MATH225 - Calculus Week 4 Test -Grade- 90% 2026 American Public University

MATH225 - Calculus Week 4 Test -Grade- 90% 2026 American Public University

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Questions 20 Questions

Limits 180 Minutes

Attempt 1

Points 100 pts possible

No due date.

90% (90 of 100)

Completed on 03/30/25 at 09:18PM

Score for this quiz: 90% ( 90 /100) Submitted Mar 30 at 9:18pm This attempt took 36 minutes.

Question 1 : 5 ptsSkip to question text.

Given the equation2x+4y=8,find dydx.Given the equation2x+4y=8,find dydx. dydx=−32dydx=−32 dydx=32dydx=32 dydx=−12dydx=−12 dydx=12dydx=12 2x + 4y = 8

Due Date

d

d

(8 d d ) 5/5 x x d d d Question 2 : 5 ptsSkip to question text.

2x 4y (8 Suppose a curve is defined by the equation(6−x)y 2=x3. What is the equation of theline tangent to the curv d + d = d ) e at (3, 3)? x x x Suppose a curve is defined by the equation(6−x)y2=x3. What is the equation of theline tangent to the curve at (3, 3)? d d d y=2x+3y=2x+3 (8 y=2x−3y=2x−3 x+ y 2d 4 d = d ) y=12x+32y=12x+32 x x x y=12x−32y=12x−32 d (6−x)y2=x36y2−xy2=x3ddx6y2−ddxxy2=ddxx312ydydx−2xydydx−y2=3x212ydydx−2xydydx=3x2+y2dy y 2+ dx(12y−2xy)=3x2+y2dydx=3x2+y212y−2xym=3(3)2+(3)212(3)−2(3) = 4 0 (3)=2y−y0=m(x−x0)y=2x−3(6−x)y2=x36y2−xy2=x3ddx6y2−ddxxy2=ddxx312ydydx−2xydydx−y2=3x21 d 2ydydx−2xydydx=3x2+y2dydx(12y−2xy)=3x2+y2dydx=3x2+y212y−2xym=3(3)2+(3)212(3)−2(3) x (3)=2y−y0=m(x−x0)y=2x−3 d 5/5 y = 4 −2 d Question 3 : 5 ptsUse the linear approximation formula to approximate e 2.1. x Use the linear approximation formula to approximate e 2.1. e2.1≈8.1661699e2.1≈8.1661699 d − − e2.1≈11e210e2.1≈11e210 y 2 1 e2.1≈e2e2.1≈e2 = = . d e2.1≈9e210e2.1≈9e210 4 2 x According to the linear approximation formula,f(x0+Δx)≈f′(x0)(Δx) +f(x0).To use the formula, you will need to know thefirst derivative.f′(x)=exTherefore:f(x0+Δx)≈f′(x0)(Δx) +f(x0)f(2+.1)≈f′(2)(.1)+f(2)f(2.1)≈e2(.1)+e2f(2.1)≈e2+10e210f(2.1)=e2.1≈11e210According to the linear approximation f ormula,f(x0+Δx)≈f′(x0)(Δx)+f(x0).To use the formula, you will need to know thefirst derivative.f′ (x)=exTherefore:f(x0+Δx)≈f′(x0)(Δx)+f(x0)f(2+.1)≈f′(2)(.1)+f(2)f(2.1)≈e2(.1)+e2f(2.1)≈e2+10e210f(2.1)=e2.1≈11e210 5/5 (2x + 4y) =

Question 4 : 5 ptsGiven cos (x + y) = sin x sin y, find dy / dx by implicit differentiation.

