+

So, uvuv +≤+

74. Let1proj, v wu = asindicatedinthefigure.Because1 w isascalarmultipleof, v youcanwrite

122 c uwwvw =+=+

Takingthedotproductofbothsideswith v produces ()22cc uvvwvvvwv ⋅=+⋅=⋅+⋅

2 , c v = because2 w and v areorthogonol.

So,22 cc uv uvv v ⋅=  = and 12 proj. c v uv wuvv v ===

75. cos cos cos becausecos1. uvuv uvuv uv uv θ θ θ θ ⋅= ⋅= = ≤≤

Section11.4TheCrossProd uctofTwoVectorsinSpace

4. 010 001 ijk jki ×== x y (k,0, k) (kk,0) (0, kk) k k k z θ w1 w2 u v

3. 010 100 ijk jik ×==−

5. 100 001 ijk ikj ×==−

6. 001 100 ijk kij ×==

7. (a)240201016 325 ijk uvijk ×=−=+−

11. 4,1,0,6,3,0=−=−uv

41060,0,6 630 ×=−== ijk uvk

()()()()()4010060 ⋅×=+−+=  ⊥× uuvuuv

()()()()6030060⋅×=−++=  ⊥× vuvvuv

12. 5,2,2,0,1,8=−=uv

5221440514,40,5 018 ×=−=+−=− ijk uvijk

()()()()()514240250 ⋅×=−++−=  ⊥× uuvuuv

()()()()()014140850 vuvvuv ⋅×=++−=  ⊥× xy 1 i k j z 1 1 1

(b) () 201016 vuuvijk ×=−×=−−+ (c) vv0 ×=

8. (a)30515169 232 ijk uvijk ×==−++

(b) () 15169 vuuvijk ×=−×=−− (c) vv0 ×=

9. (a)732173310 115 ijk uvijk ×==−−

(b) () 173310 vuuvijk ×=−×=−++ (c) vv0 ×=

10. (a)219 621 19562 ×=− =−++ ijk uv ijk

(b) () 19562×=−×=−− vuuvijk (c)0 vv×=

13. ,2 uijkvijk =++=+− 111232,3,1 211 ijk uvijk ×==−+−=−−

()()()() 121311 0 uuv uuv ⋅×=−++− =  ⊥×

()()()()() 221311 0 vuv vuv ⋅×=−++−− =  ⊥×

14. 6,2 uijvijk =+=−++ 160613 211 ijk uvijk ×==−+

()()()() 16610 uuvuuv ⋅×=+−=  ⊥×

()()()()() 26111130 vuvvuv ⋅×=−+−+=  ⊥×

15. 4,3,1 2,5,3 431141026 253 1 14,10,26 972 1 14,10,26 183 7513 ,, 939393 =− = ×=−=−−+ × =−− × =−− =−− u v ijk uvijk uv uv

16. 8,6,4 10,12,2 8646024156 10122 1 60,24,156 3622 5213 ,, 322322322 =−− =−− ×=−−=++ × = × = u v ijk uvijk uv uv

17. 325 4 32537 114 1 3,7,1 59 371 ,, 595959 =−+− =−+ ×=−−=++ × = × = uijk vijk ijk uvijk uv uv

18. () 2 46 0028 406 1 80,1,0 8 = =+ ×== × === × uk vik ijk uvj uvjj uv

19. 010 011 uj vjk ijk uvi = =+ ×== 1 uvi A =×== 20. 111 011 uijk vjk ijk uvjk =++ =+ ×==−+ 2 uvjk A =×=−+=

21. 3,2,1 1,2,3 3218,10,4 123 u v ijk uv =− = ×=−=− 8,10,418065 A uv =×=−==

22. 2,1,0 1,2,0 u v =− =−

2100,0,3 120 0,0,33 A ijk uv uv ×=−= =×==

23. ()()()() 0,3,2,1,5,5,6,9,5,5,7,2 ABCD

1,2,3 1,2,3 5,4,0 5,4,0 AB DC BC AD = = = =

Because ABDC = and, BCAD = thefigure ABCD is aparallelogram.

AB and AD areadjacentsides

12312,15,6 540 ABAD ijk ×==−− 1442253695 AABAD=×=++=

24. ()()()() 2,3,1,6,5,1,7,2,2,3,6,4 ABCD

4,8,2 4,8,2 1,3,3 1,3,3 AB DC BC AD =− =− =− =−

Because ABDC = and, BCAD = thefigure ABCD is aparallelogram.

AB and AD areadjacentsides 48218,14,20 133 3241964002230 ABAD AABAD ijk ×=−=−− =×=++=

25. ()()() 0,0,0,1,0,3,3,2,0ABC 1111 222 1,0,3,3,2,0 1036,9,2 320 36814

ABAC

ABAC AABAC ijk ==− ×==−− =×=++=

26. ()()() 2,3,4,0,1,2,1,2,0ABC 11 22 2,4,2,3,5,4 242622 354 4411

ABAC

ABAC AABAC ijk ijk =−−=−− ×=−−=−−+ =×==

27. 20 Fk =− () 1 2cos40sin40 0cos402sin40210cos40 0020 10cos407.66ft-lb PQ PQ PQ jk ijk Fi F =°+° ×=°°=−° ×=°≈

28. () 2000cos30sin30100031000 0.16 000.16 0100031000 1603 1603ft-lb PQ PQ PQ Fjkjk k ijk F i F =−°+°=−− = ×= = ×=

29. (a) () 5 15inchesfeet 4 12inches1foot 5 4 180cossin AC BC AB jk Fjk θθ == ==

(c)When1330,225180268.38 22

(d)If225sin180cos, T =+θθ 0 T = for4 225sin180costan141.34. 5 θθθθ =−

For0141.34, θ

(e)

Frompart(d),thezerois141.34, θ ≈° whenthevectorsareparallel.

30. (a)Placethewrenchinthe xy-plane,asindicatedinthefigure.

Theanglefrom AB to F is30180210°+°+=°+θθ ()() ()() ()() 18inches1.5feet

423sin210coscos210sin42cos210cossin210sin 1331 423cossin42cossin84sin 2222

and F areperpendicular.

(b)When

(c)Let84sin 84cos0when90. T dT d θ

θ = ===° Thisisreasonable.When90, θ =° theforceisperpendiculartothewrench.

31. () 100 0101 001 uvw×==

32. () 111 2101 001 uvw⋅×==−

33. () 201 0306 001 uvw⋅×==

34. () 200 1110 022 uvw⋅×==

35. () () 110 0112 101 2 V uvw uvw ⋅×== =⋅×=

36. () () 131 06672 404 72 uvw uvw V ⋅×==− =⋅×=

37. () () 3,0,0 0,5,1 2,0,5 300 05175 205 75 V u v w uvw uvw = = = ⋅×== =⋅×=

38. () () () 0,4,0 3,0,0 1,1,5 040 30041560 115 60 u v w uvw uvw V = =− =− ⋅×=−=−−= =⋅×=

39. (a) ()()() ()()() ()()() ()()() b c d h x uvwvwu wuvuvw vwuuwv vwuwuv ⋅×=×⋅ =⋅×=×⋅ =⋅×=−⋅ =⋅×=×⋅ (e) ()()() ()()() f g uwvwvu wvuuvw ⋅×=⋅× =⋅×=−×⋅

So, abcdh ==== and efg ==

40.uv0u ×=  and v areparallel. uv0u ⋅=  and v areorthogonal.

So, u or v (orboth)isthezerovector.

41. Thecrossproductisorthogonaltothetwovectors,soit isorthogonaltothe yz-plane.Itliesonthe x-axis,sinceit isoftheform,0,0. k

42. Formthevectorsfortwosidesofthetriangle,and computetheircrossproduct.

212121313131 ,,,, x xyyzzxxyyzz −−−×−−−

43. False.Ifthevectorsareorderedpairs,thenthecross productdoesnotexist.

44. False.Thecrossproductiszeroifthegivenvectorsare parallel.

45. False.Let,01,0,0,1,0,1,0,0.uvw ===− Then,,but. uvuw0vw ×=×=≠

46. True

48. 123123 ,,,,,,isascalar: uuuvvvcuv==

49.

50.

51.

233213113212213 233213113212213 uvijk uvu0 uvv0 uvuvuvuvuvuv uvuvuuvuvuuvuvu uvuvvuvuvvuvuvv ×=−−−+− ×⋅=−+−+−= ×⋅=−+−+−= So, uvu ×⊥ and uvv ×⊥

52. If u and v arescalarmultiplesofeachother, c uv = forsomescalar c ()()() uvvvvv00 ccc ×=×=×==

If, uv0 ×= thensin0. uv θ = () Assume,. u0v0 ≠≠ So,sin0,0, θθ== and u and v areparallel.So, c uv = forsomescalar c

53. sin uvuv θ ×=

If u and v areorthogonal,2 θπ = andsin1. θ = So, uvuv ×=

54. 111222333 , ,,,,,,, c ababcabcuvw ===

Section11.5LinesandPlanesinSpace

1. Theparametricequationsofaline L parallelto ,,,abc v = andpassingthroughthepoint () 111 ,, Pxyz are 1, x xat=+ 1, y ybt=+ 1 zzct =+ Thesymmetricequationsare 111 x xyyzz abc ==

2. Intheequationoftheplane ()()() 214350, xyz −+−−+= 2,4,ab== and1. c =− Therefore,thenormalvectoris2,4,1.

3. Answerswillvary.Anyplanethathasamissing xvariableinitsequationisparalleltothe x-axis.

Sample answer:35 yz−=

4. Firstchooseapoint Q inoneplane.ThenuseTheorem 11.13:

PQn D = n

where P isapointintheotherplaneand n isnormalto thatplane.

