Integrating Vector-Valued Functions - AP Calculus BC

1. A particle has velocity $\vec v(t)=\langle 2t,\cos t,3\rangle$ and position $\vec r(0)=\langle 1,0,-2\rangle$; find $\vec r(t)$.
   Answer: $\vec r(t)=\langle t^2+1,\sin t,-2+3t\rangle$ 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.
2. Find $\int_{0}^{\pi}\langle \sin t,\,\cos t\rangle\,dt$.
   Answer: $\langle 2,\,0\rangle$ 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.
3. Compute $\int \langle e^{t},\,\sec^2 t\rangle\,dt$ as a vector-valued antiderivative.
   Answer: $\langle e^{t}+C_1,\,\tan t+C_2\rangle$ This problem asks for the indefinite integral of a vector-valued function. When finding antiderivatives of vector functions, we integrate each component separately and include a constant of integration for each component. The x-component gives ∫eᵗ dt = eᵗ + C₁, and the y-component gives ∫sec² t dt = tan t + C₂. Therefore, the antiderivative is ⟨eᵗ + C₁, tan t + C₂⟩. Choice B incorrectly uses the same constant C for both components, but vector antiderivatives require independent constants for each component. The fundamental principle is that each component of a vector function can have its own arbitrary constant when finding antiderivatives.
4. Given $\vec r\,'(t)=\langle \sin t,\,2\cos t\rangle$ and $\vec r\left(\frac{\pi}{2}\right)=\langle 0,5\rangle$, find $\vec r(t)$.
   Answer: $\vec r(t)=\langle -\cos t,\,2\sin t+3\rangle$ This problem involves integrating a derivative and using a non-zero initial condition. Integrating $\vec{r}'(t) = \langle \sin t, 2\cos t \rangle$ gives $\vec{r}(t) = \langle -\cos t + C_1, 2\sin t + C_2 \rangle$. Using $\vec{r}(\pi/2) = \langle 0, 5 \rangle$: we get $-\cos(\pi/2) + C_1 = 0$, so $C_1 = 0$, and $2\sin(\pi/2) + C_2 = 5$, so $2 + C_2 = 5$, thus $C_2 = 3$. Choice C incorrectly calculates $C_2 = 5$, likely forgetting that $\sin(\pi/2) = 1$. When applying initial conditions at non-zero points, carefully evaluate the trigonometric functions at that specific point.
5. A particle has velocity $\vec v(t)=\langle 6t,\,4-3t^2\rangle$ and position $\vec r(0)=\langle 2,-1\rangle$. What is $\vec r(t)$?
   Answer: $\vec r(t)=\langle 3t^2+2,\,4t-t^3-1\rangle$ This problem requires integrating a vector-valued velocity function to find position. To find position from velocity, we integrate each component separately: the x-component gives ∫6t dt = 3t² + C₁ and the y-component gives ∫(4-3t²) dt = 4t - t³ + C₂. Using the initial condition r⃗(0) = ⟨2,-1⟩, we substitute t=0 into our integrated result to find C₁ = 2 and C₂ = -1. Therefore, r⃗(t) = ⟨3t² + 2, 4t - t³ - 1⟩. Choice D might tempt students who forget to integrate the constant 4 in the y-component, leaving just 4 instead of 4t. When integrating vector functions, integrate each component independently and then apply initial conditions to find the constants.
6. Compute $\int_0^{\pi} \langle \cos t,\,\sin t,\,2t\rangle\,dt$ as an ordered triple.
   Answer: $\left\langle 0,\,2,\,\pi^2\right\rangle$ 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.
7. Compute $\displaystyle \int_1^4 \langle 2t^{-1},\;3t^2\rangle\,dt$ for a displacement vector.​
   Answer: $\left\langle 2\ln 4,\;63\right\rangle$ This problem requires evaluating a definite integral of a vector function. For the first component: $\int_1^4 2t^{-1} dt = 2\ln|t|]_1^4 = 2\ln(4) - 2\ln(1) = 2\ln(4) - 0 = 2\ln(4)$. For the second component: $\int_1^4 3t^2 dt = t^3]_1^4 = 64 - 1 = 63$. Choice B incorrectly evaluates the second integral as 64, likely forgetting to subtract the lower limit evaluation. When computing definite integrals of vector functions, evaluate each component separately and remember to subtract the antiderivative evaluated at the lower limit.
