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Larson & Battaglia - Calculus for AP 2nd edition (Homework)

James Finch

Math - High School, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 6 / 18

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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Question
Points
1 2 3 4 5 6 7 8 9 10
0/1 1/2 1/1 1/1 0/1 2/5 0/1 0/1 1/1 0/4
Total
6/18 (33.3%)

  • Instructions

    Calculus for AP 2nd edition (Cengage Learning) by Ron Larson and Paul Battaglia is designed specifically for the AP Curriculum Framework and exam. Ron Larson and co-author Paul Battaglia developed an updated program that meets the current needs of the AP Calculus course. Exercises connect to the AP Calculus Exam and will help students increase their conceptual understanding of mathematics. The WebAssign component for this title features Fast Track to a 5: AP Test Preparation questions, video links, tutorial questions, worked-out solutions, and links to a complete eBook.

    Question 1 is a sample multiple-choice question from the Fast Track to a 5 workbook that provides preparation for the AP exam.

    Question 2 accepts equivalent answers in any form and includes randomized graphs based on the problem given.

    Question 3 illustrates the prompt given when an answer does not exist and demonstrates the corresponding grading.

    Question 4 includes an embedded video.

    Question 5 accepts equivalent equations in any form. Additionally, there is no set order in which the equations must be entered.

    Question 6 is an Advanced Placement question intended to prepare students for the Advanced Placement test. It demonstrates interval grading, which can grade any canonically equivalent interval and enforce proper notation.

    Question 7 grades the integral and enforces correct form and notation while also allowing all mathematically correct answers. This question also includes a Master It tutorial.

    Question 8 uses differential equation grading to test the validity of the answer. It accepts any correct form of the answer and runs student's responses through a series of tests to ensure the assumptions and requirements of the question are met. This questions also features a Watch It video

    Question 9 is a Just In Time question intended to address pre-requisite knowledge for this particular section.

    Question 10 is an Advanced Placement question designed to prepare students for the Advanced Placement test and includes a solution. Part (d) illustrates vector grading. 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.
    All questions selected for this Example Assignment have worked out solutions. 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.

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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.

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Your last submission is used for your score.

1. 0/1 points  |  Previous Answers LarHSCalc2 FT5.1.004. My Notes
Question Part
Points
Submissions Used
1
0/1
7/50
Total
0/1
 
Evaluate the limit, if it exists.
lim x→1 
tan−1(x)
sin−1(x) + 1
     Incorrect: Your answer is incorrect.


Solution or Explanation
(D). Evaluate the limit numerically (direct substitution).
lim x→1 
tan−1(x)
sin−1(x) + 1
 = 
π
4
π
2
 + 1
 = 
π
4
π + 2
2
 = 
π
4
2
π + 2
 = 
π
2π + 4
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2. 1/2 points  |  Previous Answers LarHSCalc2 2.2.062. My Notes
Question Part
Points
Submissions Used
1 2
0/1 1/1
5/50 7/50
Total
1/2
 
Consider the following.
Function Point
h(t) = sin t + 
1
9
et
 		    
π, 
1
9
eπ
 
(a) Find an equation of the tangent line to the graph of h at the given point.
y =
19​eπ+(−1+19​eπ)(x−π)
Incorrect: Your answer is incorrect. webMathematica generated answer key


(b) Use a graphing utility to graph the function and its tangent line at the point.

Correct: Your answer is correct.



Solution or Explanation
(a)    h(t) = sin t + 
1
9
et
h'(t) = cos t + 
1
9
et
At
π, 
1
9
eπ
: h'(π) = −1 + 
1
9
eπ
Tangent line:
y − 
1
9
eπ		 = 
−1 + 
1
9
eπ
(t − π)
y = 
−1 + 
1
9
eπ
t + 
1
9
eπ + π − 
1
9
πeπ


(b)     WebAssign Plot
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3. 1/1 points  |  Previous Answers LarHSCalc2 2.2.068. My Notes
Question Part
Points
Submissions Used
1
1/1
3/50
Total
1/1
 
Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line. (If the function has no horizontal tangent line, enter NONE.)
y = x + 6ex
(x, y) = 
NONE
Correct: Your answer is correct. webMathematica generated answer key


Solution or Explanation
y = x + 6ex
y' = 1 + 6ex cannot equal 0.
So, there are no horizontal tangents.
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4. 1/1 points  |  Previous Answers LarHSCalc2 2.2.VE.001. My Notes
Question Part
Points
Submissions Used
1
1/1
3/50
Total
1/1
 
Watch the video below then answer the question.

d
dx
 cos(x) = −sin(x).