Given cos (x + y) = sin x sin y, find dy / dx by implicit differentiation. sin(x+y)+cosxsinysin(x+y)+sinxcosysin(x+y)+cosxsinysin(x+y)+sinxcosy −cos(x+y)+cosxsinycos(x+y)+sinxcosy−cos(x+y)+cosxsinycos(x+y)+sinxcosy −sin(x+y)+cosxsinysin(x+y)+sinxcosy−sin(x+y)+cosxsinysin(x+y)+sinxcosy sin(x+y)−cosxsinysin(x+y)−sinxcosysin(x+y)−cosxsinysin(x+y)−sinxcosy

−(1+dydx)sin(x+y)=cosxsiny+sinx(dydx)cosy−sin(x+y)−(dydx)sin(x+y)−(dydx)sinxcosy=cosxsiny(−sin(x+y) −sinxcosy)dydx=sin(x+y)+cosxsinydydx=−sin(x+y)+cosxsinysin(x+y) +sinxcosy−(1+dydx)sin(x+y)=cosxsiny+sinx(dydx)cosy−sin(x+y)−(dydx)sin(x+y)−(dydx)sinxcosy=cosxsiny(−sin(x+y) −sinxcosy)dydx=sin(x+y)+cosxsinydydx=−sin(x+y)+cosxsinysin(x+y)+sinxcosy 0/5 Question 5 : 5 ptsSkip to question text.

Given the equation f(x)=3x4, find f ′′(x).Given the equation f(x)=3x4, find f ′′(x). f ″(x) = 36x 2 f ″(x) = 36x 3 f ″(x) = 12x 2 f ″(x) = 12x 3

f (x) = 3x 4 f ′(x) = 3(4)x 4 − 1 f ′(x) = 12x 3 f ″(x) = 12(3)x 3 − 1 f ″(x) = 36x 2 5/5 Question 6 : 5 ptsWhat is the slope of the line tangent to the curve ( y − 1)x 2 = y + 1 at the point (0, −1)?

What is the slope of the line tangent to the curve ( y − 1)x 2 = y + 1 at the point (0, −1)?

0 1 The slope is undefined.

None of these

First, find the derivative dy/dx. Thenreplace x and y with their values.Take the derivative of both sides of the equatio n and solve for dy/dx.x2dydx+2x(y−1)=dydxx2dydx−dydx=−2x(y−1)(x2−1)dydx=−2x(y−1)dydx=−2x(y−1)x2−1At the point (0,−1), dydx∣∣∣(0,−1)=0−1=0.First, find the derivative dy/dx. Thenreplace x and y with their values.Take the deriva tive of both sides of the equation and solve for dy/dx.x2dydx+2x(y−1)=dydxx2dydx−dydx=−2x(y−1) (x2−1)dydx=−2x(y−1)dydx=−2x(y−1)x2−1At the point (0,−1), dydx|(0,−1)=0−1=0.

5/5

Question 7 : 5 ptsUse implicit differentiation to find an equation of the line tangent to the curve x 2 + y 2 = 10 at the point (3, −1).

Use implicit differentiation to find an equation of the line tangent to the curve x 2 + y 2 = 10 at the point (3, −1). y = −3x + 8 y = 3x − 8 y = −3x + 9 y = 3x − 10

First, find the derivative dydx. Then replace x and y with their values to findthe slope and write the equation of the t angent line. Take the derivative of both sides of the equation and solve for dydx.2x+2ydydx=0 2ydydx=−2x dydx=−xyThe slope at the point (3,−1) is dydx∣∣∣ (3,−1)=−3−1=3.The equation of the tangent line is as follows: y− (−1)=3(x−3) y=3x−10.First, find the derivative dydx. Then replace x and y with their values to findthe slope and write the equation of the tangent line. Take the derivative of both sides of the equation and solve for dydx.2x+2ydydx=0 2ydydx=−2x dydx=−xyThe slope at the point (3,−1) is dydx|(3,−1)=−3−1=3.The equation of the tangent line is as follows: y−(−1)=3(x−3) y=3x−10.

5/5

Question 8 : 5 ptsSkip to question text.