8. Evaluate $\displaystyle \int_0^2 \langle 6t,\,3t^2\rangle\,dt$.
   Answer: $\langle 12,8\rangle$ This problem requires evaluating a definite integral of a vector function. We integrate $\langle 6t, 3t^2 \rangle$ component-wise: $\int_0^2 6t\,dt = [3t^2]_0^2 = 12$ and $\int_0^2 3t^2\,dt = [t^3]_0^2 = 8$. The result is $\langle 12, 8 \rangle$. Choice A incorrectly evaluates the second component as 12 instead of 8, possibly by misapplying the power rule or arithmetic error. When integrating vector functions over an interval, evaluate each component's definite integral separately and combine the results.
9. Compute $\displaystyle \int_0^1 \langle (t+1)^2,\,(t+1)^{-2}\rangle\,dt$.
   Answer: $\left\langle \frac{7}{3},\,\frac{1}{2}\right\rangle$ This problem requires integrating a vector function with composite power functions. For $\langle (t+1)^2, (t+1)^{-2} \rangle$, we can use substitution $u = t+1$, $du = dt$, changing limits from $t \in [0,1]$ to $u \in [1,2]$. Then $\int_1^2 u^2\,du = [\frac{u^3}{3}]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ and $\int_1^2 u^{-2}\,du = [-u^{-1}]_1^2 = -\frac{1}{2} - (-1) = \frac{1}{2}$. Choice B incorrectly calculates the first component as $\frac{8}{3}$, likely making an error with the bounds. When integrating vector functions with substitution, apply the substitution consistently to each component and adjust the limits accordingly.
10. Evaluate $\int_{0}^{2}\langle t^2-1,\,4t\rangle\,dt$ as a vector.
    Answer: $\left\langle \frac{2}{3},\,8\right\rangle$ This problem requires integrating a vector-valued function over a definite interval, a key skill in vector calculus. To solve, integrate each component separately, beginning with the x-component where ∫(t² - 1) dt from 0 to 2 evaluates to (8/3 - 2) or 2/3. For the y-component, ∫4t dt from 0 to 2 evaluates to 8. Combining these gives the vector ⟨2/3, 8⟩. A tempting distractor like choice B might arise from miscalculating the x-component as 5/3, but the correct evaluation is 2/3. Always integrate vector functions component-wise and evaluate definite integrals carefully for each part to ensure precision.
11. A particle has $\vec v(t)=\langle e^{t},\,\sin t,\,2t\rangle$ and $\vec r(0)=\langle 0,1,-3\rangle$; find $\vec r(t)$.
    Answer: $\vec r(t)=\langle e^{t}-1,\,-\cos t+2,\,t^{2}-3\rangle$ 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.
12. A particle has $\vec v(t)=\langle 4,\,2t-3,\,\sin t\rangle$ and $\vec r(0)=\langle -2,1,5\rangle$; find $\vec r(t)$.
    Answer: $\vec r(t)=\langle 4t-2,\,t^2-3t+1,\,-\cos t+6\rangle$ 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.
13. Compute $\displaystyle \int_0^{\pi}\langle \sin t,\;2t\rangle\,dt$ for a moving point in the plane.​​
    Answer: $\left\langle 2,\;\pi^2\right\rangle$ 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.
14. Compute the antiderivative $\int \langle t^{-2},\,\sec^2 t,\,5t^4\rangle\,dt$.
    Answer: $\left\langle -t^{-1},\,\tan t,\,t^5\right\rangle+\vec C$ This problem requires finding the indefinite integral of a vector-valued function. For the x-component, integrate $t^{-2}$ to obtain -$t^{-1}$ plus a constant. For the y-component, integrate sec² t to get tan t plus a constant. For the z-component, integrate $5t^4$ to yield $t^5$ plus a constant. A tempting distractor like choice D uses sec² t in the y-component, which fails because it confuses the integrand with its own antiderivative. Always integrate each component of a vector function separately, adding a constant vector for indefinite integrals.
