AP Calculus BC Quiz: Integrating Vector Valued Functions

What this quiz covers

This quiz focuses on Integrating Vector Valued Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\vec v(t)=\langle 3t^2+1,\,-4t+2\rangle$ (correct answer)
2. $\vec v(t)=\langle 3t^2,\,-4t\rangle$
3. $\vec v(t)=\langle 6t^2+1,\,-4t+2\rangle$
4. $\vec v(t)=\langle 3t^2+2,\,-4t+1\rangle$
5. $\vec v(t)=\langle 3t^2+1,\,-4t-2\rangle$

Explanation: This problem requires integrating a vector-valued acceleration to find velocity, a key skill in vector calculus. To solve, integrate each component separately, starting with the x-component where ∫6t dt is 3t² plus a constant. For the y-component, ∫-4 dt is -4t plus a constant. Apply the initial condition at t=0 to get constants 1 and 2, resulting in ⟨3t² + 1, -4t + 2⟩. A tempting distractor like choice B omits the constants, ignoring the initial velocity. Always integrate vector functions component-wise and incorporate initial conditions to find the specific solution.

Question 2

1. $\langle e^t+C,\,\arctan(t)+C\rangle$
2. $\langle e^t,\,\arctan(t)\rangle$
3. $\langle e^t+C_1,\,\arctan(t)+C_2\rangle$ (correct answer)
4. $\langle e^t+C_1,\,\ln(1+t^2)+C_2\rangle$
5. $\langle te^t+C_1,\,\arctan(t)+C_2\rangle$

Explanation: This problem requires finding the indefinite integral of a vector-valued function, a key skill in vector calculus. To solve, integrate each component separately, with the x-component where ∫e^t dt is e^t plus a constant C1. For the y-component, ∫1/(1 + t²) dt is arctan t plus a constant C2. This gives the antiderivative ⟨e^t + C1, arctan t + C2⟩. A tempting distractor like choice D substitutes ln(1 + t²) for arctan t, but its derivative is 2t/(1 + t²), not matching. Always integrate vector functions component-wise, recalling standard antiderivatives like arctan for 1/(1 + t²) and including separate constants.

Question 3

1. $\left\langle \frac{2}{3},\,\ln 3,\,0\right\rangle$ (correct answer)
2. $\left\langle \frac{2}{3},\,\ln 2,\,0\right\rangle$
3. $\left\langle \frac{8}{3},\,\ln 3,\,0\right\rangle$
4. $\left\langle \frac{2}{3},\,\ln 3,\,\frac{2}{\pi}\right\rangle$
5. $\left\langle \frac{2}{3},\,\frac{1}{3},\,0\right\rangle$

Explanation: This problem requires evaluating a definite integral of a vector function. For the first component: ∫₀² (t-1)² dt = ∫₀² (t² - 2t + 1) dt = [t³/3 - t² + t]₀² = 8/3 - 4 + 2 = 2/3. For the second component: ∫₀² 1/(1+t) dt = [ln|1+t|]₀² = ln 3 - ln 1 = ln 3. For the third component: ∫₀² sin(πt) dt = [-(1/π)cos(πt)]₀² = -(1/π)(cos 2π - cos 0) = 0. Choice C incorrectly evaluates the first component as 8/3, perhaps computing only the t³/3 term. When integrating polynomials, expand first and integrate term by term to avoid errors.

Question 4

1. $\vec r(t)=\langle t^2+1,\sin t,-2+3t\rangle$ (correct answer)
2. $\vec r(t)=\langle 2t^2+1,\sin t,-2+3t\rangle$
3. $\vec r(t)=\langle t^2,\sin t,-2+3t\rangle$
4. $\vec r(t)=\langle t^2+1,-\sin t,-2+3t\rangle$
5. $\vec r(t)=\langle t^2+1,\sin t,-2+t^3\rangle$

Explanation: This problem involves integrating a vector-valued velocity function to find the position function. For the x-component, integrate 2t with respect to t to obtain t² plus a constant. For the y-component, integrate cos t to get sin t plus a constant. For the z-component, integrate 3 to yield 3t plus a constant, then use the initial position at t=0 to solve for the constants, resulting in +1, +0, and -2 respectively. A tempting distractor like choice B includes 2t² +1 for the x-component, which fails because it incorrectly doubles the integral of 2t instead of properly computing t². When integrating vector functions, always perform component-wise integration and apply initial conditions to determine the constant vector.