5. 2,3,4 x tytzt =−+==+

(a) ()0,6,6:For02, x t ==−+ youhave2. t = Then () 326 y == and426. z =+= Yes, ()0,6,6liesontheline.

(b) ()2,3,5:For22, x t ==−+ youhave4. t =

Then () 34123. y ==≠ No, ()2,3,5doesnotlie ontheline.

Then () 326 y =−=− and422. z =−= Yes, ()4,6,2liesontheline.

6. 37 2 28 xy z ==+

(a) () 7,23,0:Substituting,youhave 73237 02 28 222 ==+ == Yes, () 7,23,0liesontheline.

(b) () 1,1,3:Substituting,youhave 1317 32 28 111 ==−+ −=−=− Yes, () 1,1,3liesontheline.

7. Point: () 0,0,0

Directionvector:3,1,5

Directionnumbers:3,1,5

(a)Parametric:3,,5 x tytzt===

(b)Symmetric:35 x z y ==

8. Point: () 0,0,0

Directionvector:52,,1 2 v =−

Directionnumbers:4,5,2

(a)Parametric:4,5,2 x tytzt=−==

(b)Symmetric:452 x yz ==

9. Point: () 2,0,3

Directionvector:2,4,2 v =−

Directionnumbers:2,4,2

(a)Parametric:22,4,32 x tytzt=−+==−

(b)Symmetric:23 242 xyz+− ==

10. Point: () 3,0,2

Directionvector:0,6,3 v =

Directionnumbers:0,2,1

(a)Parametric:3,2,2 x ytzt =−==+

(b)Symmetric:2,3 2 y zx =−=−

11. Point: () 1,0,1

Directionvector:32vijk =−+

Directionnumbers:3,2,1

(a)Parametric:13,2,1 x tytzt =+=−=+

(b)Symmetric:11 321 xyz ==

12. Point: () 3,5,4

Directionsnumbers:3,2,1

(a)Parametric:33,52,4 x tytzt =−+=−=+

(b)Symmetric:354 32 xy z +− ==−

13. Points: () 22 5,3,2,,,1 33 

Directionvector:17113 33 vijk =−−

Directionnumbers:17,11,9

(a)Parametric: 517,311,29 x tytzt =+=−−=−−

(b)Symmetric:532 17119 xyz−++ ==

14. Points: ()() 0,4,3,1,2,5

Directionvector:1,2,2

Directionnumbers:1,2,2

(a)Parametric:,42,32 x tytzt ==+=−

(b)Symmetric:43 22 yz x ==

15. Points: ()() 7,2,6,3,0,6

Directionvector:10,2,0

Directionnumbers:10,2,0

(a)Parametric:710,22,6 xtytz =−=−+=

(b)Symmetric:Notpossiblebecausethedirection numberfor z is0.But,youcoulddescribethe lineas72,6. 102 xy z −+ ==

16. Points: ()() 0,0,25,10,10,0

Directionvector:10,10,25

Directionnumbers:2,2,5

(a)Parametric:2,2,255 x tytzt===−

(b)Symmetric:25 225 xyz ==

17. Point: () 2,3,4

Directionvector: vk =

Directionnumbers:0,0,1

Parametric:2,3,4 x yzt ===+

18. Point: () 4,5,2

Directionvector: vj =

Directionnumbers:0,1,0

Parametric:4,5,2 xytz=−=+=

19. Point: () 2,3,4

Directionvector:32vijk =+−

Directionnumbers:3,2,1

Parametric:23,32,4 x tytzt =+=+=−

20. Point () 4,5,2

Directionvector:2vijk =−++

Directionnumbers:1,2,1

Parametric:4,52,2 x tytzt =−−=+=+

21. Point: () 5,3,4

Directionvector:2,1,3 v =−

Directionnumbers:2,1,3

Parametric:52,3,43 x tytzt=+=−−=−+

22. Point: () 1,4,3

Directionvector:5vij =−

Directionnumbers:5,1,0

Parametric:15,4,3 xtytz =−+=−=−

23. Point: () 2,1,2

Directionvector:1,1,1

Directionnumbers:1,1,1

Parametric:2,1,2 x tytzt =−=+=+

24. Point: () 6,0,8

Directionvector:2,2,0

Directionnumbers:2,2,0

Parametric:62,2,8 xtytz=−−==

25. Let () 0:3,1,2tP==−− () otheranswerspossible 1,2,0 v =− () anynonzeromultipleofiscorrect v

26. Let ()0:0,5,4tP== () otheranswerspossible

4,1,3 v =− () anynonzeromultipleofiscorrect v

27. Leteachquantityequal0: () 7,6,2 P =−− () otheranswerspossible 4,2,1 v = () anynonzeromultipleofiscorrect v

28. Leteachquantityequal0: () 3,0,3 P =− () otheranswerspossible 5,8,6 v = () anynonzeromultipleofiscorrect v

29. 11:3,2,4 L =− v and () 6,2,5 P =− on1 L

22:6,4,8 L =−− v and () 6,2,5 P =− on2 L

Thelinesareidentical.

30. 11:2,1,3 L =− v and () 1,1,0 P =− on1 L

22:2,1,3 L =− v and P noton1 L

Thelinesareparallel.

31. 11:4,2,3 L =− v and () 8,5,9 P =−− on1 L

22:8,4,6 L =−− v and () 8,5,9 P =−− on2 L

Thelinesareidentical.

32. 11:4,2,4 L = v and () 1,1,3 P =− on1 L

22:1,0.5,1 L = v and P noton2 L

Thelinesareparallel.

33. Atthepointofintersection,thecoordinatesforoneline equalthecorrespondingcoordinatesfortheotherline. So,

(i)4222, ts+=+ (ii)323, s =+ and (iii)11. ts −+=+

From(ii),youfindthat0 s = andconsequently,from (iii),0. t = Letting0, st== youseethatequation(i) issatisfiedandsothetwolinesintersect.Substituting zerofor s orfor t,youobtainthepoint () 2,3,1. () () 4Firstline 22Secondline 817717 cos 17931751

34. Byequatinglikevariables,youhave (i)3131, ts −+=+ (ii)4124, ts+=+ and (iii)241. ts+=−+

From(i)youhave, s t=− andconsequentlyfrom(ii), 1 2 t = andfrom(iii),3. t =− Thelinesdonotintersect.

35. Writingtheequationsofthelinesinparametricformyou have 3 14 xt x s = =+ 2 2 yt ys =− =−+ 1 33. zt zs =−+ =−−

Forthecoordinatestobeequal,314 ts =+ and 22. ts−=−+ Solvingthissystemyields17 7 t = and 11 7. s = Whenusingthesevaluesfor s and t,the z coordinatesarenotequal.Thelinesdonotintersect.

36. Writingtheequationsofthelinesinparametricformyou have

23 32 x t x s =− =+ 26 5 yt ys =+ =−+ 3 24. zt zs =+ =−+

Byequatinglikevariables,youhave2332, ts−=+ 265, ts+=−+ 324. ts+=−+ So,1,1 ts=−= andthepointofintersectionis () 5,4,2.

41. Point: () 3,2,2

Normalvector:23nijk =+−

()()() 2332120 23100 xyz xyz −+−−−= +−−=

42. Point: () 0,0,0

Normalvector:32 nik =−+

() () 3,6,1Firstline 2,1,4Secondline 442966 cos 4621966483

82.6 u v uv uv θ θ =− = ==== ≈°

37. 2410xyz+−−=

(a) ()()()() 7,2,1:7224110 −−−+−−−=

Pointisinplane.

(b) ()()() 5,2,2:5224210 +−−= Pointisinplane.

38. 2360 xyz++−=

(a) ()()() 3,6,2:2363260 −++−−=

Pointisinplane.

(b) ()()() 1,5,1:21531660 −−−++−−=−≠

Pointisnotinplane.