15. Evaluate $\displaystyle \int_{0}^{2}\langle (t-1)^{2},\,\frac{1}{1+t},\,\sin(\pi t)\rangle\,dt$.
    Answer: $\left\langle \frac{2}{3},\,\ln 3,\,0\right\rangle$ 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.
16. Evaluate $\displaystyle \int_1^3 \langle \frac{2}{t},\,t^2-1\rangle\,dt$.
    Answer: $\langle 2\ln 3,\,\frac{20}{3}\rangle$ This problem involves evaluating a definite integral with logarithmic and polynomial components. Integrating $\langle \frac{2}{t}, t^2-1 \rangle$ from 1 to 3: First component is $\int_1^3 \frac{2}{t}dt = [2\ln t]_1^3 = 2\ln 3 - 2\ln 1 = 2\ln 3$. Second component is $\int_1^3 (t^2-1)dt = [\frac{t^3}{3} - t]_1^3 = (9-3) - (\frac{1}{3}-1) = 6 - (-\frac{2}{3}) = \frac{20}{3}$. Choice B incorrectly calculates the second component as $\frac{26}{3}$, likely making an arithmetic error. When evaluating definite integrals of vector functions, carefully compute each component's integral and evaluate at the bounds.
17. Compute $\int \langle 2t^{-1},\,5t^4\rangle\,dt$ as a general antiderivative vector.
    Answer: $\langle 2\ln|t|+C_1,\,t^5+C_2\rangle$ 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 ∫(2/t) dt is 2 ln|t| plus a constant C1. For the y-component, ∫$5t^4$ dt is $t^5$ plus a constant C2. This gives the general antiderivative ⟨2 ln|t| + C1, $t^5$ + C2⟩. A tempting distractor like choice E incorrectly uses $5t^5$ for the y-component, forgetting to divide by 5 when integrating $t^4$. Always integrate vector functions component-wise, applying the power rule correctly and including separate constants for each dimension.
18. Find $\vec r(t)$ if $\vec r'(t)=\langle 4,\,3t^2,\,2\sin t\rangle$ and $\vec r(0)=\langle -1,5,0\rangle$.
    Answer: $\vec r(t)=\langle 4t-1,\,t^3+5,\,-2\cos t+2\rangle$ This problem involves integrating a vector-valued velocity to find position with an initial condition. For the x-component, integrate 4 to get 4t plus a constant, using r(0) to find -1. For the y-component, integrate 3t² to obtain t³ plus a constant, determining +5 from the initial value. For the z-component, integrate 2 sin t to -2 cos t plus a constant, solving for +2 using r(0). A tempting distractor like choice E uses 2 cos t - 2 for z, which fails because it has the wrong sign for the antiderivative of sin t. Integrate vector functions by handling each component independently and using initial conditions to resolve constants.
19. Given $\vec r'(t)=\langle \sin t,\,2t,\,1/t\rangle$ and $\vec r(1)=\langle 0,3,2\rangle$, find $\vec r(t)$.
    Answer: $\vec r(t)=\langle -\cos t+\cos 1,\,t^2+2,\,\ln t+2\rangle$ 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.
20. Find $\vec r(t)$ if $\vec v(t)=\langle \cos(2t),\;\sin(2t)\rangle$ and $\vec r(0)=\langle 3,1\rangle$.​​
    Answer: $\vec r(t)=\left\langle \tfrac12\sin(2t)+3,\;-\tfrac12\cos(2t)+\tfrac32\right\rangle$ This problem requires integrating velocity to find position. Integrating $\vec{v}(t) = \langle \cos(2t), \sin(2t) \rangle$: for the first component, $\int \cos(2t) dt = \frac{1}{2}\sin(2t) + C_1$; for the second component, $\int \sin(2t) dt = -\frac{1}{2}\cos(2t) + C_2$. Using $\vec{r}(0) = \langle 3, 1 \rangle$: we get $0 + C_1 = 3$ and $-\frac{1}{2} + C_2 = 1$, so $C_1 = 3$ and $C_2 = \frac{3}{2}$. Choice C forgets the chain rule factor of $\frac{1}{2}$ when integrating, showing $\sin(2t)$ and $-\cos(2t)$ instead of the correct scaled versions. When integrating trigonometric functions with linear arguments, remember to divide by the coefficient of $t$ inside the function.

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