Language Studies and Academics - Mathematics

Mathematics 2100 Packet 2 of 5 Fall 2012 Unit 2: Applications of the Derivative The material contained in this packet of notes will be covered by Test #2

MA2100 Derivative Applications Max/Min Problems Handout 2-1 Learning Objectives 2.1

Applied Maximum and Minimum Problems 8th Ed: Chapter 24.7: page 718 – 722 9th Ed: Chapter 24.7: page 720 - 724 Procedure for solving optimization problems 

Identify the quantity for which you want to find the maximum or the minimum.

Sketch a figure to summarize the given information and assign symbols to all given quantities.

Write an equation in which the variable to be minimized is the dependent value.

Determine the domain of the equation. That is, determine the values for which the problem makes sense. (Pay close attention to the endpoints.)

By means of the derivative, determine the extrema of the equation. Recall, at extrema, f '  x   0 .

Determine the exact nature of the extrema by either 1st or 2nd derivative test. Recall at a min, f '  x  changes from negative to positive, and from positive to negative at a maximum. Also, at a minimum the graph will be concave up, f ''  x   0 , and at a maximum the graph will be concave down, f ''  x   0 .

Be sure check the end points of the domain on the graph as well, even if they are not critical numbers.

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Examples 1. Find two numbers with difference of 150 such that the product of the two numbers is a minimum.

2. (p. 722, #4) A small oil refinery estimates that its daily profit P (in dollars) from refining x barrels of oil is P  x   8 x  0.02 x 2 . (a) How many barrels should be refined for maximum daily profit and (b) what is the maximum profit?

3. (p. 723, #30) The electric power P (in W) produced by a certain battery is given by the equation 144r P r  2  r  0.6  where r is the resistance in the circuit. For what value of r is the power a maximum?

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4. (p. 724, #39) A beam of rectangular cross section is to be cut from a log 1.00m in diameter. The stiffness of the beam varies directly as the width and cube of the depth. What dimensions will give the beam maximum stiffness?

5. Find the point on the parabola y ď&#x20AC;˝ 12 x 2 that is nearest to the point (4, 1).

6. A rectangular microprocessor chip is designed to have an area of 25mm2. What must be its dimensions if its perimeter is to be a minimum?

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7. A computer is programmed to display a slowly changing right triangle with its hypotenuse always equal to 12.0 cm. What are the legs of the triangle when it has its maximum area?

8. A can (cylinder shape) is to be made to hold 1L of oil. Find the dimensions that will minimize the cost of the metal to make the can. Note: 1L = 1000 cm3

9. A Sheppard wants to fence off a 500m2 rectangular field along a river. No fence is needed on the river side. The fence costs 50QR per meter on the side parallel to the river and 35QR per meter on the ends perpendicular to the river. What is the least expensive fencing possible?

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10. A rectangle is to be inscribed in a semicircle of radius 2m. What is the largest area the rectangle can have and what are its dimensions?

b , where a v2 and b are positive constants. At what speed does the airplane have the least drag?

11. The drag D on an airplane traveling at speed v is D ď&#x20AC;˝ av 2 ď&#x20AC;Ť

12. An open rectangular box with square ends must hold 6400ft3. The material for the base cost 75 cents per square foot and the material for the sides cost 25 cents per square foot. Find the dimensions that result in the smallest cost for materials.

Additional Practice: 8th Ed: page 722, #3 - #41, #43 9th Ed: page 724, #3 - #40, #43, #46

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2) a) 200 barrels b) \$800

3) r = 0.6

4) width = 0.5m 3 m depth = 2

5) (2,2)

6) 5mm  5mm

7) both equal 6 2cm .