     Correct: Your answer is correct.


Solution or Explanation
True
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5. 0/1 points  |  Previous Answers LarHSCalc2 2.5.083. My Notes
Question Part
Points
Submissions Used
1
0/1
2/50
Total
0/1
 
Find equations of both tangent lines to the graph of the ellipse
x2
36
 + 
y2
49
 = 1
 that pass through the point
(36, 0)
 not on the graph. (Enter your answers as a comma-separated list of equations.)
x+y=2, x=4
Incorrect: Your answer is incorrect. webMathematica generated answer key

Solution or Explanation
x2
36
 + 
y2
49
 = 1    (36, 0)
2x
36
 + 
2yy'
49
 = 0
y' = 
−49x
36y
−49x
36y
 = 
y − 0
x − 36
−49x(x − 36) = 36y2

But,
49x2 + 36y2 = 1764 right double arrow implies 36y2 = 1764 − 49x2. So −49x2 + 1764x = 36y2 = 1764 − 49x2 right double arrow implies x = 1.

Points on ellipse:
1, ±
7
6
35
At
1, 
7
6
35
: y' = 
−49x
36y
 = 
−49
36
7
6
35
 = − 
35
30
At
1, − 
7
6
35
: y' = 
35
30

Tangent lines:
y = − 
35
30
(x − 36) = − 
35
30
x +
6
5
35
y = 
35
30
(x − 36) = 
35
30
x −
6
5
35
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6. 2/5 points  |  Previous Answers LarHSCalc2 2.AP.016. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5
1/1 0/1 0/1 0/1 1/1
5/50 2/50 2/50 2/50 3/50
Total
2/5
 
Do not use technology.

The figure below shows the graph of the velocity, in feet per second, for a particle moving along the line
x = 4.
(a) During which time interval(s) is the particle doing the following? Enter your answer using interval notation.
(i) moving upward
((0,1)∪(225​,5))
Correct: Your answer is correct. webMathematica generated answer key

(ii) moving downward
(2, 2255​)
Incorrect: Your answer is incorrect. webMathematica generated answer key

(iii) at rest
(1,22)
Incorrect: Your answer is incorrect. webMathematica generated answer key

(b) What is the acceleration of the particle at the following times?
(i)
t = 0.61
Incorrect: Your answer is incorrect. seenKey

-2

 ft/s2

(ii)
t = 4.2
Correct: Your answer is correct. seenKey

5

 ft/s2


Solution or Explanation
(a)
(i) Because
v(t) > 0
 when
0 < t < 1
 and
4.4 < t < 5,
 the particle is moving upward on the intervals (0, 1) and (4.4, 5).

(ii) Because
v(t) < 0
 when
2 < t < 4.4,
 the particle is moving downward on the interval (2, 4.4).

(iii) Because
v(t) = 0
 when
1 < t < 2,
 the particle is at rest on the interval (1, 2).

(b)
(i) When
t = 0.61,
 the slope of the line of
v(t) = −2.
 So, the acceleration of the particle is −2 feet per second squared.

(ii) When
t = 4.2,
 the slope of the line of
v(t) = 5.
 So, the acceleration of the particle is 5 feet per second squared.
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7. 0/1 points  |  Previous Answers LarHSCalc2 4.7.060.MI. My Notes
Question Part
Points
Submissions Used
1
0/1
1/50
Total
0/1
 
Evaluate the definite integral. Use a graphing utility to verify your result.
π/4
π/8
(8 csc(2θ) − 8 cot(2θ)) dθ
woohoo
Incorrect: Your answer is incorrect. webMathematica generated answer key

Solution or Explanation
u = 2θ,
du = 2 dθ,
θ = 
π
8
 right double arrow implies u = 
π
4
,
θ = 
π
4
 right double arrow implies u = 
π
2


π/4
π/8
(8 csc(2θ) − 8 cot(2θ)) dθ		 = 4
π/2
π/4
(csc(u) − cot(u)) du
 = 4
−ln|csc(u) + cot(u)| − ln|sin(u)|
π/2
π/4
 
