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Math - High School, section 1, Fall 2019
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Sullivan et al - Calculus for AP Course 2/e (Homework)

James Finch

Math - High School, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 11 / 44

Due : Thursday, August 29, 2019 19:00 EDT

Last Saved : n/a Saving... ()

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Question

Points

1	2	3	4	5	6	7	8	9	10	11	12
–/10	–/1	–/5	–/7	–/1	–/2	–/1	–/1	–/8	–/2	–/5	–/1

Total

11/44 (25.0%)

Instructions

WebAssign Premium for Calculus for the AP Course, 2nd edition, by Michael Sullivan and Kathleen Miranda and published by Bedford, Freeman, & Worth is a flexible suite of resources designed for this textbook and for your AP® Calculus course. Combining the most widely used online homework platform with interactive content from the textbook, WebAssign Premium enhances the classroom experience for teachers and extends students’ opportunities to engage in the study of calculus.

Question 1 is a multi-part question which features grading for exact answers to find and compare slopes of secant and tangent lines for randomized values.

Question 2 exhibits implicit differentiation grading that accepts any form of the correct answer, such as using the given equation (9x² + 2y)³ = y to rewrite the expression for dy/dx.

Question 3 shows expression grading where any equivalent form of the expression is accepted while finding the derivative of a function. The second part of question grades all solutions written as a list. The prompt alerts the student to enter a comma-separated list of answers.

Question 4 highlights writing solutions in interval notation to describe motion of an object along a horizontal line requiring use of the infinity symbol or prompt to enter DNE. Either the keyboard or the calcPad can be used to enter the answer.

Question 5 demonstrates indefinite integral grading that enforces the use of C and proper use of trigonometric functions, while still allowing for any equivalent form.

Question 6 features a step-by-step tutorial that walks the student through using the washer method to find the volume of revolution of a region. It utilizes special trigonometric grading that forces students to enter simplified trigonometric functions.

Question 7 uses series grading, which allows any correct version of the power series based on the summation index.

Question 8 allows the student to enter any correct version of the parametric equations for an object that moves along an elliptical curve.

Question 9 illustrates vector grading.

Question 10 contains standard equation grading that accepts any form of the general equation of a plane, including 3D visualizations.

Question 11 highlights an AP® Practice Problem that consists of multiple parts for finding the derivative of a function and the equation of a tangent line, requiring expression grading and ordered pair format.

Question 12 demonstrates an AP® Practice Problem of multiple choice selection for applying the product rule to find the deriviative. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. –/10 points SullivanCalc2HS 1.1.051. My Notes

For

f(x) = 2x²,

do the following.

(a) Find the slope of the secant line containing the points

(3, 18)

and

(4, 32).

(No Response) $webMathematica generated answer key$

(b) Find the slope of the secant line containing the points

(3, 18)

and

(x, f(x)),

x ≠ 3.

(No Response) $webMathematica generated answer key$

(c) Create a table to investigate the slope of the tangent line to the graph of f at 3 using the result from (b).

x	2.9	2.99	2.999	→ 3 ←	3.001	3.01	3.1
m_sec	(No Response) 11.8	(No Response) 11.98	(No Response) 11.998	(No Response) 12	(No Response) 12.002	(No Response) 12.02	(No Response) 12.2

(d) On the same set of axes, graph f, the tangent line to the graph of f at the point

(3, 18),

and the secant line from (a).

Solution or Explanation

(a) The secant line containing the points

(3, 18)

and

(4, 32)

has a slope of

m_sec =

32 − 18

4 − 3

= 14.

(b) The secant line containing the points

(3, 18)

and

(x, f(x))

for

x ≠ 3

has slope of

m_sec =

2x² − 18

x − 3

2(x − 3)(x + 3)

x − 3

= 2(x + 3).

f(x)

at 3 is

lim x → 3 m_sec = 12.

x	2.9	2.99	2.999	→ 3 ←	3.001	3.01	3.1
m_sec	11.8	11.98	11.998	m_sec approaches 12	12.002	12.02	12.2

(d) The secant line from part (a) has slope 14 and passes through the point

(3, 18).

The equation of this secant line is, therefore,

y − 18 = 14(x − 3)

y = 14x − 24.

The tangent line at

x = 3

has slope 12 and also passes through the point

(3, 18);

the equation of this line is

y − 18 = 12(x − 3)

y = 12x − 18.

The figure below displays the graph of f as the solid curve, the graph of the tangent line as the dashed curve, and the graph of the secant line as the dotted curve.