8) radius = 5.419cm height = 10.839cm

9) 26.458m  18.898m Costmin  \$2645.75

10)

2  2 2  4m 2

11) V 

4

b a

12) 20 ft 16 ft

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MA2100 Derivative Applications Parametric Equations and Curvilinear Motion Handout 2-2 Learning Objectives 2.2

Curvilinear Motion 8th Ed: Chapter 24.3, pages 698 – 702 9th Ed: pages 701 – 704 In MA1101 we looked at rectilinear motion, which dealt with objects moving in a straight line (1D motion). Curvilinear motion deals with objects moving in a plane along a curved path (2D motion) and is described using the language of vectors. The motion of an object undergoing curvilinear motion is usually expressed using a parametric equation. That is, both the x and y coordinates of the points on the path are expressed independently as functions of a third variable (the parameter). In the case of movement, the parameter is usually time, t. If the path of an object is given in parametric form,  x  t  , y  t   , then the velocity components are

vx 

dx : Horizontal Velocity dt

vy 

dy : Vertical Velocity dt

Similarly the acceleration components are

ax 

dVx d 2 x  2 : Horizontal Acceleration dt dt

ay 

dVy dt

d2y : Vertical Acceleration dt 2

Since we will be dealing with 2-dimensions instead of 1-dimension (as was the case with rectilinear motion) we will need to identify both the magnitude of motion as well as the direction. Therefore both are to be calculated. The total magnitudes are calculated using Pythagorean Theorem so that v  vx2  v y2

Total Velocity

a  ax2  a y2

Total Acceleration

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The direction in both cases is calculated using the tangent trigonometric ratio, v a tan  v   y or tan  a   y . vx ax NOTE: When finding direction you should sketch the components so as to make sure you get the correct angle using the tan 1 function!

Examples 1. Find the velocity and direction of motion when t  2 of an object moving such that its position at time t is given by 1  2t , t 2  3t  . dx  2  Vx  2   2 dt dy Vy   2t  3  V2  2   1 dt Vx 

 V  22  12 

5

1    tan 1    26.6o  0.4643rad 2

2. Find the magnitude and direction of acceleration when t  2 for an object that is moving such that its x- and y-coordinates of position are given by x  t 3 and y  1  t 2 . Vx  3t 2  ax  6t  ax  2   12 Vy  2t  a y  2  a y  2   2 a  122   2   2 37 2

 12     tan 1    80.5o  279.5o  4.8782rad  2 

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3. A helicopter is flying at 18.0 m s and at an altitude of 120m when a rescue marker is released from it. The marker maintains a horizontal velocity and follows a path given by y  120  0.0151x 2 . Find the magnitude and direction of the velocity and acceleration of the marker 3.00s after release.

t 3

x  t   18t

Vx  18 ms

y  t   120  0.015 x 2  t   120  4.86t 2

ax  0 sm2 Vy  t  

dy  9.72t  Vy  3  29.16 ms dt

dVy 2  34.3mm V  3  182   29.16 s 2 9.722s 2  aay3t  0dt  9.72 sm2  9.72 18   o  5.7299rad V  tan 1    328.3 3  o 270   a29.16 rad 2

4. A rocket moving vertically, if the only force on it is due to gravity and its mass is decreasing at a constant rate r, its velocity as a function of time given by v  t   vo  g t  k ln 1  mr to where vo is the initial velocity, mo is the

initial mass, and k is a constant. Determine the expression for the acceleration. a t  

dV k  g  dt 1  mrto

 rt   mo

 kr g   mo  rt

V  5 ms , V  26.6o

2) a  2 37 sm ,  a  350.5o 2

4)

3) V  34.3 ms , V  301.7 o a  9.72 ,  a  270 m s2

o

a

kr g mo  rt

Additional Practice: 8th Ed: Page 702, #3 - #27 9th Ed: Page 704, #3 - #27 9 of 30

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MA2100 Derivative Applications Electrical Application Handout 2-3 Learning Objectives 2.3

Electrical Applications In the study of electricity, current, i (measured in Amperes), is defined as a measure of the amount of charge, q (measured in coulombs), that passes a given point in a circuit per unit of time. Since we know that derivative represents a rate of change, we can write the following equation, Current : i

dq dt

Power (measured in Watts) is defined as the rate of change in the amount of work (measured in Joules). Hence we define the equation, Power : dW P dt Examples di , where L is the dt inductance (measured in Henry’s, H ). If the current through a 0.040H inductor is i  1.60t 2 . Find the voltage at t  0.50sec .