Question 5

1. $\left\langle 8,\,8,\,e^2-1\right\rangle$ (correct answer)
2. $\left\langle 12,\,8,\,e^2-1\right\rangle$
3. $\left\langle 8,\,4,\,e^2-1\right\rangle$
4. $\left\langle 8,\,8,\,e^2\right\rangle$
5. $\left\langle 4,\,8,\,e^2-1\right\rangle$

Explanation: This problem requires evaluating a definite integral of a vector-valued function from 0 to 2. For the x-component, integrate 3t² to get t³ evaluated from 0 to 2, yielding 8. For the y-component, integrate 4 to obtain 4t evaluated from 0 to 2, resulting in 8. For the z-component, integrate e^t to e^t evaluated from 0 to 2, giving e² - 1. A tempting distractor like choice B shows 12 for the x-component, which fails because it mistakenly uses 4t³/4 instead of t³ for the antiderivative of 3t². Remember, for vector integrals, compute each component's definite integral separately and combine them into the resulting vector.

Question 6

1. $\left\langle 2,\,\pi^{2},\,3\pi\right\rangle$ (correct answer)
2. $\left\langle 0,\,\pi^{2},\,3\right\rangle$
3. $\left\langle 2,\,2\pi,\,3\pi\right\rangle$
4. $\left\langle -2,\,\pi^{2},\,3\pi\right\rangle$
5. $\left\langle 2,\,\frac{\pi^{2}}{2},\,3\pi\right\rangle$

Explanation: This problem involves evaluating a definite integral of a vector-valued function. We integrate each component separately over the interval [0, π]: ∫₀^π sin t dt = [-cos t]₀^π = -cos π - (-cos 0) = -(-1) - (-1) = 2. For the second component: ∫₀^π 2t dt = [t²]₀^π = π² - 0 = π². For the third component: ∫₀^π 3 dt = [3t]₀^π = 3π - 0 = 3π. Choice C incorrectly evaluates the second component as 2π instead of π², confusing the antiderivative of 2t. When integrating vector functions, apply the fundamental theorem of calculus to each component separately.

Question 7

1. $\vec r(t)=\langle e^{t},\,-\cos t+1,\,t^{2}-3\rangle$
2. $\vec r(t)=\langle e^{t}-1,\,-\cos t,\,t^{2}-3\rangle$
3. $\vec r(t)=\langle e^{t}-1,\,-\cos t+2,\,t^{2}-3\rangle$ (correct answer)
4. $\vec r(t)=\langle e^{t}-1,\,\cos t+1,\,2t^{2}-3\rangle$
5. $\vec r(t)=\langle e^{t},\,-\cos t+2,\,t^{2}-3\rangle$

Explanation: This problem requires finding position from velocity with an initial condition. Integrating the velocity components: ∫e^t dt = e^t + C₁, ∫sin t dt = -cos t + C₂, and ∫2t dt = t² + C₃. Applying r⃗(0) = ⟨0, 1, -3⟩: e^0 + C₁ = 0 gives C₁ = -1; -cos 0 + C₂ = 1 gives C₂ = 2; and 0² + C₃ = -3 gives C₃ = -3. Choice A incorrectly calculates C₂ = 1 instead of C₂ = 2, missing that -cos 0 = -1. The crucial step in initial value problems is carefully evaluating each component at t = 0 to find the correct constants.

Question 8

1. $\left\langle 0,\,0,\,\pi^2\right\rangle$
2. $\left\langle 0,\,2,\,\pi^2\right\rangle$ (correct answer)
3. $\left\langle 2,\,0,\,\pi^2\right\rangle$
4. $\left\langle 0,\,0,\,2\pi\right\rangle$
5. $\left\langle 0,\,2,\,2\pi\right\rangle$

Explanation: This problem requires computing a definite integral of a vector-valued function from 0 to π. For the x-component, integrate cos t to sin t evaluated from 0 to π, yielding 0. For the y-component, integrate sin t to -cos t evaluated from 0 to π, resulting in 2. For the z-component, integrate 2t to t² evaluated from 0 to π, giving π². A tempting distractor like choice A shows 0 for y, which fails because it neglects the sign change in evaluating -cos t at the limits. When performing definite integrals on vectors, integrate and evaluate each component separately to form the final vector.

