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Question

At time t, the position of a particle moving in the xy-plane is given by the parametric equations x(t) and y(t), where dx dt = t2 and dy dt = 3t+1 . Which of the following gives the total distance traveled by the particle over the time interval [0,2] ?

A. 0 2 2t+3 dt
B. 0 2 4t2+9 dt
C. 0 2 t2+3t+1 dt
D. 0 2 t4+9t2+6t+1 dt

Explanation

The total distance traveled by a particle moving along a parametric curve on an interval is equal to the length of the parametric curve on the interval. The derivative functions are given, so substitute dx dt , dy dt , and the endpoints of the interval into the arc length formula to get t1 t2 ( dx dt ) 2 + ( dy dt ) 2 dt = 0 2 ( t2 ) 2 + ( 3t+1 ) 2 dt = 0 2 t4 + 9t2 + 6t + 1 dt .

Question

Which of the following gives the length of the curve defined by the parametric equations x(t) = sin t and y(t) = e2t from t = 0 to t = 3 ?

A. 0 3 sin2t + e2t dt
B. 0 3 sin2t + e4t dt
C. 0 3 cos2t + 4e2t dt
D. 0 3 cos2t + 4e4t dt

Explanation

To find the integral that gives the length of the curve defined by the given parametric equations, first differentiate the given functions to find expressions for dx dt and dy dt .

The function for y is a composite function of the form eu, where u = 2t, so apply the chain rule to differentiate.

x(t) = sin t Given parametric functions y(t) = e2t
dx dt =cost Differentiate dy dt =e2t2
Rewrite dy dt dy dt =2e2t

Substitute the endpoints of the interval and the resulting derivatives into the arc length formula and simplify.

t1 t2 ( dx dt ) 2 + ( dy dt ) 2 dt Parametric arc length
0 3 (cost) 2 + (2e2t) 2 dt Substitute derivatives and limits of integration
0 3 cos2t + 4e4t dt Square terms: (e2t)2 = e2t ⋅ 2

Therefore, the length of the curve defined parametrically by x(t) = sin t and y(t) = e2t from t = 0 to t = 3 is given by 0 3 cos2t + 4e4t dt .

Question

A particle moves in the xy-plane so that its position for t0 is given by the parametric equations x(t) = sin 3t and y(t) = -cos5t . Find the total distance traveled by the particle over the time interval 0t1 .

A. 0.983
B. 1.517
C. 1.918
D. 2.089

Explanation

The total distance traveled by a particle moving along a parametric curve described by x(t) and y(t) on an interval [t1,t2] is equal to the length of the parametric curve (arc length) on that interval.

Length of parametric curve

The integrand in the arc length formula contains the derivatives of the parametric equations x(t) and y(t), so differentiate each given parametric equation.

The equations for x(t) and y(t) are both composite functions of the form sin u and cos u, respectively, so use the chain rule to differentiate.

x(t)=sin3t Given equations x(t)=-cos5t
x(t) =cos 3t 3 2 t-12 Use chain rule to differentiate y(t) =sin 5t 5 2 t-12
x(t) = 3cos3t 2 t Simplify y(t) = 5sin5t 2 t

Now substitute x(t) = 3cos3t 2 t and y(t) = 5sin5t 2 t and the endpoints of the given interval of time t1=0 and t2=1 into the parametric curve length formula. Then evaluate the resulting definite integral.

t1 t2 ( x(t) ) 2 + ( y(t) ) 2 dt Length of a parametric curve
0 1 ( 3cos3t 2 t ) 2 + ( 5sin5t 2 t ) 2 dt Substitute derivatives and limits of integration
2.089 Evaluate

Therefore, the total distance traveled by the particle over the time interval 0 ≤ t ≤ 1 is 2.089.

(Choice A) 0.983 may result from using x(t) and y(t), instead of their derivatives, in the formula for the length of a parametric curve.

(Choice B) 1.517 may result from not squaring x(t) and y(t) in the formula for the length of a parametric curve.

(Choice C) 1.918 may result from mistakenly using the formula for the length of a non-parametric function ( a b 1 + ( dy dx ) 2 dx ) , but the given function is parametric.

