Integrating Vector-Valued Functions
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AP Calculus BC › Integrating Vector-Valued Functions
A particle’s acceleration is $\vec a(t)=\langle 6t,,-4\rangle$ with $\vec v(0)=\langle 1,2\rangle$; find $\vec v(t)$.
$\vec v(t)=\langle 3t^2+1,,-4t-2\rangle$
$\vec v(t)=\langle 3t^2+2,,-4t+1\rangle$
$\vec v(t)=\langle 3t^2,,-4t\rangle$
$\vec v(t)=\langle 3t^2+1,,-4t+2\rangle$
$\vec v(t)=\langle 6t^2+1,,-4t+2\rangle$
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.
Compute $\int \langle e^t,,\frac{1}{1+t^2}\rangle,dt$ as an antiderivative vector.
$\langle e^t,,\arctan(t)\rangle$
$\langle te^t+C_1,,\arctan(t)+C_2\rangle$
$\langle e^t+C_1,,\ln(1+t^2)+C_2\rangle$
$\langle e^t+C,,\arctan(t)+C\rangle$
$\langle e^t+C_1,,\arctan(t)+C_2\rangle$
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.
A particle has $\vec v(t)=\langle 4,,2t-3,,\sin t\rangle$ and $\vec r(0)=\langle -2,1,5\rangle$; find $\vec r(t)$.
$\vec r(t)=\langle 4t-2,,t^2-3t+1,,\cos t+4\rangle$
$\vec r(t)=\langle 4t-2,,2t^2-3t+1,,-\cos t+6\rangle$
$\vec r(t)=\langle 4t-2,,t^2-3t+1,,-\cos t+6\rangle$
$\vec r(t)=\langle 4t+2,,t^2-3t+1,,-\cos t+6\rangle$
$\vec r(t)=\langle 4t-2,,t^2-3t-1,,-\cos t+6\rangle$
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.
If $\vec r'(t)=\langle t-2,,\sec^2 t\rangle$ and $\vec r(0)=\langle 0,1\rangle$, what is $\vec r(t)$?
$\vec r(t)=\langle \frac{1}{2}t^2-2t,,\tan t\rangle$
$\vec r(t)=\langle \frac{1}{2}t^2-2t,,\tan t+1\rangle$
$\vec r(t)=\langle \frac{1}{2}t^2-2t,,\sec^2 t+1\rangle$
$\vec r(t)=\langle \frac{1}{2}t^2-2t+1,,\tan t+1\rangle$
$\vec r(t)=\langle t^2-2t,,\tan t+1\rangle$
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.
Find $\int_{0}^{\pi}\langle \sin t,,\cos t\rangle,dt$.
$\langle 2,,2\rangle$
$\langle 0,,2\rangle$
$\langle 0,,0\rangle$
$\langle -2,,0\rangle$
$\langle 2,,0\rangle$
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.
Evaluate $\displaystyle \int_{0}^{2}\langle(t-1)^{2},,\frac{1}{1+t},,\sin(\pi t)\rangle,dt$.
$\left\langle \frac{2}{3},,\ln 2,,0\right\rangle$
$\left\langle \frac{2}{3},,\ln 3,,0\right\rangle$
$\left\langle \frac{2}{3},,\ln 3,,\frac{2}{\pi}\right\rangle$
$\left\langle \frac{2}{3},,\frac{1}{3},,0\right\rangle$
$\left\langle \frac{8}{3},,\ln 3,,0\right\rangle$
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.
A particle has velocity $\vec v(t)=\langle 2t,,3\cos t,,4e^{t}\rangle$ and position $\vec r(0)=\langle 1,-2,0\rangle$; find $\vec r(t)$.
$\vec r(t)=\langle t^{2}+1,,3\sin t-2,,4e^{t}-4\rangle$
$\vec r(t)=\langle t^{2},,3\sin t,,4e^{t}-4\rangle$
$\vec r(t)=\langle t^{2}+1,,3\cos t-2,,4e^{t}-4\rangle$
$\vec r(t)=\langle 2t+1,,3\sin t-2,,4e^{t}-4\rangle$
$\vec r(t)=\langle t^{2}+1,,3\sin t-2,,4e^{t}\rangle$
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.
A particle has velocity $\vec v(t)=\langle 2t,\cos t,3\rangle$ and position $\vec r(0)=\langle 1,0,-2\rangle$; find $\vec r(t)$.
$\vec r(t)=\langle t^2+1,\sin t,-2+t^3\rangle$
$\vec r(t)=\langle t^2+1,\sin t,-2+3t\rangle$
$\vec r(t)=\langle 2t^2+1,\sin t,-2+3t\rangle$
$\vec r(t)=\langle t^2,\sin t,-2+3t\rangle$
$\vec r(t)=\langle t^2+1,-\sin t,-2+3t\rangle$
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.
Evaluate $\int_0^2 \langle 3t^2,,4,,e^t\rangle,dt$ as a vector.
$\left\langle 8,,8,,e^2-1\right\rangle$
$\left\langle 4,,8,,e^2-1\right\rangle$
$\left\langle 12,,8,,e^2-1\right\rangle$
$\left\langle 8,,4,,e^2-1\right\rangle$
$\left\langle 8,,8,,e^2\right\rangle$
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.
Evaluate $\displaystyle \int_{0}^{\pi}\langle \sin t,,2t,,3\rangle,dt$.
$\left\langle 0,,\pi^{2},,3\right\rangle$
$\left\langle -2,,\pi^{2},,3\pi\right\rangle$
$\left\langle 2,,2\pi,,3\pi\right\rangle$
$\left\langle 2,,\frac{\pi^{2}}{2},,3\pi\right\rangle$
$\left\langle 2,,\pi^{2},,3\pi\right\rangle$
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.