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AP Calculus BC Practice Quiz: Defining and Differentiating Vector-Valued Functions

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

If a vector-valued function is defined by r(t) = <3t^2 - 1, 4t^3 + 2>, what is r'(t)?

All Questions (7)

If a vector-valued function is defined by r(t) = <3t^2 - 1, 4t^3 + 2>, what is r'(t)?

A) <6t, 12t^2>

B) <3t, 4t^2>

C) <6t - 1, 12t^2 + 2>

D) <t^3 - t, t^4 + 2t>

Correct Answer: A

Based on citation [3038] and [3039], the derivative of a vector-valued function is found by taking the derivative of each component with respect to t. The derivative of the first component, 3t^2 - 1, is 6t using the power rule. The derivative of the second component, 4t^3 + 2, is 12t^2. Therefore, r'(t) = <6t, 12t^2>.

Let r(t) be a vector-valued function defined by r(t) = <cos(t), 5sin(t)>. Which of the following is r'(t)?

A) <sin(t), -5cos(t)>

B) <sin(t), 5cos(t)>

C) <-sin(t), 5cos(t)>

D) <-sin(t), -5cos(t)>

Correct Answer: C

To find the derivative of the vector-valued function, we differentiate each component function. The derivative of cos(t) is -sin(t), and the derivative of 5sin(t) is 5cos(t). Thus, r'(t) = <-sin(t), 5cos(t)>. This extends standard derivative rules to vector-valued functions as noted in [3039].

The position of a particle is given by the vector-valued function p(t) = <e^(3t), ln(t^2)>. What is the velocity vector, p'(t)?

A) <3e^(3t), 1/t^2>

B) <e^(3t), 2/t>

C) <3e^(3t), 2/t>

D) <3e^(3t), 1/t>

Correct Answer: C

The derivative is calculated component-wise [3038]. The derivative of e^(3t) requires the chain rule: e^(3t) * d/dt(3t) = 3e^(3t). The derivative of ln(t^2) also requires the chain rule: (1/t^2) * d/dt(t^2) = (1/t^2) * (2t) = 2/t. Therefore, p'(t) = <3e^(3t), 2/t>.

Given the vector-valued function r(t) = <t^2 sin(t), 4t^3>. Find r'(t).

A) <2t cos(t), 12t^2>

B) <2t sin(t) + t^2 cos(t), 12t^2>

C) <2t sin(t) - t^2 cos(t), 12t^2>

D) <t^2 cos(t), 12t^2>

Correct Answer: B

The derivative of the second component, 4t^3, is 12t^2. The derivative of the first component, t^2 sin(t), requires the product rule [3039]: d/dt(uv) = u'v + uv'. Let u = t^2 and v = sin(t). Then u' = 2t and v' = cos(t). The derivative is (2t)(sin(t)) + (t^2)(cos(t)). Thus, r'(t) = <2t sin(t) + t^2 cos(t), 12t^2>.

Let r(t) = <sin(t^3), e^(-2t)>. What is r'(t)?

A) <cos(t^3), -2e^(-2t)>

B) <3t^2 cos(t^3), e^(-2t)>

C) <3t^2 cos(t^3), -2e^(-2t)>

D) <cos(t^3) * 3t^2, -e^(-2t)>

Correct Answer: C

This problem requires applying the chain rule to both components [3039]. For the first component, the derivative of sin(u) is cos(u) * u', so the derivative of sin(t^3) is cos(t^3) * 3t^2. For the second component, the derivative of e^u is e^u * u', so the derivative of e^(-2t) is e^(-2t) * (-2). Therefore, r'(t) = <3t^2 cos(t^3), -2e^(-2t)>.

If r(t) = <sqrt(4t+1), 1/(t+1)>, find the value of the derivative r'(2).

A) <1/3, -1/9>

B) <2/3, -1/9>

C) <1/3, 1/9>

D) <2/3, 1/9>

Correct Answer: B

First, find the derivative r'(t) [3038]. The first component (4t+1)^(1/2) has a derivative of (1/2)(4t+1)^(-1/2) * 4 = 2/sqrt(4t+1). The second component (t+1)^(-1) has a derivative of -1(t+1)^(-2) * 1 = -1/(t+1)^2. So, r'(t) = <2/sqrt(4t+1), -1/(t+1)^2>. Now, substitute t=2: r'(2) = <2/sqrt(4(2)+1), -1/(2+1)^2> = <2/sqrt(9), -1/(3)^2> = <2/3, -1/9>.

The motion of a particle in the xy-plane is described by the vector-valued function r(t) = <t / (t^2 + 1), tan(t)>. What is the velocity vector of the particle at t=0?

A) <0, 1>

B) <1, 0>

C) <1, 1>

D) <1, sec^2(1)>

Correct Answer: C

First, find the derivative r'(t) [3038]. The first component requires the quotient rule [3039]: [(1)(t^2+1) - (t)(2t)] / (t^2+1)^2 = (1-t^2)/(t^2+1)^2. The derivative of the second component, tan(t), is sec^2(t). So, r'(t) = <(1-t^2)/(t^2+1)^2, sec^2(t)>. Now, evaluate at t=0: r'(0) = <(1-0^2)/(0^2+1)^2, sec^2(0)> = <1/1^2, (1/cos(0))^2> = <1, (1/1)^2> = <1, 1>.