Question 9

1. $\vec r(t)=\langle 4\sin t-1,\,-5\cos t+8\rangle$ (correct answer)
2. $\vec r(t)=\langle 4\sin t+1,\,-5\cos t+8\rangle$
3. $\vec r(t)=\langle 4\cos t-1,\,5\sin t+3\rangle$
4. $\vec r(t)=\langle 4\sin t-1,\,5\cos t+3\rangle$
5. $\vec r(t)=\langle -4\sin t-1,\,-5\cos t+8\rangle$

Explanation: This problem involves integrating trigonometric velocity components to find position. Integrating $\vec{v}(t) = \langle 4\cos t, 5\sin t \rangle$ gives $\vec{r}(t) = \langle 4\sin t + C_1, -5\cos t + C_2 \rangle$. Using $\vec{r}(0) = \langle -1, 3 \rangle$: we get $4\sin(0) + C_1 = -1$, so $C_1 = -1$, and $-5\cos(0) + C_2 = 3$, so $C_2 = 8$. Choice D incorrectly has $5\cos t$ instead of $-5\cos t$, forgetting the negative sign when integrating sine. Remember that $\int \sin t\,dt = -\cos t + C$, and apply initial conditions after integration.

Question 10

1. $\langle 5\sin t + C_1,\,4\cos t + C_2\rangle$ (correct answer)
2. $\langle 5\sin t + C,\,4\cos t + C\rangle$
3. $\langle 5\cos t + C_1,\, -4\sin t + C_2\rangle$
4. $\langle -5\sin t + C_1,\,4\cos t + C_2\rangle$
5. $\langle 5\sin t + C_1,\,-4\cos t + C_2\rangle$

Explanation: This problem asks for the indefinite integral of a vector-valued trigonometric function. Integrating component by component: ∫5cos t dt = 5sin t + C₁ and ∫(-4sin t) dt = 4cos t + C₂. The antiderivative is ⟨5sin t + C₁, 4cos t + C₂⟩. Choice E incorrectly shows -4cos t for the y-component, which would result from forgetting that integrating -sin t gives +cos t (the negative sign is already accounted for). When integrating vector functions, carefully track signs and remember that each component needs its own integration constant.

Question 11

1. $\left\langle 0,\;\pi^2\right\rangle$
2. $\left\langle 2,\;\pi^2\right\rangle$ (correct answer)
3. $\left\langle -2,\;\pi^2\right\rangle$
4. $\left\langle 2,\;2\pi\right\rangle$
5. $\left\langle 1,\;\pi\right\rangle$

Explanation: This problem asks us to evaluate a definite integral of a vector-valued function. We integrate each component separately: $\int_0^\pi \sin t dt = [-\cos t]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2$. For the second component: $\int_0^\pi 2t dt = [t^2]_0^\pi = \pi^2 - 0 = \pi^2$. Choice A incorrectly evaluates the first integral as 0, likely confusing the integral of sine over a full period with this half-period integral. When integrating vector functions, evaluate each component's definite integral separately using the fundamental theorem of calculus.

Question 12

1. $\vec r(t)=\langle 4t-2,\,t^2-3t+1,\,\cos t+4\rangle$
2. $\vec r(t)=\langle 4t-2,\,t^2-3t+1,\,-\cos t+6\rangle$ (correct answer)
3. $\vec r(t)=\langle 4t+2,\,t^2-3t+1,\,-\cos t+6\rangle$
4. $\vec r(t)=\langle 4t-2,\,2t^2-3t+1,\,-\cos t+6\rangle$
5. $\vec r(t)=\langle 4t-2,\,t^2-3t-1,\,-\cos t+6\rangle$

Explanation: This problem requires integrating a three-dimensional vector-valued velocity to find position, a key skill in vector calculus. To solve, integrate each component separately, starting with the x-component where ∫4 dt is 4t plus a constant. For the y-component, ∫(2t - 3) dt is t² - 3t plus a constant; for the z-component, ∫sin t dt is -cos t plus a constant. Apply the initial condition at t=0 to find constants -2, 1, and 6, yielding ⟨4t - 2, t² - 3t + 1, -cos t + 6⟩. A tempting distractor like choice A uses +4 for z instead of +6, possibly from miscalculating -cos(0) + C = 5 as C=4 instead of 6. Always integrate vector functions component-wise, even in higher dimensions, and verify constants with initial conditions.

