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AP Calculus BC Practice Quiz: Integrating 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

A particle moves in the xy-plane with a velocity vector `v(t) = <2t, 3t^2>`. If the particle is at the point `(2, 0)` when `t=1`, what is the position vector `r(t)`?

All Questions (7)

A particle moves in the xy-plane with a velocity vector `v(t) = <2t, 3t^2>`. If the particle is at the point `(2, 0)` when `t=1`, what is the position vector `r(t)`?

A) <t^2 + 1, t^3 - 1>

B) <t^2 + 2, t^3>

C) <2, 6t>

D) <t^2, t^3>

Correct Answer: A

To find the position vector `r(t)`, integrate the velocity vector `v(t)`. `∫v(t) dt = ∫<2t, 3t^2> dt = <t^2 + C_x, t^3 + C_y>`. Use the initial condition `r(1) = <2, 0>` to solve for the constants: `1^2 + C_x = 2` implies `C_x = 1`, and `1^3 + C_y = 0` implies `C_y = -1`. Therefore, the particular solution is `r(t) = <t^2 + 1, t^3 - 1>`. [cite: 3045]

The velocity of a particle is given by `v(t) = <cos(t), -sin(t)>`. At time `t=0`, the position of the particle is `(0, 2)`. What is the position vector `r(t)`?

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

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

C) <sin(t), cos(t) + 1>

D) <sin(t) + 1, cos(t)>

Correct Answer: C

The position vector `r(t)` is the integral of the velocity vector `v(t)`. Integrating `v(t) = <cos(t), -sin(t)>` gives `r(t) = <sin(t) + C_x, cos(t) + C_y>`. Using the initial condition `r(0) = <0, 2>`, we have `sin(0) + C_x = 0`, which gives `C_x = 0`, and `cos(0) + C_y = 2`, which gives `1 + C_y = 2`, so `C_y = 1`. Thus, `r(t) = <sin(t), cos(t) + 1>`. [cite: 3045, 3046]

A particle's rate of change of position is described by the vector `r'(t) = <e^(2t), 1/t>`. If the particle's position at `t=1` is `(e^2/2, 5)`, find the `y`-component of the position vector `r(t)` for `t > 0`.

A) ln(t) + 5

B) -1/t^2 + 6

C) ln(t)

D) e^(2t)/2

Correct Answer: A

To find the `y`-component of the position vector, we integrate the `y`-component of the rate vector: `y(t) = ∫(1/t) dt = ln|t| + C`. Using the initial condition `y(1) = 5`, we get `ln(1) + C = 5`, which simplifies to `0 + C = 5`, so `C = 5`. Therefore, for `t > 0`, the `y`-component is `ln(t) + 5`. [cite: 3045, 3046]

The velocity of a particle moving in a plane is given by `v(t) = <3, 2t*cos(t^2)>`. If the particle is at `(4, 1)` at `t=0`, what is its position vector `r(t)`?

A) <3t + 4, sin(t^2) + 1>

B) <3t + 4, -sin(t^2) + 1>

C) <3, 2cos(t^2) - 4t^2*sin(t^2)>

D) <3t, sin(t^2)>

Correct Answer: A

To find the position vector `r(t)`, integrate `v(t)` component-wise. The integral of the x-component is `∫3 dt = 3t + C_x`. The integral of the y-component, `∫2t*cos(t^2) dt`, requires a u-substitution with `u = t^2`, resulting in `sin(t^2) + C_y`. So, `r(t) = <3t + C_x, sin(t^2) + C_y>`. Using the initial condition `r(0) = <4, 1>`, we find `C_x = 4` and `C_y = 1`. Thus, `r(t) = <3t + 4, sin(t^2) + 1>`. [cite: 3045, 3046]

Which of the following statements best describes the process of finding a particular solution for a particle's position vector, `r(t)`, given its velocity vector, `v(t)`, and an initial condition `r(t_0)`?

A) Differentiate each component of `v(t)` and then solve for the constants of differentiation using `r(t_0)`.

B) Integrate each component of `v(t)` with respect to `t` to find a general solution, then use the initial condition `r(t_0)` to solve for the vector constant of integration.

C) Evaluate `v(t)` at `t_0` and add it to the initial position `r(t_0)`.

D) The methods for finding integrals of real-valued functions cannot be applied to vector-valued functions; a different technique is required.

Correct Answer: B

The position vector `r(t)` is the antiderivative of the velocity vector `v(t)`. Therefore, one must integrate each component of `v(t)`. This integration introduces a constant of integration for each component (a vector constant). The initial condition `r(t_0)` is then used to determine the specific values of these constants, yielding the particular solution. This aligns with the principle that methods for real-valued functions extend to vector-valued functions. [cite: 3045, 3046]

A particle has velocity `v(t) = <t*e^t, 1/(t+1)>`. Given that its position at `t=0` is `(-1, 0)`, what is the position vector `r(t)` for `t > -1`?

A) <t*e^t - e^t, ln(t+1)>

B) <e^t + t*e^t, -1/(t+1)^2>

C) <t*e^t - e^t + 1, ln(t+1) - 1>

D) <t*e^t - e^t, ln(t+1)>

Correct Answer: D

Integrate `v(t)` to find `r(t)`. The x-component `∫t*e^t dt` requires integration by parts, yielding `t*e^t - e^t + C_x`. The y-component is `∫(1/(t+1)) dt = ln(t+1) + C_y`. So, `r(t) = <t*e^t - e^t + C_x, ln(t+1) + C_y>`. Using `r(0) = <-1, 0>`, we get `0*e^0 - e^0 + C_x = -1`, so `-1 + C_x = -1` and `C_x = 0`. For the y-component, `ln(0+1) + C_y = 0`, so `0 + C_y = 0` and `C_y = 0`. Thus, `r(t) = <t*e^t - e^t, ln(t+1)>`. [cite: 3045, 3046]

The velocity of a particle is given by `v(t) = <sin(t), 2t>`. What is the displacement of the particle from `t=0` to `t=π`?

A) <0, π^2>

B) <2, π^2>

C) <-2, π^2>

D) <cos(π), 2>

Correct Answer: B

Displacement is the definite integral of velocity: `∫[a,b] v(t) dt`. We calculate `∫[0,π] <sin(t), 2t> dt = <∫[0,π] sin(t) dt, ∫[0,π] 2t dt>`. The x-component is `[-cos(t)]` from 0 to `π`, which is `-cos(π) - (-cos(0)) = -(-1) - (-1) = 1 + 1 = 2`. The y-component is `[t^2]` from 0 to `π`, which is `π^2 - 0^2 = π^2`. The displacement vector is `<2, π^2>`. [cite: 3046]