Given the equationxy=5,find dydx.Given the equationxy=5,find dydx. dydx=−yxdydx=−yx dydx=yxdydx=yx dydx=5−yxdydx=5−yx dydx=5+yxdydx=5+yx xy = 5 d d (xy) (5 d = d ) x x d d d (5 y+ x xd y d =d ) x x x d y + y x d = 0 x d y = x d −y x d − y =y .

d x x

5/5

Question 9 : 5 ptsSkip to question text.

A pebble is dropped into a small pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a rate of 1 foot per second. When this radius is 4 feet, at what rate is the total area of the disturbed water increasing?

16π ft2/sec 8π ft2/sec 16π ft/sec 8π ft/sec

Let r be the radius of the outermost circle.drdt=1Find dAdt when r=4. A=πr2dAdt=2πrdrdtdAdt=2π(4)(1).dAdt=8π f t2/secLet r be the radius of the outermost circle.drdt=1Find dAdt when r=4. A=πr2dAdt=2πrdrdtdAdt=2π(4)(1).dAdt=8π f t2/sec 5/5

Question 10 : 5 ptsA ladder 25 ft long is leaning against a vertical wall. If the base of the ladder is pulled horizontally away from the wall at 3 ft / sec, how fast is the top of the ladder sliding down the wall when the base is 15 ft from the wall?

A ladder 25 ft long is leaning against a vertical wall. If the base of the ladder is pulled horizontally away from the wall at 3 ft / sec, how fast is the top of the ladder sliding down the wall when the base is 15 ft from the wall? −4/9 ft/sec −9/4 ft/sec −9/10 ft/sec −10/9 ft/sec

Let x be the distance up the wall.Let y be the distance from the wall.dydx=3Find dx/dt when y=15.x2+y2=252x2+152=252x2=400x=20x2+y2=2522xdxdt+2ydydt=02(20)dxdt+2(15)(3)=0dxdt=−2(15) (3)2(20)=−94 ft/secLet x be the distance up the wall.Let y be the distance from the wall.dydx=3Find dx/dt when y=15.x2 +y2=252x2+152=252x2=400x=20x2+y2=2522xdxdt+2ydydt=02(20)dxdt+2(15)(3)=0dxdt=−2(15)(3)2(20)=−94 ft/sec 5/5 Question 11 : 5 ptsUse the linear approximation formula to approximate

27.1−−−−√3.27.13.

Use the linear approximation formula to approximate 27.1−−−−√3.27.13.

27.1−−−−√3≈80927027.13≈809270 27.1−−−−√3≈327.13≈3 27.1−−−−√3≈81127027.13≈811270 27.1−−−−√3≈3.003699127.13≈3.0036991

According to the linear approximation formula, f (x0 + Δx) ≈ f ′(x0)(Δx) + f (x0).

To use the formula, you will need to know the first derivative. f ′(x) = x (−2/3) /3 Therefore: f(x0+Δx)≈f′(x0)(Δx)+f(x0)f(27+.1)≈f′(27)(.1)+f(27)f(27.1)≈13⋅9(.1)+3f(27.1)≈1270+81027027.1−−− −√3≈811270f(x0+Δx)≈f′(x0)(Δx)+f(x0)f(27+.1)≈f′(27)(.1)+f(27)f(27.1)≈13⋅9(.1)+3f(27.1)≈1270+81027027.13≈811270 5/5 Question 12 : 5 ptsSkip to question text.

Use the linear approximation formula toapproximate 16.2−−−−√.Use the linear approximation formula toapproximate 16.2.

16.2−−−−√≈1594016.2≈15940 16.2−−−−√≈416.2≈4 16.2−−−−√≈1614016.2≈16140 16.2−−−−√≈4.024922416.2≈4.0249224 According to the linear approximation formula,f(x0+Δx)≈f′(x0)(Δx)+f(x0).According to the linear approximation formula, f(x0+Δx)≈f′(x0)(Δx)+f(x0).