39. Point: () 1,3,7

Normalvector:0,1,0 nj==

()()() () 0113070 30 xyz y −+−+−−= −=

40. Point: () 0,1,4

Normalvector:0,0,1 nk==

()()() 0001140 40 xyz z −+++−= −=

()()() 3000200 320 xyz xz −−+−+−= −+=

43. Point: () 1,4,0

Normalvector:2,1,2 v =−−

()()() 2114200 2260 xyz xyz +−−−−= −−+=

44. Point: () 3,2,2

Normalvector:43 vijk =+− ()()() 432320 4380 xyz xyz −+−−−= +−−=

45. Let u bethevectorfrom ()0,0,0to ()2,0,3:2,0,3 u =

Let u bethevectorfrom ()0,0,0to ()3,1,5:3,1,5 v −−=−−

Normalvectors:2033,19,2 315 ijk uv×==−−

()()() 30190200 31920 xyz xyz −−−−−= −−=

46. Let u bethevectorfrom () 3,1,2to () 2,1,5: 1,2,3 u =−

Let u bethevectorfrom () 3,1,2to () 1,2,2: 2,1,4 v =−−−

Normalvector:

1235,10,551,2,1 214 ijk uv×=−=−−=−−

()()() 132120 210 xyz xyz −++−−= +−+=

47. Let u bethevectorfrom () 1,2,3to ()3,2,1:22uik =−

Let v bethevectorfrom () 1,2,3to ()1,2,2:24vijk−−=−−−

Normalvector:

() () 1 2101434 241 ijk uvijk ×−=−=−+

()()() 4132430 434100 xyz xyz −−−+−= −+−=

48. () 1,2,3,Normalvector: () ,110,10vi xx =−=−=

49. () 1,2,3,Normalvector: () ,130,30 vk zz =−=−=

50. Theplanepassesthroughthethreepoints ()() () 0,0,0,0,1,0,3,0,1.

Thevectorfrom ()0,0,0to () 0,1,0: uj =

Thevectorfrom ()0,0,0to ()3,0,1:3vik =+

Normalvector:0103 301 ijk uvik ×==−

30xz−=

51. Thedirectionvectorsforthelinesare2, uijk =−++ 34. vijk =−+−

Normalvector: ()2115 341 ijk uvijk ×=−=−++

Pointofintersectionofthelines: () 1,5,1

()()()1510 50 xyz xyz ++−+−= ++−=

Section 11.5 Lines and Planes in Space 1129

52. Thedirectionofthelineis2. uijk =−+ Chooseany pointontheline, ()0,4,0,forexample,     andlet v bethe vectorfrom ()0,4,0tothegivenpoint () 2,2,1: 22 vijk =−+

Normalvector:2112 221 ijk uvik ×=−=−

()()2210 20 xz xz −−−= −=

53. Let v bethevectorfrom ()1,1,1to () 2,2,1: 32 vijk =++

Let n beavectornormaltotheplane 233:23 xyz nijk−+==−+

Because v and n bothlieintheplane P,thenormal vectorto P is 312711 231 ijk vnijk ×==−−

()()() 72121110 71150 xyz xyz −+−−−= +−−=

54. Let v bethevectorfrom ()3,2,1to () 3,1,5: 6 vjk =−−

Let n bethenormaltothegivenplane: 672 nijk =++

Because v and n bothlieintheplane P,thenormal vectorto P is: () 01640366 672 220183 ijk vnijk ijk ×=−−=−+ =−+

()()() 203182310 20183270 xyz xyz −−−+−= −+−=

55. Let ui = andlet v bethevectorfrom () 1,2,1to ()2,5,6:77 vijk =++

Because u and v bothlieintheplane P,thenormal vectorto P is: ()100777 177 ijk uvjkjk ×==−+=−−

()()210 10 yz yz

−+=

56. Let uk = andlet v bethevectorfrom ()4,2,1to ()3,5,7:736 vijk =−++ Because u and v bothlieintheplane P,thenormalvectorto P is:

()0013737 736 ijk uvijij ×==−−=−+

()() 34720 37260 xy xy −+−= +−=

57. Let () ,,x yz beequidistantfrom ()2,2,0and () 0,2,2. ()()() ()()() 222222 222222 220022 44444444 4848 0Plane xyzxyz xxyyzxyyzz xz xz −+−+−=−+−+− −++−++=+−++−+ −+=−+

58. Let () ,,x yz beequidistantfrom () 1,0,2and () 2,0,1. ()()() ()()() 222222 222222 102201 21444421 245425 220 0Plane xyzxyz xxyzzxxyzz xzxz xz

59. Let () ,,x yz beequidistantfrom ()3,1,2and () 6,2,4.

()()() ()()() 222222 222222 312624 692144123644816 62414124856 1864420 932210Plane xyzxyz xxyyzzxxyyzz xyzxyz xyz xyz ++−+−=−+++−

60. Let () ,,x yz beequidistantfrom ()5,1,3and () 2,1,6 ()()() ()()() 222222 222222 513216 1025216944211236 102635421241 1441860 72930Plane xyzxyz xxyyzzxxyyzz xyzxyz xyz

61. Firstplane:15,2,8 =−− n and () 0,3,0 P = onplane

Secondplane:21 15,6,243 =−=− nn and P noton plane

Parallelplanes (Note:Theequationsarenotequivalent.)

62. Firstplane:12,1,3 =− n and () 4,0,0 P = onplane

Secondplane:21 8,4,124=−= nn and P noton plane.

Parallelplanes (Note:Theequationsarenotequivalent.)

63. Firstplane:13,2,5 =− n and () 0,0,2 P = onplane

Secondplane:21 75,50,12525=−= nn and P onplane

Planesareidentical.

(Note:Theequationsareequivalent.)

64. Firstplane:11,4,1 =−− n and () 6,0,0 P =− onplane

Secondplane:21 555 222 ,10, =−−= nn and P onplane

Planesareidentical.

(Note:Theequationsareequivalent.)

65. (a)132nijk =+− and242 nijk =−+ 12 12 76 cos 14216 nn nn θ ===

65.91 θ  ≈°

(b)Thedirectionvectorforthelineis

() 214272. 321 1 ijk nnjk ×=−=+

Findapointofintersectionoftheplanes. 64214 420 714 2 xyz xyz x x +−= −+= = =

Substituting2for x inthesecondequation,youhave 422 yz −+=− or21. zy=− Letting1, y = a pointofintersectionis () 2,1,1.

2,1,12 x ytzt==+=+

66. (a)12,1,1 =− n and26,3,2 =− n

12 12 13136 cos 64942 40.70 θ θ ===  ≈° nn nn

(b)Thedirectionvectorforthelineis

() 1221152. 632 ijk nnij ×=−=+

Findapointofintersectionoftheplanes. 6336 6324 510 2 xyz xyz z z −++= −+= = =

Substituting2for z inthefirstequation,youhave 20 xy −+= or2. yx = Letting0, x = apointof intersectionis () 0,0,2.

5,10,2xtytz=== or,2,2 xtytz===

67. (a)13,1,1 =− n and24,6,3 = n

12 12 99671 cos 1161671

69.67 θ θ ===  ≈° nn nn

(b)Thedirectionvectorforthelineis

123119522. 463 ×=−=−−+ ijk nnijk

Findapointofintersectionoftheplanes. 186642 4632 22944 xyz xyz xz −+= ++= +=

Let0,22442 zxx ==  = and () 32071. yy−+=  =−

Apointofintersectionis () 2,1,0. 29,15,22 x tytzt=−=−−=

68. (a)163, nijk =−+ 25 nijk =−++ 12 12 42138 cos 4627207 1.684596.52 nn nn θ θ === ≈≈°

(b)Thedirectionvectorforthelineis

1263116,31,3. 115 ijk nn×=−=−−

Findapointofintersectionoftheplanes. 635 55 xyz xyz −+= −++=   635 663030 33135 xyz xyz yz −+= −++= +=

Let () 9,244,9,2. 416,931,23 yzx x tytzt =−=  =−  =−−=−−=+

69. Thenormalvectorstotheplanesare 15,3,1,n =− 21,4,7, n = 12 12 cos0. nn nn θ ==

So,2 θπ = andtheplanesareorthogonal.

70. Thenormalvectorstotheplanesare 13,1,4,n =− 29,3,12. n =−−

Because213,nn =− theplanesareparallel,butnotequal.

71. Thenormalvectorstotheplanesare 136, nijk =−+ 25, nijk =+−

12 12 53641382138 cos. 4627414207 nn nn θ ⋅−− ====

So, 2138 arccos83.5. 207 θ  =≈°

72. Thenormalvectorstotheplanesare 132, nijk =+− 242, nijk =−+

12 12 382766 cos. 1421426 nn nn θ ====

1 z =

28 xy+=

=≈°

So, 6 arccos65.9. 6 θ

73. Thenormalvectorstotheplanesare11,5,1 n =−− and 25,25,5. n =−− Because215,nn = theplanesare parallel,butnotequal.

74. Thenormalvectorstotheplanesare 122,0,1,4,1,8,nn=−=

1 12 cos0 2 nn nn θ == So, 2 π θ = andtheplanesareorthogonal.

75. 2 y ≤−

42612 xyz++=

3626 xyz++=

82. 24 xyz−+=

83. Writingtheequationofthelineinparametricformand substitutingintotheequationoftheplaneyouhave: 72,4,15 x tytzt=−+=+=−+ ()()() 7234156 66 ttt −+++−−+= =

Theequationisvalidforall t

Thelineliesintheplane.

84. Writingtheequationofthelineinparametricformand substitutingintotheequationoftheplaneyouhave: ()() 14,2,36 1 214325, 2 x tytzt ttt =+==+ ++=−=

Substituting1 2 t =− intotheparametricequationsfor thelineyouhavethepointofintersection () 1,1,0.

Thelinedoesnotlieintheplane.

85. Writingtheequationofthelineinparametricformand substitutingintotheequationoftheplaneyouhave: 13,12,3 x tytzt =+=−−=+ ()() 21331210,110, tt ++−−=−= contradiction

So,thelinedoesnotintersecttheplane.