 = 4
−ln(1 + 0) − ln(1) + ln
2
 + 1
 + ln
2
2
 
 = 4
ln
2
 + 1
 + ln
2
2
 
 = 4 ln
1 + 
2
2

Need Help? Master It

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8. 0/1 points  |  Previous Answers LarHSCalc2 5.3.008. My Notes
Question Part
Points
Submissions Used
1
0/1
1/50
Total
0/1
 
Find the general solution of the differential equation. (Enter your solution as an equation.)
xy' = 14y
7
Incorrect: Your answer is incorrect. webMathematica generated answer key

Solution or Explanation
xy' = 14y
dy
y
 = 
14dx
x
ln(y) = 14ln(x) + ln(C) = ln(Cx14)
y = Cx14
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9. 1/1 points  |  Previous Answers LarHSCalc2 5.4.JIT.002. My Notes
Question Part
Points
Submissions Used
1
1/1
2/50
Total
1/1
 
Find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
 
x2 − 64
x
 dx
C−ln(x64)+x22​
Correct: Your answer is correct. webMathematica generated answer key

Solution or Explanation
x2 − 64
x
 dx		 = 
x − 
64
x
 dx
 = 
x2
2
 − 64 ln(|x|) + C
 = 
x2
2
 − ln(x64) + C
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10. 0/4 points  |  Previous Answers LarHSCalc2 9.AP.008. My Notes
Question Part
Points
Submissions Used
1 2 3 4
0/1 0/1 0/1 0/1
1/50 1/50 1/50 1/50
Total
0/4
 
A particle moving along a curve in the xy-plane is at position
(x(t), y(t))
 at time t, where
dx
dt
 = e−t2 + 1 − 2
and
dy
dt
 = 3
25 − t2
for
0 ≤ t ≤ 5.
 At time
t = 0,
 the particle is at
(2, 3).
(a) Find the speed of the object at time
t = 4.
 (Round your answer to three decimal places.)
Incorrect: Your answer is incorrect. seenKey

9.220

(b) Find the total distance traveled on the interval
[0, 5].
 (Round your answer to three decimal places.)
Incorrect: Your answer is incorrect. seenKey

59.725

(c) Find
y(4).
 (Round your answer to three decimal places.)
Incorrect: Your answer is incorrect. seenKey

55.774

(d) There is a point corresponding to a value of t in the interval
[0, 5]
 at which the tangent line to the curve is vertical. Find the acceleration vector at this point. (Round your answer to three decimal places.)
a(t) =
?
Incorrect: Your answer is incorrect. webMathematica generated answer key


Solution or Explanation
(a)    At
t = 4,
 the speed is
dx
dt
2
 
 + 
dy
dt
2
 
 = 
e−t2 + 1 −2
2
 
 + 
3
25 − t2
2
 
 
 = 
e−(4)2 + 1 −2
2
 
 + 
3
25 − (4)2
2
 
 
≈9.220.

(b)    
5
dx
dt
2
 
 + 
dy
dt
2
 
0
 dt = 
5  
e−t2 + 1 −2
2
 
 + 
3
25 − t2
2
 
 dt ≈ 59.725
0


(c)    
y(4) = y(0) + 
4
dy
dt
 dt
0
 = 3 + 
4 3
25 − t2
 dt
0
 = 3 + 3
1
2
t
25 − t2
 + 25 arcsin
t
5
4
0
 ≈ 55.774

(d)    
dy
dx
 = 
3
25 − t2
e−t2 + 1 − 2
 is undefined when
e−t2 + 1 − 2 = 0
e−t2 + 1 = 2
−t2 + 1 = ln(2)
−t2 = ln(2) − 1
t = ±
1 − ln(2)
 
t≈0.5539.

v(t) = 
x'(t), y'(t)
 = 
e−t2 + 1 − 2, 3
25 − t2

a(t) = 
−2te−t2 + 1, 3
1
2
(25 − t2)−1/2(−2t)
 = 
−2te−t2 + 1, − 
3t
25 − t2
a(
1 − ln(2)
) = 
−2
1 − ln(2)
 e−(√1 − ln(2))2 + 1, 
−3
1 − ln(2)
25 − (
1 − ln(2)
)2
 ≈ 
−2.216, −0.334
At
t = 
1 − ln(2)
 ≈ 0.5539,
a(t) = 
−2.216, −0.334
.
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