Question 13

1. $\vec r(t)=\langle \frac{1}{2}t^2-2t,\,\tan t\rangle$
2. $\vec r(t)=\langle \frac{1}{2}t^2-2t,\,\tan t+1\rangle$ (correct answer)
3. $\vec r(t)=\langle t^2-2t,\,\tan t+1\rangle$
4. $\vec r(t)=\langle \frac{1}{2}t^2-2t,\,\sec^2 t+1\rangle$
5. $\vec r(t)=\langle \frac{1}{2}t^2-2t+1,\,\tan t+1\rangle$

Explanation: This problem requires integrating a vector-valued derivative with an initial condition, a key skill in vector calculus. To solve, integrate each component separately, starting with the x-component where ∫(t - 2) dt is (1/2)t² - 2t plus a constant. For the y-component, ∫sec² t dt is tan t plus a constant. Apply the initial condition at t=0 to find constants 0 and 1, giving ⟨(1/2)t² - 2t, tan t + 1⟩. A tempting distractor like choice C omits the 1/2 in the x-component, treating it as t² instead. Always integrate vector functions component-wise, ensuring correct coefficients from the power rule and using initial conditions properly.

Question 14

1. $\langle 0,\,2\rangle$
2. $\langle 2,\,0\rangle$ (correct answer)
3. $\langle -2,\,0\rangle$
4. $\langle 2,\,2\rangle$
5. $\langle 0,\,0\rangle$

Explanation: This problem requires evaluating a definite integral of a trigonometric vector function, a key skill in vector calculus. To solve, integrate each component separately, with the x-component where ∫sin t dt from 0 to π is [-cos t]_0^π = 2. For the y-component, ∫cos t dt from 0 to π is [sin t]_0^π = 0. This yields the vector ⟨2, 0⟩. A tempting distractor like choice C might negate the x-component to -2, possibly from reversing the limits in evaluation. Always integrate vector functions component-wise and evaluate trigonometric integrals carefully over the given interval.

Question 15

1. $\vec r(t)=\langle t^{2}+1,\,3\sin t-2,\,4e^{t}\rangle$
2. $\vec r(t)=\langle t^{2}+1,\,3\sin t-2,\,4e^{t}-4\rangle$ (correct answer)
3. $\vec r(t)=\langle 2t+1,\,3\sin t-2,\,4e^{t}-4\rangle$
4. $\vec r(t)=\langle t^{2},\,3\sin t,\,4e^{t}-4\rangle$
5. $\vec r(t)=\langle t^{2}+1,\,3\cos t-2,\,4e^{t}-4\rangle$

Explanation: This problem requires integrating a vector-valued velocity function to find position. To find position from velocity, we integrate each component separately: ∫2t dt = t² + C₁, ∫3cos t dt = 3sin t + C₂, and ∫4e^t dt = 4e^t + C₃. Using the initial condition r⃗(0) = ⟨1, -2, 0⟩, we find the constants: C₁ = 1, C₂ = -2, and C₃ = -4. Choice A incorrectly omits the constant C₃ = -4 in the third component, forgetting that 4e^0 = 4. The key strategy for vector integration is to integrate each component independently, then apply initial conditions to find all integration constants.

Question 16

1. $\left\langle 3,\,\frac{e^{2}-1}{2},\,\frac{2}{3}\right\rangle$ (correct answer)
2. $\left\langle 3,\,e^{2}-1,\,\frac{2}{3}\right\rangle$
3. $\left\langle 6,\,\frac{e^{2}-1}{2},\,\frac{2}{3}\right\rangle$
4. $\left\langle 3,\,\frac{e^{2}-1}{2},\,\frac{3}{2}\right\rangle$
5. $\left\langle 3,\,\frac{e^{2}}{2},\,\frac{2}{3}\right\rangle$

Explanation: This problem asks for the definite integral of a vector function. For the first component: ∫₀¹ 6t dt = [3t²]₀¹ = 3. For the second component: ∫₀¹ e^(2t) dt = [(1/2)e^(2t)]₀¹ = (1/2)(e² - e⁰) = (e² - 1)/2. For the third component: ∫₀¹ √t dt = ∫₀¹ t^(1/2) dt = [(2/3)t^(3/2)]₀¹ = 2/3. Choice B incorrectly evaluates the second component as e² - 1, forgetting the factor of 1/2 that comes from the chain rule when integrating e^(2t). Always remember to include the reciprocal of the inner function's derivative when integrating composite functions.