86. Writingtheequationofthelineinparametricformand substitutingintotheequationoftheplaneyouhave:

()() 42,13,25 54231317,0 x tytzt ttt =+=−−=−+ ++−−==

Substituting0 t = intotheparametricequationsforthe lineyouhavethepointofintersection () 4,1,2. Thelinedoesnotlieintheplane.

87. Point: () 0,0,0 Q

Plane:23120 xyz++−=

Normaltoplane:2,3,1 n =

Pointinplane: () 6,0,0 P

Vector6,0,0 PQ =−

147 PQ D n n ⋅ ===

88. Point: () 0,0,0 Q

Plane:590 xyz+−−=

Normaltoplane:5,1,1 n =−

Pointinplane: () 0,9,0 P

Vector0,9,0 PQ =− 9 3 27 PQ D n n ===

89. Point: () 2,8,4 Q

Plane:25 xyz++=

Normaltoplane:2,1,1 n =

Pointinplane:0,0,5 P

Vector:2,8,1 PQ =− 11116 66 PQ D n n ===

90. Point: () 1,3,1 Q

Plane:34560 xyz−+−=

Normaltoplane:3,4,5 n =−

Pointinplane: () 2,0,0 P

Vector:1,3,1 PQ 20 22 50 PQ D n n ===

91. Thenormalvectorstotheplanesare11,3,4 n =− and 21,3,4. n =− Because12, nn = theplanesare parallel.Chooseapointineachplane.

() 10,0,0 P isapointin3410. xyz−+=

() 6,0,0 Q isapointin346. xyz−+=

4,0,0, PQ =− 1 1 4226 2613 PQ D n n ===

92. Thenormalvectorstotheplanesare12,7,1 = n and 22,7,1. = n Because12, = nn theplanesareparallel. Chooseapointineachplane.

() 0,0,13 P isapointin2713. xyz++=

() 0,0,9 Q isapointin279. xyz++=

0,0,4 PQ =  1 1 426 549 PQ D === n n

93. Thenormalvectorstotheplanesare13,6,7 n =− and 26,12,14. n =−− Because212,nn =− theplanesare parallel.Chooseapointineachplane.

() 0,1,1 P isapointin3671. xyz −++=

25 ,0,0 6 Q    isapointin6121425. xyz−−= 1 1 25 ,1,1 6 272272794 94294188 PQ PQ D n n =− ⋅ ====

94. Thenormalvectorstotheplanesare11,6,2 =− n and 2 1 2,3,1. =− n Because122, = nn theplanesare parallel.Chooseapointineachplane.

() 3,0,0 P isapointin623. xyz −++=

() 0,0,4 Q isapointin1234. xyz −++=

3,0,4 PQ = 1 1 5541 4141 PQ D === n n

95. 4,0,1 u =− isthedirectionvectorfortheline.

() 1,5,2 Q isthegivenpoint,and () 2,3,1 P isonthe line.

3,2,3 PQ =−

3232,9,8 401 PQ ijk u ×=−=−−− 1492533 1717 PQ D u u × ===

96. 2,1,2 u = isthedirectionvectorfortheline.

() 1,2,4 Q isthegivenpoint,and () 0,3,2 P isapoint ontheline () let0. t =

1,1,2 PQ =

1120,2,1 212 ijk u PQ ×==− 55 93 u u PQ D × ===

97. 1,1,2 u =−− isthedirectionvectorfortheline.

() 2,1,3 Q isthegivenpoint,and () 1,2,0 P isontheline () let0intheparametricequationsfortheline. t =

3,1,3 PQ =−−

3131,9,4 112 PQ ijk u ×=−−=−−−

1811698773 114633 u u PQ D × ++ ===== ++

98. 0,3,1 u = isthedirectionvectorfortheline.

() 4,1,5 Q isthegivenpoint,and () 3,1,1 P isontheline. 1,2,4 PQ =−

12414,1,3 031 PQ ijk u ×=−=−− 2 1419206103515 911055 PQ D u u × = ++ ==== +

99. Thedirectionvectorfor1 L is11,2,1. v =−

Thedirectionvectorfor2 L is23,6,3. v =−−

Because213,vv =− thelinesareparallel.

Let () 2,3,4 Q tobeapointon1 L and () 0,1,4 P apoint on2 L 2,2,0. PQ = 2 uv = isthedirectionvectorfor2 L

22206,6,18 363 PQ ijk v ×==−− 2 2 36363243962266 93695433 PQ D v v × = ++ ==== ++

100. Thedirectionvectorfor1 L is16,9,12. v =−

Thedirectionvectorfor2 L is24,6,8. v =−

Because12 3 2,vv = thelinesareparallel.

Let () 3,2,1 Q tobeapointon1 L and () 1,3,0 P apoint on2 .4,5,1.LPQ =−

2 uv = isthedirectionvectorfor2 L

245134,36,44 468 PQ ijk v ×=−= 2 2 222 343644 163664 4388109731813 1162929 PQ D v v × = ++ = ++ ===

105. 0.230.146.85zxy=++

(a) Theapproximationsareclosetotheactualvalues.

(b)If x and y bothincrease,thensodoes z

106. Ononesideyouhavethepoints ()() 0,0,0,6,0,0,and () 1,1,8. 1600486 118 ijk njk ==−−

101. Exactlyoneplanecontainsthepointandline.Selecttwo pointsonthelineandobservethatthreenoncolinear pointsdetermineauniqueplane.

102. Thereareaninfinitenumberofplanesorthogonaltoa givenplaneinspace.

103. Yes,Considertwopointsononeline,andathirddistinct pointonanotherline.Threedistinctpointsdeterminea uniqueplane.

104. (a)0 axbyd++= matches(iv).Theplaneis paralleltothe z-axis.

(b)0 axd+= matches(i).Theplaneisparalleltothe yz-plane.

(d)0 axczd++= matches(iii).Theplaneis paralleltothe y-axis.

Ontheadjacentsideyouhavethepoints ()() 0,0,0,0,6,0,and () 1,1,8. 2060486 118 ijk nik ==+ 12 12 361 cos 234065 1 arccos89.1 65 nn nn θ θ === =≈° Year200920102011201220132014 z(Approx)18.9319.4620.3121.1021.5822.62

107. 11:6; L xt =+ 18, yt =− 13 zt =+

22:1, L xt =+ 22, yt =+ 22 zt =

(a)At0, t = thefirstinsectisat () 16,8,3P andthesecondinsectisat () 21,2,0.P

Distance ()()() 222 618230708.37=−+−+−=≈ inches

(b)Distance ()()() ()() 2222222 121212562353070,010 xxyyzzttttt =−+−+−=+−+−=−+≤≤

(c)Thedistanceisneverzero.

(d)Usingagraphingutility,theminimumdistanceis5incheswhen3 t = minutes.

108. Firstfindthedistance D fromthepoint () 3,2,4 Q totheplane.Let () 4,0,0 P beontheplane. 2,4,3 n =− isthenormaltotheplane.

7,2,42,4,314812181829 4169292929 n n PQ D −+− ===== ++

Theequationofthespherewithcenter ()3,2,4andradius182929is ()()()222324 324. 29 xyz ++−+−=

109. Thedirectionvector v ofthelineisthenormaltothe plane,3,1,4. v =−

Theparametricequationsofthelineare53, x t =+ 4,yt =− 34. zt =−+

Tofindthepointofintersection,solvefor t inthe followingequation: ()()() 4 13 35344347 268 ttt t t +−−+−+= = =

Pointofintersection: () () () () 444774823 13131313131353,4,34,, +−−+=−

110. Thenormaltotheplane,2,1,3 n =−− isperpendicular tothedirectionvector () 2,4,0 v = ofthelinebecause 2,1,32,4,00. −−⋅=

So,theplaneisparalleltotheline.Tofindthedistance betweenthem,let () 2,1,4 Q beonthelineand () 2,0,0 P ontheplane.4,1,4. PQ =−− 4,1,42,1,3191914 4191414 PQ D n n = === ++

111. 251211113 314 ijk uvijk ×=−=−−−

Directionnumbers:21,11,13 21,111,413 x tytzt==+=+

112. Theunknownline L isperpendiculartothenormalvector 1,1,1 n = oftheplane,andperpendiculartothedirection vector1,1,1. u =− So,thedirectionvectorof L is 1112,2,0. 111 ijk v ==−

Theparametricequationsfor L are12, x t =− 2,yt = 2. z =

113. True

114. False.Theymaybeskewlines. () SeeSectionProject.

115. True

116. False.Forexample,thelines, x t = 0, y = 1 z = and 0, x = , yt = 1 z = arebothparalleltotheplane 0, z = butthelinesarenotparallel.