Things to remember:
The length of a parametric curve defined by x(t) and y(t) from t1 to t2 is given by:

t1 t2 ( x(t) ) 2 + ( y(t) ) 2 dt

Question

Which of the following is the result of applying the nth term test to the series n=1 n n+1 ?

A. The series converges.
B. The series diverges.
C. The test is inconclusive.

Explanation

The nth term test cannot determine if a series converges, so eliminate Choice A.

Take the limit as n approaches infinity of the general term of the given series. If the limit does not equal 0, then the series diverges. If the limit equals 0, then the test is inconclusive.

n=1 n n+1 Given series
lim n n n+1 Take limit at ∞ of general term

The limit at infinity of a rational function is equal to the limit of the ratio of the highest-degree terms (largest powers of n) in the numerator and denominator.

lim n n n+1 = lim n n n

The limit of the simplified function is equal to the limit of the original function, so evaluate and determine the result of the nth term test.

lim n n n Resulting limit
lim n 1 Cancel factors of n
1 Evaluate limit of constant

The limit at infinity of the general term of the given series does not equal 0, so the nth term test shows that the given series diverges.

Question

The series n=1 1 np/2 converges for which of the following values of p?

  • p = 1
  • p = 2
  • p = 3
A. I only
B. III only
C. I and II
D. II and III

Explanation

The given series n=1 1 np/2 is a p-series with exponent p2 . A p-series converges when its exponent is greater than 1, so plug in the given values of p to determine which converge.

p-series-table

Of the given values, only p = 3 results in an exponent (1.5) greater than 1. Therefore, the given series converges for Equation III only.

Question

The series n=1 1 converges.

A. True
B. False

Explanation

The terms in the given infinite series increase by 1 as n increases by 1. Therefore, the partial sums increase without bound, so the series diverges.

To verify, examine the partial sums to find a general expression for k terms and take the limit as k approaches infinity.

General Series

The partial sum of the given series for k terms is Sk = k. Calculate the limit as k approaches infinity to find the sum of the series.

n=1 1 = lim k Sk Sum of series equals infinite limit of partial sums
n=1 1 = lim k k Substitute Sk = k
n=1 1 = Evaluate limit

The sum of the given series is infinite, so it diverges.

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Everything you need to pass the AP Calc BC exam

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1800+ Exam-Level Questions
Hone your skills with AP exam-level questions that match the difficulty of the actual exam.
Create practice tests focused on the topics you need to boost your score strategically.
Watch yourself improve as you practice with performance tracking.
Concentrate on the topics you must master to customize your study plan.
Tailor your study sessions to fit your schedule. Choose your available days and study time, and get a personalized plan that keeps you on track.

Best Value!

AP Calculus BC
Review Course

Starting at $99

1800+ Exam-Level Questions
Hone your skills with AP exam-level questions that match the difficulty of the actual exam.
Create practice tests focused on the topics you need to boost your score strategically.
Watch yourself improve as you practice with performance tracking.
Concentrate on the topics you must master to customize your study plan.
Simple and focused, our study guides integrate smoothly with video lessons and question bank for a well-rounded study experience.
Our check-for-understanding questions ensure you grasp key concepts before you tackle advanced AP practice questions from our QBank.
Tailor your study sessions to fit your schedule. Choose your available days and study time, and get a personalized plan that keeps you on track.
Led by subject matter experts, our video lessons simplify difficult topics with easy-to-understand, step-by-step teaching animations.

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Frequently Asked Questions (FAQs)

Seasoned AP educators and subject matter experts develop our specialized AP Calculus BC practice questions and detailed answer explanations. Questions align with the latest College Board® content.
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Yes. If you have a strong grasp of the core concepts covered in your AP Calc BC class, diving straight into the practice question bank and focusing on content you feel the least confident about is the ideal method to get a top score.
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To simulate the real exam experience, set your AP Calculus BC practice exam to include 60 questions and limit yourself to 90 minutes.
We simplify complex concepts with vivid illustrations and detailed answer explanations to boost your confidence for the AP Calc BC exam. Our scientifically backed exam simulations use active learning to develop your critical thinking skills, boost your retention rate, and instill confidence as you approach test day. Plus, our practice QBank mirrors the official AP Calc BC exam questions and is organized by unit, topic, and subtopic to help you master the material you need.
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