1. Voltage in a circuit, is given by the equation v  L

2. If the work done (in Joules) in moving an object is W  5t 4  2t , find the power (in joules/sec) when t  2sec .

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3. Find the current i, in a circuit, as a function of time if the charge, q, is given by q  2t 2  t . Also, find the current when t  5.0sec .

4. Find the Power in watts generated by an electric motor if work done is W  10t 3  3t 4 (joules) at t  2sec .

2) P  2   162W

3) i  5.0   21A

4) P  2   216W

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MA2100 Derivative Applications Related Rates Handout 2-4 Learning Objectives 2.4

Related Rates 8th Ed: Chapter 24.4, p. 703 – 707 9th Ed: Chapter 24.4, p. 705 – 708 Any two variables that vary with time and also have a known relationship can have the time rate of change of one (derivative) expressed in terms of the time rate of change for the other. In other words if we have an equation relating 2 or more quantities we can express the derivative of one variable in terms of the derivative of the other(s). Procedure for solving related rate problems  

 

Make a clearly labeled sketch to summarize all given information. Write an equation relating the variables whose rates of change are either given or to be determined. If necessary, use the geometry of the situation to eliminate, by substitution, any unnecessary variables. (That is those variables whose rates of change are neither given nor to be determined.) Use the chain rule to implicitly differentiate both sides of the equation with respect to t. And solve for the unknown rate. Substitute the given numerical information into the resulting equation.

Examples 1. The voltage of a certain thermocouple as a function of the temperature is given by E  2.800T  0.006T 2 . If the temperature is increasing at the rate of 1.00 minC , how fast is the voltage increasing when T  100 oC ? o

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2. A spherical balloon is being inflated such that its volume increases at the m3 . Find the rate at which the radius is increasing constant rate of 2.00 min when it is 3.00m. Vsphere  43  r 3 

3. According to Boyle’s Law, whenever a volume of gas is compressed (at a constant temperature) the pressure and volume are related by the equation PV  C , where C is a constant. Find the rate at which the volume is changing at the point where the volume is 600cm3 , and where

kPa . the pressure is P  t *   150kPa and decreasing at a rate of 20 min

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Notes:  a) A common error is to substitute the given numerical information (for the given rates of change) at too early a stage.  b) You must know the basic geometric formulae for a circle, triangle, rectangle, cylinder and prism.  c) In problems with a right triangle try :  Pythagorean theorem  Trigonometric functions  Similar triangles

4. A rectangular swimming pool 16m by 10m is being filled at the rate m3 . How fast is the height of the water rising? of 0.8 min

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5. A radar tracks an oncoming plane that is at an angle of  . If x defines the dx horizontal distance between the radar and the plane, find in terms dt d . of dt

6. A conical tank (with vertex down) is 5m across at the top and 6m deep. If m3 , how fast is the water level rising water flows into the tank at a rate of 3 min in the tank when the water is 4m deep? Vcone  13  r 2 h 

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7. A circular oil slick of uniform thickness is caused by a spill of 10m3 of oil. The thickness of the oil slick is decreasing at a rate of 1mm/h, as the oil spreads out. At what rate is the radius of the oil slick increasing when the radius is 12m?

8. As a spherical raindrop falls it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area. Show that the radius decreases at a constant rate.  Asphere  4 r 2 

Additional Practice: 8th Ed: page 706, #3 - #36 9th Ed: page 706, #7 - #44 17 of 30

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MA2100 Derivative Applications Newton’s Method Handout 2-5 Learning Objectives 2.5