117. False.Forexample,planes7115 xyz+−= and 5241 xyz+−= arebothperpendiculartoplane 233, xyz−+= butarenotparallel.

118. True

Section11.6SurfacesinSpace

1. Quadricsurfacesarethethree-dimensionalanalogsof conicsections.

2. Inthe xz-plane,2 zx = isaparabola. Inthree-space,2 zx = isacylinder.

3. Thetraceofasurfaceistheintersectionofthesurface withaplane.Youfindatracebysettingonevariable equaltoaconstant,suchas0 x = or2. z =

4. No.Forexample,2220 xyz++= isasinglepoint and221 xy+= isarightcircularcylinder.

5. Ellipsoid Matchesgraph(c)

6. Hyperboloidoftwosheets Matchesgraph(e)

7. Hyperboloidofonesheet Matchesgraph(f)

8. Ellipticcone Matchesgraph(b)

9. Ellipticparaboloid Matchesgraph(d)

10. Hyperbolicparaboloid Matchesgraph(a)

11. 229yz+=

The x-coordinateismissingsoyouhavearightcircular cylinderwithrulingsparalleltothe x-axis.Thegenerating curveisacircle.

Section 11.6 Surfaces in Space 1137

12. 26yz+=

The x-coordinateismissingsoyouhaveaparabolic cylinderwiththerulingsparalleltothe x-axis.The generatingcurveisaparabola. 13. 22 22 44 1 14 xy

The z-coordinateismissingsoyouhaveanelliptic cylinderwithrulingsparalleltothe z-axis.Thegenerating curveisanellipse.

14. 22 22 25 1 2525 yz yz −= −=

The x-coordinateismissingsoyouhaveahyperbolic cylinderwithrulingsparalleltothe x-axis.Thegenerating curveisahyperbola.

15. 222 41 xyz−−=

Hyperboloidoftwosheets

xy-trace:2241 xy−= hyperbola

yz-trace:none

xz-trace:2241 xz−= hyperbola

16. 222 1 162525 xyz++=

Ellipsoid

xy-trace: 22 1 1625 xy+= ellipse

xz-trace: 22 1 1625 xz += ellipse

yz-trace:2225 yz+= circle

17. 222 2 22 16164 441 4 xyz y xz −+= −+=

Hyperboloidofonesheet

xy-trace: 2 2 41 4 y x −= hyperbola

xz-trace: () 22 41 xz+= circle

yz-trace: 2 2 41 4 y z += hyperbola

18. 22 4 zxy =+

Ellipticparaboloid

xy-trace:point () 0,0,0

xz-trace:2 zx = parabola

yz-trace:2 4 zy = parabola

19. 222 1 141 xyz++=

Ellipsoid

xy-trace: 22 1 14 xy+= ellipse

xz-trace:221 xz+= circle

yz-trace: 22 1 41 yz+= ellipse

20. 2 221 4 y zx−−=

Hyperboloidoftwosheets

xy-trace:none

xz-trace:221 zx−= hyperbola

yz-trace: 2 21 4 y z −= hyperbola

10: z =± 22 1 936 xy+= ellipse

21. 2 22 9 y zx=+

Ellipticcone

xy-trace:point () 0,0,0

xz-trace: zx =±

yz-trace: 3 y z =±

When1, z =± 2 21 9 y x += ellipse

22. 322 zyx =−+

Hyperbolicparaboloid

xy-trace: yx =±

xz-trace:2 1 3 zx =

yz-trace:2 1 3 zy =−

23. 220 xyz−+=

Hyperbolicparaboloid

xy-trace: yx =±

xz-trace:2 zx =−

yz-trace:2 zy = 1: y =± 12zx =−

24. 222 22 x yz=+

EllipticCone

xy-trace:2 x y =±

xz-trace:2 x z =±

yz-trace:point: () 0,0,0

25. 220xyz−+=

Ellipticparaboloid

xy-trace:2 yx =

xz-trace: () 220, point0,0,0 xz+=

yz-trace:2 yz = 1: y = 221xz+=

26. 222 222 818182 9941 xyz yzx −++= +−=

Hyperboloidofonesheet

xy-trace:22941 yx−= hyperbola

yz-trace:22991 yz+= circle

xz-trace:22941 zx−= hyperbola

27. Thesehavetobetwominussignsinordertohavea hyperboloidoftwosheets.Thenumberofsheetsisthe sameasthenumberofminussigns.

28. Yes.Everytraceisanellipse(orcircleorpoint).

29. No.Seethetableonpages800and801.

30. 22 zxy =+

(a)Youareviewingtheparaboloidfromthe x-axis: () 20,0,0

(b)Youareviewingtheparaboloidfromabove,butnot onthe z-axis: () 10,10,20

(c)Youareviewingtheparaboloidfromthe z-axis: () 0,0,20

(d)Youareviewingtheparaboloidfromthe y-axis: () 0,20,0

31. () 222 x zry+=    and () 5,zryy == so

22225. x zy+=

32. () 222 x zry+=    and () 3,zryy =± so

229. x zy+=

33. () 222 x yrz+=    and () 13 2, y rzz== so 22234. x yz+=

34. () 222 x yrz+=    and () , z x rze== so

222 z x ye+=

35. () 222 yzrx +=    and () 2 , yrx x == so 2 2222 2 24 yzyz x x  +=  += 

36. () 222 yzrx +=    and () 12 24,zrxx ==− so ()222 1 44 yzx +=−  222444.xyz++=

37. () 22 2 22 20 2 xyz x yz +−= +=

Equationofgeneratingcurve:2yz = or2 x z =

38. 222 cos x zy+=

Equationofgeneratingcurve:cosorcos x yzy==

39. ( ) 2 2222 5858 yzxx +=−=−

Equationofgeneratingcurve:2 58 yx =− or 582zx =−

+=−=−

40. 222 2 2222 6221 11 33 22 xyz yzxx ++=

Equationofgeneratingcurve:2 1 3 2 yx =− or 12 3 2 zx =−

42. [] 0 2 0 2sin 2sincos2 Vyydy yyy

43. 22 24 x y z =+

(a)When2 z = wehave 22 2, 24 x y =+ or 22 1 48 x y =+

Majoraxis:2842 = Minoraxis:244 = 2222,4,2cabcc=−==

Foci: () 0,2,2 ±

(b)When8 z = wehave 22 8, 24 x y =+ or 22 1. 1632 x y =+

Majoraxis:23282 =

Minoraxis:2168 = 2321616,4 cc=−==

Foci: () 0,4,8 ± 44. 22 24 x y z =+

(a)When4 y = youhave 2 4, 2 x z =+ () 12 44. 2 zx  −=

Focus: 9 0,4, 2  

(b)When2 x = youhave () 2 2 2,42. 4 y zzy =+−=

Focus: () 2,0,3

45. If () ,,x yz isonthesurface,then ()() 2222 2222 22 22 4444 8 yxyz yyxyyz xzy +=+−+ ++=+−++ +=

Ellipticparaboloid Tracesparallelto xz-planearecircles.

46. If () ,,x yz isonthesurface,then () 2222 2222 22 22 4 816 8162 88 zxyz zxyzz xy zxyz =++− =++−+ =++  =++

Ellipticparaboloidshiftedup2units.Tracesparallelto xy-planearecircles.

47. 222 2221396339633950 xyz++=

48.

Letting, x at = youobtainthetwointersectinglines , x at = , ybt =− 0 z = and, x at = 222 ,2. ybtabzabtab =+=+

50. Equatingtwicethefirstequationwiththesecondequation: 222222 2644826432 4832 346,aplane xyzyxyzx yx xy +−+−=+−−−

51. TheKleinbottle does not havebothan“inside”andan “outside.”Itisformedbyinsertingthesmallopenend throughthesideofthebottleandmakingitcontiguous withthetopofthebottle.

Section11.7CylindricalandSphericalCoordinates

1. Thecylindricalcoordinatesystemisanextensionofthe polarcoordinatesystem.Inthissystem,apoint P in spaceisrepresentedbyanorderedtriple () ,,.rz θ () , r θ isapolarrepresentationoftheprojectionof P inthe xy-plane,and z isthedirecteddistancefrom () , r θ to P

2. Thepointis2unitsfromtheorigin,inthe xz-plane,and makesanangleof30° withthe z-axis.

3. ()7,0,5,cylindrical cos7cos07 sin7sin00

4. () 2,,4, π cylindrical

()2,0,4,rectangular

5. 3,,1, 4

6. 3 6,,2,cylindrical 2

()0,6,2,rectangular

7. 7 4,,3,cylindrical

()23,2,3,rectangular

9. ()0,5,1,rectangular

22 055

cylindrical

10. () 6,23,1,rectangular () 2 2 62336124843 r =+=+==

11. () 2,2,4,rectangular () 22 2222 r =+−= () 2 tanarctan1 24 π θθ =  =−=− 4 z =− 22,,4, 4 π 

cylindrical

12. () 3,3,7,rectangular () () 22 331832 3 tanarctan1 34 7 r z π θθ =+−== =  =−=− = 32,,7, 4 π

cylindrical

13. () 1,3,4,rectangular () 2 2 132 3 tanarctan3 13 4 r z π θθ =+= =  == = 2,,4, 3 π

cylindrical

14. () 23,2,6,rectangular 1244 15 arctan 36 6 r z π θ =+=

=−=

= 4,,6, 6 π 

cylindrical

15. 4 z = istheequationincylindricalcoordinates. (plane)