Find the derivative and plug into the formula. f(x0+Δx)≈f′(x)(Δx)+f(x0)f(16+.2)≈f′(16)(.2)+f(16)f(16.2)≈1216−−√(.2)+4f(16.2)≈280+4f(16.2)≈16140f(x0+Δx)≈f′(x) (Δx)+f(x0)f(16+.2)≈f′(16)(.2)+f(16)f(16.2)≈1216(.2)+4f(16.2)≈280+4f(16.2)≈16140 5/5 Question 13 : 5 ptsWhat is the equation of the line tangent to the curve x 2 + y 2 = 25 at the point (0, −5)?

What is the equation of the line tangent to the curve x 2 + y 2 = 25 at the point (0, −5)? x=5 x = −5 y=5 y = −5

Start by finding the derivative implicitly.x 2+y2=25ddxx2+ddxy2=ddx252x+2ydydx=0dydx=−xyPlug in the point you n eed.dydx=−0−5=0Plug the slope and point into the point-slope form of a line.y+5=0(x−0)y=−5So that equation is yo ur tangent line.Start by finding the derivative implicitly.x2+y2=25ddxx2+ddxy2=ddx252x+2ydydx=0dydx=−xyPlug in th e point you need.dydx=−0−5=0Plug the slope and point into the point-slope form of a line.y+5=0(x−0)y=−5So that equat ion is your tangent line.

5/5

Question 14 : 5 ptsSkip to question text.

Given y=3x2, find dydx.Given y=3x2, find dydx. dydx=3dydx=3 dydx=3xdydx=3x dydx=2xdydx=2x dydx=6xdydx=6x y=3xddx(y)=ddx(3x2)dydx=3[ddx(x2)]dydx=3(2x)dydx=6xy=3xddx(y)=ddx(3x2)dydx=3[ddx(x2)]dydx=3(2x)dydx=6x 5/5 Question 15 : 5 ptsSkip to question text.

A rope is attached to a boat at water level and a woman on a dock is pulling on the rope at the rate of 50 ft / min. If the woman’s hands are 16 ft above the water level, how fast is the boat approaching the dock when the amount of rope out is 20 ft?

Your Answer: 83.33 ft/min Correct Answer(s):

0/5 Question 16 : 5 ptsSkip to question text.

Given the equationx2+y2=9,find dydx.Given the equationx2+y2=9,find dydx. dydx=xydydx=xy

dydx=−xydydx=−xy dydx=yxdydx=yx dydx=−yxdydx=−yx x2 + y2 = 9 5/5 d d (x 2 + y (9 d 2 ) = d ) Question 17 : 5 ptsSkip to question text. x x Given sin3(cosxy)=kwhere k is some constant,find dydx.Given sin3(cosxy)=kwhere k is some constant,find dydx. d d d x2

y2

(9

d + d = d ) x x x d y 2x + = 2y ⋅ 0 d x d y 2y = ⋅ −2x d x

dydx=yxdydx=yx dydx=−yxdydx=−yx dydx=xydydx=xy dydx=−xydydx=−xy None of these

sin3(cosxy)=kddxsin3(cosxy)=ddx(k)3sin2(cosxy)ddx(sin(cosxy))=03sin2(cosxy)cos(cosxy)ddx(cosxy)=0 3sin2(cosxy)cos(cosxy) (−sinxy)ddx(xy)=0−3sin2(cosxy)cos(cosxy)sinxy(xddxy+yddxx)=0−3sin2(cosxy)cos(cosxy)sinxy(xdydx+ y)=0−3xsin2(cosxy)cos(cosxy)sinxydydx=3ysin2(cosxy)cos(cosxy)sinxydydx=3ysin2(cosxy)cos(cosxy)sin xy−3xsin2(cosxy)cos(cosxy)sinxy=−yxsin3(cosxy)=kddxsin3(cosxy)=ddx(k)3sin2(cosxy)ddx(sin(cosxy))=0 3sin2(cosxy)cos(cosxy)ddx(cosxy)=03sin2(cosxy)cos(cosxy)