Newton’s Method 8th Ed: Chapter 24.2, pages 695 – 698 9th Ed: Chapter 24.2, pages 697 – 700 In MA1700 and MA1101 we looked at two basic methods for finding the roots of an equation, f  x   0 . The first was the quadratic formula that gives the two roots of a quadratic equation,

f  x   ax 2  bx  c  x 

b  b 2  4ac . 2a

The second method involved recognizing some common patterns and factoring, x 2  a 2   x  a  x  a 

x3  a 3   x  a   x 2  ax  a 2 

ax3  bx 2  cx  x  ax 2  bx  c  Beyond these very specific examples, there is little we can do to find the exact roots of a higher degree polynomial or functions involving ln  f  x   , e f  x  ,sin  x  , In such cases we have to look for a numerical approximation to the roots. One such method is Newton’s Method. Newton’s Method is an iterative process that approximates the roots of an equation to as many significant figures as are required. The process of iteration involves the following loop: 1. Make a guess as to the value of the root, x0 . 2. Use the initial guess to get a better approximation, x1 . 3. Return to step 1, make x0  x1 . Keep repeating until enough significant digits are found. The key step in the process is Step 2, using one guess to get a better guess. The way that Newton’s Method gets the better approximation is by finding the x-

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intercept of the tangent line to the curve at  x0 , f  x0   . Recall the point-slope formula for a tangent line through the point  a, b  ,

 y  b   f '  x  x a  x  a  . Newton’s Method is as following. 1. Make a guess as to the root,  a  x0 , b  f  xo   2. Find the x-intercept of the tangent line, x1 . (Let y  0 , find xint ) 3. Find the point on curve with x  x1 ,  x1 , f  x1  

4. This point then becomes the new guess and we return to Step 2. Two iterations of Newton’s method are graphed below.

Figure 1: Two iterations of Newton’s method Translating this method into mathematical notation we get the general formula for th the  n  1 approximation to the root (see derivation at end of handout),

Newton's Method : xn 1  xn 

f ( xn ) f '( xn )

When performing the iterations you will find it useful to keep track of the results using the following table, n

xn

f ( xn ) =

f '( xn ) =

xn1  xn 

f ( xn ) f '( xn )

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Consider: Referring to Figure 1, why it is important that the initial guess be close to the actual root that you are seeking? 1. (a) Show that f  x   x 2  4 x  2 has a root between -4 and -3 and (b) find

the root to 4 decimal places.

n

xn

f ( xn )

f '( xn )

xn1  xn 

f ( xn ) f '( xn )

0 1 2 3 4 5 6 2. The function f  x   x 4  x3  3 x 2  x  4 has a root between 2 and 3. Find the root, to 4 decimal places, using Newton’s Method.

n

xn

f ( xn )

f '( xn )

xn1  xn 

f ( xn ) f '( xn )

0 1 2 3 4 5 6

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3. The altitude h (in m) of a rocket is given by h  t   2t 3  84t 2  480t  10 , where t is the time (in s). When does the rocket hit the ground? Find the answer to five decimal places.

n

xn

f ( xn )

f '( xn )

xn1  xn 

f ( xn ) f '( xn )

0 1 2 3 4 5 6 4. The curve y  x3  3 x  1 has one positive solution. Make an initial estimate for the positive root and then use Newton’s method to approximate the solution by four iterations. What is the root? How many decimal places are significant?

n

xn

f ( xn )

f '( xn )

xn1  xn 

f ( xn ) f '( xn )

0 1 2 3 4 5 6

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Additional Practice: 8th Ed: page 698, #2 - #16, #19- #21 9th Ed: page 700, #2 - #16, #19- #21 Optional: Derivation of Newton’s Method

The derivation of Newton’s method is quite straightforward. Starting with an initial guess as to the root,  x0 , f  xo   , find the equation of the tangent line, where  a, b    x0 , f  xo   ,

 y  b   f '  a  x  a  ,

y  f  xo   f '  xo  x  xo  .