16. 9,rectangularequation cos9

9sec,cylindricalequation x r r θ θ = = =

17. 22225,xyz+−= rectangularequation

2225,rz−= cylindricalequation

18. 22 2 11,rectangularequation 11,cylindricalequation zxy zr =+− =−

19. 2 , yx = rectangularequation

() 2 2 sincos sincos

sectan,cylindricalequation rr r r θθ θθ θθ = = =⋅

20. 22 2 8,rectangularequation 8cos 8cos,cylindricalequation xyx rr r θ θ += = =

21. () 22 22 222 10,rectangularequation sin10 sin10,cylindricalequation yz rz rz θ θ =− =− +=

22. 222 22 30,rectangularequation 30,cylindricalequation xyzz rzz ++−= +−=

23. 22 22 3,cylindricalequation 3 9,rectangularequation r xy xy = += +=

24. 2, z =− cylindricalequation 2, z =− rectangularequation

25. 6,cylindricalequation tan 6 1 3 3 30,rectangularequation y x y x xy xy

θ π = = = = −= 26. 22 2 22 2,cylindricalequation 2 40,rectangularequation z r z xy z xy = += +−= 27. 22 222 5,cylindricalequation 5,rectangularequation rz xyz += ++=

28. 22 2 cos,cylindricalequation ,rectangularequation zr zx θ = =

29. () 2 22 22 22 4sin,cylindricalequation 4sin 4 444 24,rectangularequation r rr xyy xyy xy θ θ = = += +−+= +−=

30. () 2 22 22 22 2cos,cylindricalequation 2cos 2 20 11,rectangularequation r rr xyx xyx xy θ θ = = += +−= −+=

31. ()4,0,0,rectangular 222 4004 ρ =++= tan00 y x θθ ==  = arccos0 2 π φ == 4,0,,spherical 2 π

  spherical x y z 321 9 123456

32. ()4,0,0,rectangular () 222 4004 tan00 arccosarccos0 2 4,0,,spherical 2 y x z ρ θθ π φ ρ π =−++= ==  = ===   

33. ()2,23,4,rectangular () () 222 223442 ρ =−++= 23 tan3 2 2 3 y x θ π θ ===− = 1 arccosarccos 24 2 42,,,spherical 34 z π φ ρ ππ ===

34. ()5,5,2,rectangular ()() () 222 55252213 ρ =−+−+== 5 tan1 54 y x π θθ ===  = 226 arccosarccosarccos 21326 z φ ρ === 26 213,,arccos,spherical 426 π  

35. ()3,1,23,rectangular 31124 ρ =++= 1 tan 3 6 y x θ π θ == = 3 arccosarccos 26 z π φ ρ === 4,,, 66 ππ

36. ()1,2,1,rectangular () () 222 1216

tan2arctan2 1 arccosarccos 6 y x z ρ θθπ

37. 4,,, 64 ππ  

41. 5,,,spherical 412 ππ

sincos5sincos0.915 124 x ππ ρφθ ==≈

sinsin5sinsin0.915 124 y ππ ρφθ ==≈

cos5cos4.830 12 z π ρθ ==≈

() 0.915,0.915,4.830,rectangular

42. 3 7,,,spherical 49 ππ  

cos4cos22 4 6,2,22,rectangular x

spherical ()

sincos4sincos6 46

sinsin4sinsin2 46

38. () 6,,,spherical 2

cos6cos0 2 6,0,0,rectangular x

sincos6sincos6 2

sinsin6sinsin0 2

3 sincos7sincos1.693 94 x ππ ρφθ ==≈− 3 sinsin7sinsin1.693 94 y ππ ρφθ ==≈

cos7cos6.578 9 z π ρφ ==≈

()1.693,1.693,6.578,rectangular

43. 2, y = rectangularequation sinsin2 2csccsc,sphericalequation ρφ θ ρφθ = =

44. 6,rectangularequation z = cos6 6sec,sphericalequation ρφ ρφ = =

45. 222 2 49, 49 7, xyz ρ ρ ++= = = rectangularequation sphericalequation

sincos12sin0cos0 4

sinsin12sin0sin0 4

cos12cos012

0,0,12,rectangular

39. () 12,,0,spherical 4

40. () 9,,,spherical 4

0,0,9,rectangular x y z

sincos9sincos0 4

sinsin9sinsin0 4

cos9cos9

46. () 222 2222 222 2 30,rectangularequation 4 4cos 14cos 1 cos 2 3,conesphericalequation xyz xyzz ρρφ φ φ π φ +−= ++= = = = =

47. 2216,xy+= rectangularequation

4csc,sphericalequation

48. 13, x = rectangularequation sincos13 13cscsec,sphericalequation

49.

50.

51. 1,sphericalequation ρ = 2211,rectangularequation xyz

52.

54. 2,sphericalequationπ φ = 222 222 cos 0 0,rectangularequation z xyz z xyz z φ = ++ = ++ = xy-plane

55. 4cos,sphericalequationρφ = () 222 222 222 222 4 40 24,0,rectangularequation z xyz xyz xyzz xyzz ++= ++ ++−= ++−=≥

56. 2sec,sphericalequation cos2 ρφ ρφ = = 2,rectangularequation z =

57. 22 22 csc,sphericalequation sin1 1 1,rectangularequation xy xy ρ φ ρφ = = += +=

58. 4cscsec,sphericalequation 4 sincos sincos4 ρφφ φθ ρφθ = = = 4,rectangularequation x =

59. 4,,0,cylindrical 4 π

22 22 404 4 arccosarccos0 2 4,,,spherical 42 z rz

60. 3,,0,cylindrical 4 π

22 22 303 4 0 arccosarccos 92 z rz

42 ππ

spherical

61. 6,,6,cylindrical 2

62. 4,,4, 3 π

63. () 12,,5, π cylindrical 22 22 12513 5 arccosarccos 13 5 13,,arccos,spherical 13 z rz

64. 22 22 4,,3,cylindrical 2 435 2 3 arccosarccos 5 3 5,,arccos,spherical 25 z

65. 10,,,spherical 62 10sin10 2 6 10cos0 2 10,,0,cylindrical 6 r z

66. 4,,,spherical 182 4sin4 2 18 4cos0 2 4,,0,cylindrical 18 r

67. 6,,,spherical 63 6sin33 3 6 6cos3 3 33,,3,cylindrical 6

68. 5 5,,,spherical 6 5sin0 5 6 5cos5 5 0,,5,cylindrical 6

69. 7 8,,,spherical 66 8sin4 6 7 6 83 8cos 62 7 4,,43,cylindrical 6

π π θ π π

70. 3 7,,,spherical 44 7sin372 42 4 372 7cos 42 7272,,,cylindrical242 r z

71. 5 r =

Cylinder

Matchesgraph(d)

72. 4 π θ = Plane

Matchesgraph(e)

73. 5 ρ = Sphere

Matchesgraph(c)

74. 4 π φ =

Cone

Matchesgraph(a)

75. 222 , rzxyz =+=

Paraboloid

Matchesgraph(f)

76. 4sec,cos4 z ρφρφ ===

Plane

Matchesgraph(b)

77. c θ = isahalf-planebecauseoftherestriction0. r ≥

78. (a)Thesurfaceisacone.Theequationis(i)

222 4 9 x yz+=

Incylindricalcoordinates,theequationis

222 22 4 9 4 9 2 3 x yz rz rz += = =

and Spherical Coordinates

(b)Thesurfaceisahyperboloidofonesheet.The equationis(ii)2222. xyz+−=

Incylindricalcoordinates,theequationis

222 22 22 2 2 2. xyz rz rz +−= −= =+

79. 22227 xyz++= (a)2227 rz+= (b)22733 ρρ =  =

80. () 4222 x yz+= (a)2242 rzrz =  = (b) () 22222222 4sincossinsincos ρφ θ ρφ θ ρφ += 22 2 4sincos, 1 tan, 4 11 tanarctan 22 φφ φ φφ = = =  =

81. 22220 xyzz++−= (a) () 22222011rzzrz +−=  +−= (b) () 22cos0 2cos0 2cos ρρφ ρρφ ρ φ −= −= =

82. 22 x yz+= (a)2rz = (b)22 2 2 sincos sincos cos sin csccot ρφρφ ρφφ φ ρ φ ρφφ = = = =

83. 224 x yy+= (a)24sin,4sin rrr==θθ (b) () 22sin4sinsin sinsin4sin0 4sin sin 4sincsc ρφρφ θ ρφρφθ θ ρ φ ρ θ φ = −= = =

84. 2245xy+=

(a)245 r = or35 r = (b)222222 22 sincossinsin45 sin45 35csc

85. 229xy−=

95. Rectangular

96. Cylindrical:

97. Spherical 46

98. Cylindrical

99. Cylindricalcoordinates:

100. Sphericalcoordinates:

101. False. ()() ,,0,0,1rz θ = and ()() ,,0,,1rzθπ = representthesamepoint ()() ,,0,0,1.xyz =

102. True () exceptfortheorigin.

103. sin,1 sin 1 zr yy zy r θ θ == ====

Thecurveofintersectionistheellipseformedbythe intersectionoftheplane zy = andthecylinder1. r =

104. 2seccos22plane 4sphere z ρφρφ ρ =  =  = =

Theintersectionoftheplaneandthesphereisacircle.