Viewing Saved Work Revert to Last Response

2. –/1 points SullivanCalc2HS 3.2.023. My Notes

Find

y' =

using implicit differentiation.

(4x² + 5y)³ = y

= (No Response) $webMathematica generated answer key$

Solution or Explanation

Assume that y is a differentiable function of x and differentiate both sides of the given equation with respect to x. This yields

((4x² + 5y)³)

3(4x² + 5y)²

(4x² + 5y)

3(4x² + 5y)²

(4x²) +

(5y)

3(4x² + 5y)²

8x + 5

(15(4x² + 5y)² − 1)

−24x(4x² + 5y)²

−

24x(4x² + 5y)²

15(4x² + 5y)² − 1

Viewing Saved Work Revert to Last Response

3. –/5 points SullivanCalc2HS 4.2.065. My Notes

For the function f, do the following.

f(x) = 3x⁴ − 6x³ − 60x² + 126x

(a) Find the derivative

f '.

f '(x) = (No Response) $webMathematica generated answer key$

(b) Use technology to find the critical numbers of f. (Enter your answers as a comma-separated list. A computer algebra system is recommended.)

x = (No Response) $webMathematica generated answer key$

Describe the behavior of f suggested by the graph at each critical number. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

The graph suggests that f has local minima at

x = (No Response) $webMathematica generated answer key$

and local maxima at

x = (No Response) $webMathematica generated answer key$ .

Solution or Explanation

Let

f(x) = 3x⁴ − 6x³ − 60x² + 126x.

(a) Then

f '(x) = 12x³ − 18x² − 120x + 126.

(b) The polynomial function f is differentiable everywhere, so the critical numbers of f occur where

f '(x) = 0.

Using technology,

12x³ − 18x² − 120x + 126 = 0

when

x = −3,

x = 1,

and

x =

Therefore,

−3,

1, and

are the critical numbers of f.

(c) The figure below displays the graph of f. The graph suggests that f has a local minimum at

−3,

a local maximum at 1, and a local minimum at

Viewing Saved Work Revert to Last Response

4. –/7 points SullivanCalc2HS 4.4.027. My Notes

The distance s of an object from the origin at time

t ≥ 0

(in seconds) is given. The motion is along a horizontal line with the positive direction to the right.

s(t) = t² − 2t + 3

(a) Determine the intervals during which the object moves to the right and the intervals during which it moves to the left. (Enter your answer using interval notation. If an answer does not exist, enter DNE.)

moving to the right		(No Response) $webMathematica generated answer key$
moving to the left		(No Response) $webMathematica generated answer key$

(b) When does the object reverse direction? (If an answer does not exist, enter DNE. Enter your answers as a comma-separated list.)

t = (No Response) $webMathematica generated answer key$

(c) When is the velocity of the object increasing and when is it decreasing? (Enter your answer using interval notation. If an answer does not exist, enter DNE.)

increasing		(No Response) $webMathematica generated answer key$
decreasing		(No Response) $webMathematica generated answer key$

(d) Draw a figure to illustrate the motion of the object.

(e) Draw a figure to illustrate the velocity of the object.

Solution or Explanation

Let

s(t) = t² − t + 3.

(a) The object moves to the right when

v(t) = s'(t) > 0

and moves to the left when

v(t) = s'(t) < 0.

Now,

v(t) = s'(t) = 2t − 2 = 2(t − 1)

so 1 is the only critical number of

s(t).

When

t < 1, v(t) < 0.

Therefore, the object moves to the right on the interval

(1, ∞)

and moves to the left on the interval

(0, 1).

(b) The object reverses direction at

t = 1.

a(t) = v'(t) > 0

and is decreasing when

a(t) = v'(t) < 0.

Now

a(t) = v'(t) = 2 > 0

for all

t ≥ 0.

Therefore, the velocity is increasing on the interval

(0, ∞).

(d) The figure below illustrates the motion of the object.

(e) The figure below illustrates the velocity of the object.

Viewing Saved Work Revert to Last Response

5. –/1 points SullivanCalc2HS 5.5.045. My Notes

Find the indefinite integral. (Use C for the constant of integration.)

tan(7x)

sec²(7x) dx

(No Response) $webMathematica generated answer key$

Solution or Explanation

Let

u = tan(7x),

then

du = 7 sec²(7x) dx,

(sec(7x))² dx =

We substitute and obtain

tan(7x)

sec²(7x) dx

4u^1/2

u^1/2 du

u^3/2 + C

(tan(7x))^3/2 + C.