Next we find the x-intercept of the tangent line, which will become the new guess as to the root. The x-intercept then will be the point  x, y    x1 , 0  , which when we substitute into the equation of the tangent we get, 0  f  xo   f '  xo  x1  xo 

Finally we simply solve for the x-intercept and we get Newton’s method, f '  xo   x1  xo   f  xo   f '  xo  f '  xo 

 f  xo   x1  xo f '  xo  x1  xo 

.

f  xo  f '  xo 

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MA2100 Differentials Handout 2-6 Learning Objectives 3.0

Differentials 8th Ed: Chapter 24.8, pages 725 – 728 9th Ed: Chapter 24.8, pages 727– 730 3.1: Definition of Differential When we began our study of derivatives in MA1101 we approximated the slope of a tangent line by the secant line so that we had, y M tangent  M secant  . x For the actual derivative we took the limit as x  0 and defined the derivative with the notation, y dy . y '  M tangent  lim  x 0 x dx In this section will look at the derivative from the point of view of a ratio,

 dy  .  dx 

While related, dy is not the same as y (change in y). dy is the infinitesimal y we get when we change x by an infinitesimal amount. As such we give dy the special name, “differential of y”. Likewise, we refer to dx as the “differential of x”. Starting with the function y  f  x  we can define the relationship between dy and dx through the derivative,

dy , dx which when we rearrange we get the differential of y , dy  f '( x)dx f ' x 

Example 1: Find the differential of y for each of the following.

a)

y  4 x3  5 x

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b)

P   4T 3  5T 

c)

y

2

2 x2  1 4e x

3.2: Graphical / Geometric Interpretations From a graphical point of view you might say that “ dy is simply y for two very closely spaced points” and you would essentially be correct since dy refers to the tangent line and y refers to the secant line, which we know become the same when we take the limit. However, at any point before the limit, we can discuss and compare dy and y .

x  0  dy  y

Sometimes y may be difficult to calculate and we may choose to approximate it with dy , which can be easier to calculate. To do so, simply set x (change in x for the curve) equal to dx (change in x for the tangent line), x  dx , And calculate by, y  dy  f '  x  dx for a “small” value of x.

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Example 2: Calculate y and dy for y  x3  2 x for x  3 and x  0.1 .

Example 3: Find the difference between y and dy for f  x   2 x3  5 x 2  7

when x  3 and x  0.1 .

3.3.1: Effect of small change in independent variable The differential gives an approximation as to the change in the dependent variable given a small change in the independent variable. That is, df is the change in the function when x is changed by dx . Example 4: (a) Find the exact change in the area of a circle, A , due to a change in its radius, r . (b) Find the differential approximation dA to the change in the area of the circle due to a change r in its radius. (c) Compare A and dA for r  1 and  r  0.01 .

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Example 5: (a) Use differentials to approximate the increase in the surface area of a soap bubble, dS , when its radius increases from 3.000 inches to

3.025 inches. (b) Also evaluate the actual change S .  Asphere  4 r 2 

3.3.2: Error Estimation Differentials are also useful in measuring the error in measurements: Absolute or Propagated Error: Eabsolute   f  d f Relative Error: Erelative 

f d f  f f

Example 6: The radius of a sphere is given as 12.4  0.2cm find the maximum

absolute error in the volume. Vsphere  43  r 3 

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Example 7: The volume of a cube has an edge length of 2.0  0.005 ft . (a) Find the approximate absolute error in the volume. (b) Find the approximate value of the relative error.

Example 8: The inside radius of a beach ball is 26cm. If the material used to make the ball is 0.05cm thick, approximate the amount of material needed to produce 5000 beach balls. [ Vsphere  43  r 3 ]

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IMPORTANT NOTE: When dealing with angle measurements all angles must be expressed in RADIANS! Example 9: A surveyor stands 100m from the base of a building and measures the angle of elevation in order to approximate the height of the building. a) What is the formula for the height as a function of the angle? (Draw a diagram and include the given information) b) Suppose the angle is found to be 67o. If the surveyor believes that the error in the angle is 1.5o, what is the approximate error in the height calculation?

Additional Practice: 8th Ed: page 728, #5 - #20, #25 - #30 9th Ed: page 729, #5 - #20, #25 - #30

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MA2100

Unit 2 notes