ReviewExercisesforChapter11

1. ()()() 1,2,4,1,5,4PQR===

(a)41,123,1 51,424,2 PQ PR u v ==−−=− ==−−=

(b)3,42 =−=+ uijvij

(d)333,14,25,5 uv −+=−−+=−

2. ()()() 2,1,5,1,2,4PQR =−−=−=

(a) ()() ()() 52,117,0 22,414,5 PQ PR ==−−−−−= ==−−−−= u v

(b)7,45 ==+ uivij

(d),5 337,04,517 uv −+=−+=−

3. () () cossin 8cos60sin60 13 84434,43 22 vvij ij ijij =+θθ =°+°  =+=+=   

4. cossin 11 cos225sin225 22 2222 , 4444 vvi+vj ij ij = θθ =°+° =−−=−−

5. () 0,4,5:5,4,0zyx===−−

6. 3 y = describesaplaneparalleltothe xz-planeand passingthrough () 0,3,0.

7. ()()() 222 213653 99422 d =−−+−+− =++=

8. () () ()() () 22 2 421115 3641656214 d =−−+−−+−−− =++==

9. ()()() ()()() 2222 222 3264 32616 xyz xyz −+++−= −+++−=

10. Center: () 040640 ,,2,3,2222 +++  =  

Radius: ()()() ()()() 222 222 20302449417 23217xyz −+−+−=++= −+−+−=

11. ()() ()() 222 232 4469449 239 xxyyz xyz −++−++=−++ −+−+=

Center: () 2,3,0

Radius:3

12. ()() () ()()() 22 2 222 102569 44342594 5324 xxyy zz xyz −++++ +−+=−+++ −+++−=

Center: () 5,3,2

Radius:2

13. (a),(d)

15. 1 3 2 1 5,2,330,2,16,6,2 2 5,2,30,6,33,3,1 8,5,1 z =−++ =−−++−− =−−++−− =− uvw

16. () () 20 1 2 1 5,2,30,2,16,6,2 2 1 1,10,4 2 1 ,5,2 2 uvwz uvw z −+−= =−+ =−−+−− =−− =−−

17. 13,64,914,2,10 v =−−−+=− 53,34,612,1,5 =−−−+=−− w

Because2, −=wv thepointslieinastraightline.

18. 85,54,573,1,2 =−−+−=−− v 115,64,376,10,4 =−+−=− w

Because v and w arenotparallel,thepointsdonotliein astraightline.

19. Unitvector:2,3,5235 ,, 38383838 == u u

20. 6,3,28482416 86,3,2,, 497777 =−=−

21. 5,0,0,4,4,0,2,0,6PQR ===

(b) () 42,41,732,5,10 v =−−−−−=− (c)2510 =+− vijk

14. (a),(d)

x y 123 45 5 3 1 2 3 2 9 10 8 z (2,1,3) (4,4,7) (2,5,10) x y 1 3 4 2 1 5 6 1 6 5 7 8 z (3,3,8) (3,5,8) (6,2,0) v

(b)36,32,803,5,8 v =−−−−=−−

(a)1,4,0 PQ ==− u 3,0,6 PR =− v=

(b) ()()()() 1340063 ⋅=−−++=uv

22. 2,1,3,0,5,1,5,5,0PQR =−==

(a)2,6,2 PQ ==−− u 3,6,3 PR ==− v

(b) ()()()()()() 23662336⋅=−++−−=uv

23. [] 33525cossin 442 ππ  =+=−+   uijij

28. 1,1,1,2,0,2=−=uv

vijij

22 2cossin3 33 ππ  =+=−+

() 52 13 2 ⋅=+uv 2525 5132 22 =+==+= uv ()() () 5221326 cos 524 θ + + === uv uv

(a)26arcos0.262 412 π θ + ==≈

(b)15 θ ≈°

24. () 1,0,3 2,2,1 1 1910 4413 1 cos 310 θ =− =− ⋅=−

(a) 1 arccos1.465 310 θ

(b)83.9 θ =°

25. 7,2,3,1,4,5uv=−=−

Because0, ⋅=uv thevectorsareorthogonal.

26. 3,0,931,0,33 =−=−−=− uv

Thevectorsareparallel.

27. 4,2,3,4==uv

== v uv wuv v

(a)12 proj 20 3,4 25 41216 3,4, 555

(b)21 121686 4,2,, 5555 wuw=−=−=−

== v uv wuv v

(a)12 proj 4 2,0,21,0,1 8

(b)211,1,11,0,10,1,0 wuw=−=−−=−

29. Therearemanycorrectanswers. Forexample:6,5,0. v =±−

30. ()() cos758cos30 3003ft-lb WPQPQ θ =⋅==° = FF

31. (a)4369267 521 

ijk uvijk

×==−+−

(b) () 9267×=−×=−+ vuuvijk

ijk uvijk

×==+−

32. (a)021112 134

(b) () 112×=−×=−−+ vuuvijk

33. 2108324852 468

  ×=−=++    ijk uvijk 60324377×==uv

Unitvector: 1 8,12,13 377

34. 3,1,5,2,4,1=−=−uv

  ×=−=+−    ijk uvijk () 22219710 510 A =×=++− = uv

31519710 241

35. 40 =− Fk () 3 94in.ft = () 3333 cos60sin60 488 PQ =°+°=+jkjk

Themomentof F about P is 333 015 88 0040 MPQ=×==− ijk Fi

Torque15ft-lb =

=⋅⋅===

uvw

36. () () 210 0212510 012 V

37. 93,110,626,11,4 v =−−−=

(a)Parametricequations: 36,11,24 x tytzt=+==+

(b)Symmetricequations:32 6114 xyz ==

38. 81,104,539,6,2 v =+−−=

(a)Parametricequations: 19,46,32 x tytzt=−+=+=+

(b)Symmetricequations:143 962 xyz+−− ==

39. () 6,8,2 P =−− 0,1,0 ==vj

6,8,2xytz=−=−+=

40. Directionnumbers: () 1,1,1,1,1,1 1,2,3 P = v 1,2,3 x tytzt =+=+=+

41. ()()() [] 3,4,2,3,4,1,1,1,2 0,8,1,4,5,4 08127432 454 ijk nijk PQR PQPR PQPR =−−=−=− =−=− =×=−=−−−

()()() 273443220 2743233 xyz xyz −+−+−−= ++=−

42. 3 nijk =−+ ()()() 3213110 380 xyz xyz +−−+−= −++=

43. Thetwolinesareparallelastheyhavethesamedirection numbers,2,1,1.Therefore,avectorparalleltothe planeis2. vi+jk =−+ Apointonthefirstlineis () 1,0,1andapointonthesecondlineis ()1,1,2.The vector23 uijk =−− connectingthesetwopointsis alsoparalleltotheplane.Therefore,anormaltothe planeis () 211 213 2422. ijk vu ijij ×=− =−−=−+

Equationoftheplane: ()120 21 xy xy −+= +=

44. Let52,12,313,3,2 v =−+−= bethedirection vectorforthelinethroughthetwopoints.Let 2,1,1 n =− bethenormalvectortotheplane.Then

3325,7,3 211 ijk vn×==−− isthenormaltotheunknownplane. ()()() 5571330 573270 xyz xyz −−+−−−= −+−+=

45. () 1,0,2 Q point 2366 xyz−+= Apoint P ontheplaneis () 3,0,0. 2,0,2 PQ =−

2,3,6 n =− normaltoplane

8 7 n n PQ D ==

46. () 3,2,4 Q point

2510 xyz−+=

Apoint P ontheplaneis () 5,0,0. 2,2,4 PQ =−−

2,5,1 n =− normaltoplane 1030 303 n n PQ D ===

47. Thenormalvectorstotheplanesarethesame, () 5,3,1. n =−

Chooseapointinthefirstplane () 0,0,2. P Choosea pointinthesecondplane, () 0,0,3. Q 0,0,5 5535 35357 n n PQ PQ D =− ====

48. () 5,1,3 Q point

1,2,1 u =−− directionvector () 1,3,5 P pointonline 6,2,2 PQ =−−− 6222,8,14 121 ijk u PQ ×=−−−=−−

49. 236xyz++= Plane

Intercepts: ()()() 6,0,0,0,3,0,0,0,2,

50. 2 yz =

Becausethe x-coordinateismissing,youhavea cylindricalsurfacewithrulingsparalleltothe x-axis. Thegeneratingcurveisaparabolainthe yz-coordinate plane.

51. 1 2 yz = Planewithrulingsparalleltothe x-axis.