Viewing Saved Work Revert to Last Response

6. –/2 points SullivanCalc2HS 6.2.017-022a.XP.Tut. My Notes

Consider the region bounded by the y-axis and the graphs of the functions

y = 9 sin(x)

and

y = 9 cos(x)

(a) Using the washer method, find the volume of revolution of this region about the x-axis. (Use a trigonometric identity.)
(No Response) $webMathematica generated answer key$ units³

(b) What difficulties are presented with attempting to find the volume of revolution about the y-axis?

One cannot write down an explicit integral that represents this volume, because it is impossible to write the relevant curves as a function of y on the given interval. The volume is given by

81π

π/4

cos²(x) − sin²(x)

dx,

and this integral is difficult to integrate. correct

The volume is given by

9√2/2

sin⁻¹

	2

+ π

cos⁻¹

	2

9√2/2

and these integrals are difficult to evaluate explicitly. The volume is given by

9√2/2

sin⁻¹(y)

	2

cos⁻¹(y)

	2

9√2/2

and these integrals are difficult to evaluate explicitly. The volume is given by

81π

π/4

cos⁻¹(y) + sin⁻¹(y)

dy,

and this integral is difficult to integrate.

Solution or Explanation

y = 9 sin(x) and y = 9 cos(x) on

(a)

outer radius f(x) = 9 cos(x); inner radius g(x) = 9 sin(x)

Volume V

π/4	(9 cos(x))² − (9 sin(x))² dx

0

81π

π/4

cos²(x) − sin²(x)

81π

π/4

cos(2x) dx

81π

sin(2x)

	π/4
	0

81π

− 0

81π

units³

(b)

y = 9 cos(x) → x = cos⁻¹

; y = 9 sin(x) → x = sin⁻¹

We need to break this up into 2 integrals.

$V = \underbrace{\pi \int_{0}^{\frac{{\color{red}9}\sqrt{2}}{2}}\left(\sin^{-1}\left(\frac{y}{{\color{red}9}}\right)\right)^2\,dy}_\text{rotate region (1)} + \underbrace{\pi \int_{\frac{{\color{red}9}\sqrt{2}}{2}}^{{\color{red}9}}\left(\cos^{-1}\left(\frac{y}{{\color{red}9}}\right)\right)^2\,dy}_\text{rotate region (2)}$
Note: Difficult to integrate!

Additional Materials

Tutorial

Viewing Saved Work Revert to Last Response

7. –/1 points SullivanCalc2HS 8.9.012. My Notes

Assuming the function can be represented by a power series, find the Maclaurin expansion of the function.

f(x) = cos(5x)

∞

k = 0

(No Response) $webMathematica generated answer key$

Solution or Explanation

To express

f(x) = cos(5x)

as a Maclaurin series, we begin by evaluating

f(x)

and its derivatives at

x = 0.

f(x)	=	cos(5x)	f(0)	=	1
f '(x)	=	−5 sin(5x)	f '(0)	=	0
f ''(x)	=	−5² cos(5x)	f ''(0)	=	−5²
f '''(x)	=	5³ sin(5x)	f '''(0)	=	0
f ⁽⁴⁾(x)	=	5⁴ cos(5x)	f ⁽⁴⁾(0)	=	5⁴

For derivatives of even order,

f ⁽²ⁿ⁾(0) = (−1)ⁿ5²ⁿ.

For derivatives of odd order,

f ^{(2n + 1)}(0) = 0.

So the Maclaurin series is

f(x)

∞

k = 0

f ^(k)(0)

x^k

1 −

5²x²

5⁴x⁴

−

(−1)ⁿ5²ⁿx²ⁿ

2n!

∞

k = 0

(−1)^k(5)^2kx^2k

(2k)!

Viewing Saved Work Revert to Last Response

8. –/1 points SullivanCalc2HS 9.1.055. My Notes

Find parametric equations for an object that moves along the ellipse

x²

y²

= 1,

where the parameter is time (in seconds) if the following is true. (Enter your answers as a comma-separated list of equations. Let

0 ≤ t ≤ 5.)

The motion begins at

(3, 0),

is counterclockwise, and requires 5 seconds for 1 revolution.

(No Response) $webMathematica generated answer key$

Solution or Explanation

For counterclockwise orientation, we choose

x(t) = 3 cos(ωt),

y(t) = 4 sin(ωt).