d −2 − y x x = = . d 2y y x (−sinxy)ddx(xy)=0−3sin2(cosxy)cos(cosxy)sinxy(xddxy+yddxx)=0−3sin2(cosxy)cos(cosxy)sinxy(xdydx+y)=0−3xsin2(cosxy )cos(cosxy)sinxydydx=3ysin2(cosxy)cos(cosxy)sinxydydx=3ysin2(cosxy)cos(cosxy)sinxy−3xsin2(cosxy)cos(cosxy)sinxy=−y x 5/5 Question 18 : 5 ptsSkip to question text.

Given y=x4, find dydx.Given y=x4, find dydx. dydx=14dydx=14 dydx=4dydx=4 dydx=d(14)dydx=d(14)

dydx=x4dydx=x4

Since y=x4, we can find dydx by taking thederivative of both sides with respect to x.y=x4ddx(y)=ddx(x4)dydx=14⋅dd x(x)dydx=14⋅d(x)dx=14Since y=x4, we can find dydx by taking thederivative of both sides with respect to x.y=x4ddx(y)= ddx(x4)dydx=14⋅ddx(x)dydx=14⋅d(x)dx=14 5/5

Question 19 : 5 ptsUse the linear approximation method from calculus to approximate the square root.

3.9−−−√3.9

Use the linear approximation method from calculus to approximate the square root.

3.9−−−√3.9 3.9−−−√≈81403.9≈8140 3.9−−−√≈43.9≈4 3.9−−−√≈79403.9≈7940 3.9−−−√≈1.974843.9≈1.97484

Let f(x)=x−−√.To find the approximate value of f(3.9),use the tangent line to the graph at f(4)and find its value at 3.9 . Find the first derivative of f(x).f(x)=x−−√⇒f′(x)=12x−−√Evaluate f′(4).f′(4)=124–√=14Since f′(4)=14 and f(4)=2, the equation of the tangent line is given byy−2=14(x−4).Simplify.y=14x−1+2y=14x+1Evaluate the function for x=3.9.y =14(3.9)+1=3.94+1=3940+4040=7940Therefore, 3.9−−−√≈7940. Let f(x)=x.To find the approximate value of f(3.9),u se the tangent line to the graph at f(4)and find its value at 3.9. Find the first derivative of f(x).f(x)=x⇒f′(x)=12xEvaluate f′( 4).f′(4)=124=14Since f′(4)=14 and f(4)=2, the equation of the tangent line is given byy−2=14(x−4).Simplify.y=14x−1+2y =14x+1Evaluate the function for x=3.9.y=14(3.9)+1=3.94+1=3940+4040=7940Therefore, 3.9≈7940.

5/5 Question 20 : 5 ptsSkip to question text.

Suppose x2+y2=16.Find d2ydx2.Suppose x2+y2=16.Find d2ydx2. d2ydx2=4d2ydx2=4 d2ydx2=−xyd2ydx2=−xy d2ydx2=16y3d2ydx2=16y3 d2ydx2=−16y3d2ydx2=−16y3 x2+y2=16ddxx2+ddxy2=ddx162x+2ydydx=0dydx=−xyddx(dydx)=ddx(−xy)d2ydx2=−yddx(x) −xddxyy2=−y−xdydxy2=−y−x(−xy)y2=−y2+x2yy2=−x2+y2y3=−16y3x2+y2=16ddxx2+ddxy2=ddx162x+2ydydx=0dydx =−xyddx(dydx)=ddx(−xy)d2ydx2=−yddx(x)−xddxyy2=−y−xdydxy2=−y−x(−xy)y2=−y2+x2yy2=−x2+y2y3=−16y3 5/5

Quiz Score: 90 out of 100