Question 17

1. $\vec r(t)=\langle -\cos t+\cos 1,\,t^2+2,\,\ln t+2\rangle$ (correct answer)
2. $\vec r(t)=\langle -\cos t+\cos 1,\,t^2+3,\,\ln t+2\rangle$
3. $\vec r(t)=\langle \cos t-\cos 1,\,t^2+2,\,\ln t+2\rangle$
4. $\vec r(t)=\langle -\cos t+\cos 1,\,2t^2+2,\,\ln t+2\rangle$
5. $\vec r(t)=\langle -\cos t+\cos 1,\,t^2+2,\,\ln t\rangle$

Explanation: This problem involves integrating a vector-valued derivative to find the original position function with a given initial condition. For the x-component, integrate sin t to get -cos t plus a constant, then use r(1) to find +cos 1. For the y-component, integrate 2t to obtain t² plus a constant, determining +2 from the initial value. For the z-component, integrate 1/t to yield ln t plus a constant, solving for +2 using r(1). A tempting distractor like choice E omits the +2 in the z-component, failing because it neglects the constant needed to satisfy the initial condition at t=1. In general, integrate vector functions component by component and use initial conditions to find the constants for each.

Question 18

1. $\vec r(t)=\left\langle \tfrac12 e^{2t},\,-\tfrac12 t^{-2},\,\tfrac23 t^{3/2}\right\rangle+\vec C$ (correct answer)
2. $\vec r(t)=\left\langle 2e^{2t},\,-\tfrac12 t^{-2},\,\tfrac23 t^{3/2}\right\rangle+\vec C$
3. $\vec r(t)=\left\langle \tfrac12 e^{2t},\,\tfrac12 t^{-2},\,\tfrac23 t^{3/2}\right\rangle+\vec C$
4. $\vec r(t)=\left\langle \tfrac12 e^{2t},\,-\tfrac12 t^{-2},\,\tfrac32 t^{3/2}\right\rangle+\vec C$
5. $\vec r(t)=\left\langle \tfrac12 e^{2t},\,-\tfrac12 t^{-2},\,2\sqrt t\right\rangle+\vec C$

Explanation: This problem involves finding an antiderivative of a vector-valued function. For the x-component, integrate e^{2t} to (1/2) e^{2t} plus a constant. For the y-component, integrate t^{-3} to - (1/2) t^{-2} plus a constant. For the z-component, integrate √t or t^{1/2} to (2/3) t^{3/2} plus a constant. A tempting distractor like choice E uses 2 √t for z, which fails because it omits the 2/3 coefficient from proper power rule integration. For antiderivatives of vector functions, apply integration rules to each component individually and include the constant vector.

Question 19

1. $\vec v(t)=\langle 4t-1,\,-3t^2+2\rangle$ (correct answer)
2. $\vec v(t)=\langle 4t+1,\,-3t^2+2\rangle$
3. $\vec v(t)=\langle 4t-1,\,-6t^2+2\rangle$
4. $\vec v(t)=\langle 2t-1,\,-3t^2+2\rangle$
5. $\vec v(t)=\langle 4t,\,-3t^2\rangle$

Explanation: This problem involves integrating acceleration to find velocity for a vector-valued function. Integrating the acceleration components: ∫4 dt = 4t + C₁ for the x-component and ∫(-6t) dt = -3t² + C₂ for the y-component. Using the initial velocity v⃗(0) = ⟨-1, 2⟩, we find C₁ = -1 and C₂ = 2. Therefore, v⃗(t) = ⟨4t - 1, -3t² + 2⟩. Choice C incorrectly shows -6t² in the y-component, which would result from forgetting to apply the power rule correctly when integrating -6t. Remember to integrate each component of the acceleration vector independently and then apply initial conditions.

Question 20

1. $\langle -t^{-2},\,\frac{2}{3}t^{3/2}\rangle+\langle C_1,C_2\rangle$ (correct answer)
2. $\langle -2t^{-2},\,\frac{2}{3}t^{3/2}\rangle+\langle C_1,C_2\rangle$
3. $\langle -t^{-2},\,\frac{3}{2}t^{3/2}\rangle+\langle C_1,C_2\rangle$
4. $\langle t^{-2},\,\frac{2}{3}t^{3/2}\rangle+\langle C_1,C_2\rangle$
5. $\langle -t^{-2}+C,\,\frac{2}{3}t^{3/2}+C\rangle$

Explanation: This problem requires integrating a vector function with power functions. For $\langle 2t^{-3}, \sqrt{t} \rangle = \langle 2t^{-3}, t^{1/2} \rangle$, we integrate: $\int 2t^{-3}dt = 2 \cdot \frac{t^{-2}}{-2} + C_1 = -t^{-2} + C_1$ and $\int t^{1/2}dt = \frac{t^{3/2}}{3/2} + C_2 = \frac{2}{3}t^{3/2} + C_2$. Choice E incorrectly uses a single constant $C$ for both components instead of separate constants $C_1$ and $C_2$. When integrating vector functions, each component gets its own arbitrary constant, written as $\langle C_1, C_2 \rangle$.