52. cos y z =

Becausethe x-coordinateismissing,youhavea cylindricalsurfacewithrulingsparalleltothe x-axis. Thegeneratingcurveiscos. y z =

(6,0,0) (0,3,0) z

53. 22 21 169 xy z ++= Ellipsoid

xy-trace: 22 1 169 xy+= xz-trace: 2 21 16 x z += yz-trace: 2 21 9 y z += x y 6 3 3 (0,0,2)

54. 222 161690 xyz+−= Cone

xy-trace:point () 0,0,0

xz-trace: 4 3 x z =±

yz-trace: 4 3 y z =±

22 4,9zxy=+=

57. 224.xz+= Cylinderofradius2about y-axis

58. 2216.yz+= Cylinderofradius4about x-axis

xy 33 2 33 4

55. 22 2 22 2 1 169 1 916 xy z yx z −+=− −−=

Hyperboloidoftwosheets

xy-trace: 22 1 916 yx −=

xz-trace:None

yz-trace: 2 21 9 y z −=

56. 222 1 254100 xyz+−=

Hyperboloidofonesheet

xy-trace: 22 1 254 xy+=

xz-trace: 22 1 25100 xz −=

yz-trace: 22 1 4100 yz −=

59. 22 zy = revolvedabout y-axis () 222 22 2 2 2 zy x zryy xzy =± +=  =  +=

60. 231 xz+= revolvedaboutthe x-axis 12 3 x z = () 2 22212 3, x yzrx

Cone

61. ()3,3,5,rectangular (a) () 2 2 331223 r =−+== 3 tan 33 π θθ =  =− 5 z =− 23,,5,cylindrical 3 π

(b) () () 22 2 33537 ρ =−++−= 3 tan 33 π θθ =−  =− 5 arccosarccos 37 z

537 37,,arccos,spherical 337 π

62. ()8,8,1,rectangular

(a)228882 r =+= 8 tan1 84 π θθ ==  = 1 z = 82,,1,cylindrical 4 π  

(b)222881129 ρ =++= 8 tan1 84 π θθ ==  = 1 arccosarccos 129 z φ ρ == 129 129,,arccos,spherical 4129 π  

63. ()5,,1,cylindrical π cos5cos5 sin5sin0 1 xr yr z θπ θπ ===− === =

()5,0,1,rectangular

66. 8,,,spherical 63 ππ

64. 2,,3,cylindrical 3 π 

cos2cos1 3 sin2sin3 3 3 xr yr z π θ π θ

1,3,3,rectangular

65. 4,,,spherical 4 π π

sinsin4sinsin0 4 cos4cos22 4 x y z π

sincos4sincos22 4

()22,0,22,rectangular

() 6,23,4,rectangular

67.

(a)Cylindrical:

68. 22216 xyz

(a)Cylindrical:2216

(b)Spherical:4

72. 9sec,sphericalequationρθ = cos9 9,rectangularequation z

ProblemSolvingforChapter11

3. Labelthefigureasindicated. Fromthefigure,youseethat 1111 and. 2222 SPRQSRPQ abab =−==+=

Becauseand, SPRQSRPQ == PSRQ isaparallelogram.

4. Labelthefigureasindicated.

()() 220,because ab ba abbaba PR SQ =+ =− +⋅−=−= ab = inarhombus.

5. (a)0,1,1 u = isthedirectionvectoroftheline determinedby1 P and2 P 1 2,0,10,1,1 2 1,2,2332 222 u u PQ D × = −× = ===

6. ()() 00nnPPPP+⊥− Figureisasquare. So,0 PP n  = andthepoints P formacircleofradius n intheplanewithcenterat0 . P

(b)Theshortestdistancetotheline segment is 12,0,15.PQ =−=

7. (a) () 1221 0 0 1 22 z Vzdzπππ  ===   

Note: 1 2(base)(altitude) () 11 1 22 ==ππ

(b) () () () 22 22 22 22 :sliceat 1 xy zzc ab xy cacb +== += At, zc = figureisellipseofarea ()()cacbabcππ = 22 0 0 22 k k abcabk Vabcdc ππ π  =⋅==   

8. (a) () 3 2223 0 0 4 22 33 r r x Vrxdxrxr πππ  =−=−=

(b)Atheight0, zd=> 222 222 22222 2222 1 1 xyd abc x ydcd abcc ++= +=−= () () 22 222222 221. xy acdbcd cc += Area () () () 222222 22 222 acdbcd ab cd ccc π π

() 22 02 3 2 2 0 2 24 33 c c ab Vcddd c abdcdabc c π π π =−

=−=  

9. FromExercise54,Section11.4, ()()()() .uvwzuvzwuvwz ×××=  ×⋅ 

10. () 1 3,1,21;4,3, 2 x tytztQs =−+=+=−=

(a)2,1,4 u =− directionvectorforline

() 3,1,1 P =− pointonline

1,2,1PQs=+ ()() 121 214 7625 PQs ss ijk u ijk ×=+ =−+−−+ ()() 22 76225 21 u u PQ ss D × −+−−+ ==

(b)

Theminimumis2.2361at1. Ds≈=−

(c)Yes,thereareslantasymptotes.Using, s x = youhave () () 222 1555 ()5101102221211 21212121 Dsxxxxxx =++=++=++→±+ () 105 1 21 ys=±+ slantasymptotes.

11. (a)2sinρφ = Torus

(b)2cos Sphere ρ φ

12. (a)2cos r θ = Cylinder

Hyperbolicparaboloid

13. (a) () cos0sin0 uuijui =+= Downwardforce wj =−

sin 1cos uT T

(b)Frompart(a),tanandsec. ==θθuT

Domain:090 θ ≤≤° (c)

(d)

(e)Bothareincreasingfunctions. (f) 22 limandlim. Yes.Asincreases,bothandincrease.

14. (a)Thetension T isthesameineachtowline. () () () ()6000cos20cos20sin20sin20 2cos20

60003192.5lb 2cos20 iij i TT T T =°+−+°+−° =°  =≈ °

(b)Asinpart(a),60002cos i T θ = 3000 cos T θ  = Domain:090 θ <<° (c) (d)

(e)As θ increases,thereislessforceappliedinthedirectionofmotion.

15. Let, θα β =− theanglebetween u and v. Then () sin. uvvu uvuv αβ ×× −==

Forcos,sin,0 u αα = andcos,sin,0,1 vuv ββ === and ()cossin0sincoscossin cossin0 ijk vuk. ββαβαβ αα ×==−

So, () sinsincoscossin. vu αβαβαβ −=×=−

16. (a)LosAngeles: () 4000,118.24,55.95 −°°

(b)LosAngeles: ()() ()() () ()() 4000sin55.95cos118.24 4000sin55.95sin118.24 4000cos55.95 ,,1568.2,2919.7,2239.7 x y z xyz =°−° =°−° =° ≈−−

RiodeJaneiro: () 4000,43.23,112.90 −°°

RiodeJaneiro: ()() ()() () ()()

4000sin112.90cos43.23 4000sin112.90sin43.23 4000cos112.90 ,,2684.7,2523.8,1556.5 x y z xyz =°−° =°−° =° ≈−−

91.17or1.59radians uv uv θ θ −+−−+− ==≈− ≈°

(d) () 40001.596360 srθ ==≈ miles

(e)ForBostonandHonolulu:

a.Boston: () 4000,71.06,47.64 −°°

Honolulu: () 4000,157.86,68.69 −°°

4000sin47.64cos71.06

4000sin47.64sin71.06

4000cos47.64

b.Boston: () () ()

959.4,2795.7,2695.1 x y z =°−° =°−° =°

4000sin68.69sin157.86

4000cos68.69

Honolulu: () () () 4000sin68.69cos157.86

3451.7,1404.4,1453.7 x y z =°−° =°−° =°

c. ()()()()()() ()() 959.43451.72795.71404.42695.11453.7 cos0.28329 40004000

73.54or1.28radians

d. () 5120miles srθ ==40001.28≈

17. FromTheorem11.13andTheorem11.7(6)youhave

18. Assumeoneof,,, abc isnotzero,say a Chooseapoint inthefirstplanesuchas () 1,0,0.da Thedistance betweenthispointandthesecondplaneis

19. 221xy+= cylinder 2 zy = plane

Introduceacoordinatesystemintheplane2. zy = Thenew u-axisistheoriginal x-axis.

Thenew v-axisistheline2,0. zyx==

Thentheintersectionofthecylinderandplanesatisfies theequationofanellipse:

20. Essay

SECTION PROJECTS

Chapter 1, Section 5, page 94 Graphs and Limits of Trigonometric Functions

(a)Onthegraphof f,itappearsthatthe y-coordinatesofpointslie ascloseto1asdesiredaslongasyouconsideronlythosepoints withan x-coordinateneartobutnotequalto0.

(b)Useatableofvaluesof x and () f x thatincludesseveralvaluesof x near0.Checktoseeifthecorrespondingvaluesof () f x arecloseto1.Inthiscase,because f isanevenfunction,onlypositivevaluesof x areneeded.

x 0.50.10.010.001

f (x)0.95890.99831.00001.0000

(c)Theslopeofthesinefunctionattheoriginappearstobe1. (Itisnecessarytouseradianmeasureandhavethesameunit oflengthonbothaxes.)

(d)InthenotationofSection1.1,0 c = and. cxx +Δ= Thus,sec sin0 . 0 x m x = Thisformulahasavalueof0.998334if0.1; x = sec0.999983m = if0.01. x = Theexactslopeofthetangentlineto g at ()0,0issec 00 sin limlim1. xx x m x →→ ==

(e)Theslopeofthetangentlinetothecosinefunctionatthepoint ()0,1is0.Theanalyticalproofisasfollows:

(f)Theslopeofthetangentlinetothegraphofthetangentfunctionat ()0,0is: () () sec 00 tan00sin11 limlimlim11. cos1 xxx xx m xxxΔ→∞Δ→Δ→

Chapter 2, Section 5, page 151 Optical Illusions

(a)222 220 x yC xyy x y y += ′ += ′ =− atthepoint () 3 3,4, 4 y ′ =−