If 1 revolution takes 5 seconds, the period is

2π

= 5,

ω =

2π

The parametrization is

x(t) = 3 cos

2π

y(t) = 4 sin

2π

To begin at

(3, 0)

the interval on t would be

0 ≤ t ≤ 5.

Viewing Saved Work Revert to Last Response

9. –/8 points SullivanCalc2HS 10.3.091. My Notes

The points

A = (−7, 0),

B = (−1, −3),

C = (4, 1),

and

D = (1, 9)

are the vertices of a parallelogram ABCD.

(a) If ABCD is translated by

v =

5, 4

find the vertices of the new parallelogram

A'B'C'D'.

A'	=	(No Response) $webMathematica generated answer key$
B'	=	(No Response) $webMathematica generated answer key$
C'	=	(No Response) $webMathematica generated answer key$
D'	=	(No Response) $webMathematica generated answer key$

(b) If ABCD is translated by

find the vertices of the new parallelogram

A'B'C'D'.

A'	=	(No Response) $webMathematica generated answer key$
B'	=	(No Response) $webMathematica generated answer key$
C'	=	(No Response) $webMathematica generated answer key$
D'	=	(No Response) $webMathematica generated answer key$

Solution or Explanation

(a) The new vertices are

−7, 0

5, 4

−2, 4

−1, −3

5, 4

4, 1

5, 4

9, 5

1, 9

5, 4

6, 13

(b) Since

v =

, 2

the new vertices are

−7, 0

, 2

−

, 2

−1, −3

, 2

, −1

4, 1

, 2

, 3

1, 9

, 2

, 11

Viewing Saved Work Revert to Last Response

10. –/2 points SullivanCalc2HS 10.6.041. My Notes

Consider the following.

P₁ = (0, 0, 0); P₂ = (1, 7, −1); P₃ = (−1, 1, 0).

(a) Find the general equation of the plane containing the points

P₁,

P₂,

and

P₃.

(No Response) $webMathematica generated answer key$

(b) Graph the plane. (A computer algebra system is recommended.)

Solution or Explanation

(a) The vectors

v = P₁P₂

= i + 7j − k

and

w = P₁P₃

= −i + j

both lie on the plane, so that a normal N to the plane is

N = v × w =

i	j	k
1	7	−1
−1	1	0

= i

7	−1
1	0

− j

1	−1
−1	0

+ k

1	7
−1	1

= i + j + 8k.

Using the point

(0, 0, 0)

and the normal vector N, the general equation of the plane is

1(x − 0) + 1(y − 0) + 8(z − 0) = 0, or x + y + 8z = 0.

(b)

Viewing Saved Work Revert to Last Response

11. –/5 points SullivanCalc2HS 2.3.AP.010. My Notes

For the function

f(x) = x² + 6,

find the following.

(a)

Find

f '(1).

(No Response)

(b)

Find an equation of the tangent line to the graph of f at

x = 1.

y = (No Response) $webMathematica generated answer key$

(c)

Find

f '(−6).

(No Response)

-12

(d)

Find an equation of the tangent line to the graph of f at

x = −6.

y = (No Response) $webMathematica generated answer key$

(e)

Find the point of intersection of the two tangent lines found in (b) and (d).

(x, y) =

(No Response) $webMathematica generated answer key$

Solution or Explanation

(a)

f '(x)

(x² + 6) = 2x

f '(1)

2(1) = 2

(b)

f(1) = (1)² + 6 = 7

The point of tangency is (1, 7).

y − 7	=	2(x − 1)
y	=	2x − 2 + 7
	=	2x + 5

(c)

f '(−6) = 2(−6) = −12

(d)

f(−6) = (−6)² + 6 = 42

The point of tangency is

(−6, 42).

y − 42	=	−12(x + 6)
y	=	−12x − 72 + 42
	=	−12x − 30

(e)

Solve the system

y	=	2x + 5
y	=	−12x − 30

2x + 5

−12x − 30

14x

−35

= −

−

+ 5 = 0

The intersection point is

−

, 0

Viewing Saved Work Revert to Last Response

12. –/1 points SullivanCalc2HS 2.4.AP.004. My Notes

y = x⁷e^x,

then what does

equal?

7x⁶ + e^x

7x⁶e^x

x⁶e^x(x + 7)

7x⁶e^x(x + 1)

Solution or Explanation

(x⁷e^x) =

(x⁷)

e^x +

e^x

x⁷ = 7x⁶e^x + e^xx⁷ = x⁶e^x(7 + x) = x⁶e^x(x + 7)

Viewing Saved Work Revert to Last Response

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