Defining and Differentiating Vector-Valued Functions
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AP Calculus BC › Defining and Differentiating Vector-Valued Functions
A particle moves with $\mathbf{r}(t)=\langle \ln t,\ t^2\cos t,\ \sqrt{t}\rangle$ for $t>0$; find $\mathbf{r}'(t)$.
$\langle -\tfrac{1}{t},\ 2t\cos t-t^2\sin t,\ \tfrac{1}{\sqrt{t}}\rangle$
$\langle \tfrac{1}{t},\ 2t\cos t+t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$
$\langle \ln t,\ 2t\cos t-t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$
$\langle \tfrac{1}{t},\ 2t\cos t-t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$
$\langle \tfrac{1}{t},\ 2\cos t-t\sin t,\ \sqrt{t}\rangle$
Explanation
This problem involves differentiating a vector-valued function to determine the particle's velocity. The x-component is $\ln t$, with derivative $\frac{1}{t}$ for $t > 0$. The y-component is $t^2 \cos t$, and using the product rule, its derivative is $2t \cos t - t^2 \sin t$. The z-component is $\sqrt{t}$, or $t^{1/2}$, which differentiates to $(1/2) t^{-1/2}$ or $\frac{1}{2 \sqrt{t}}$. A tempting distractor is choice A, which has $+t^2 \sin t$ instead of $-t^2 \sin t$, possibly from forgetting the negative sign in the derivative of $\cos t$. In general, to find the derivative of a vector-valued function, apply the appropriate differentiation rules to each component function independently.
A drone’s position is $\mathbf{r}(t)=\langle t^3-2t,\ \sin t,\ e^{2t}\rangle$ meters; which vector gives $\mathbf{r}'(t)$?
$\langle 3t^2-2,\ \sin t,\ 2e^{t}\rangle$
$\langle 3t^2-2t,\ \cos t,\ e^{2t}\rangle$
$\langle 3t^2-2,\ \cos t,\ 2e^{2t}\rangle$
$\langle 3t^2-2,\ -\sin t,\ 2e^{2t}\rangle$
$\langle t^3-2t,\ \cos t,\ 2e^{2t}\rangle$
Explanation
This problem requires differentiating a vector-valued function to find the velocity vector of the drone. The x-component is $t^3$ - 2t, and its derivative is $3t^2$ - 2 using the power rule. The y-component is sin t, which differentiates to cos t. The z-component is $e^{2t}$, and its derivative is $2e^{2t}$ by the chain rule for exponentials. A tempting distractor is choice B, which has -sin t for the y-component, likely from mistakenly differentiating cos t instead of sin t. In general, to find the derivative of a vector-valued function, apply the appropriate differentiation rules to each component function independently.
A drone’s position is $\mathbf{r}(t)=\langle t^2-3t,\ \sin t,\ e^{2t}\rangle$ meters; what is $\mathbf{r}'(t)$?
$\langle 2t-3,\ \cos t,\ 2e^{2t}\rangle$
$\langle 2t-3,\ -\sin t,\ 2t e^{2t}\rangle$
$\langle t^2-3t,\ \cos t,\ 2e^{2t}\rangle$
$\langle 2t-3,\ \sin t,\ e^{2t}\rangle$
$\langle 2t+3,\ \cos t,\ e^{2t}\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find $r'(t)$, differentiate each component separately. The first component $t^2 - 3t$ differentiates to $2t - 3$ using the power rule. The second component $\sin t$ becomes $\cos t$, while the third $e^{2t}$ differentiates to $2e^{2t}$ via the chain rule. A tempting distractor like choice C integrates the first component instead of differentiating it, leading to an incorrect antiderivative. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
A boat’s path is $\mathbf{r}(t)=\langle \sec t,\ \csc t,\ \sin(2t)\rangle$; determine $\mathbf{r}'(t)$.
$\langle \sec^2 t,\ -\csc^2 t,\ 2\cos(2t)\rangle$
$\langle \sec t\tan t,\ -\csc t\cot t,\ 2\sin(2t)\rangle$
$\langle \sec t\tan t,\ -\csc t\cot t,\ \cos(2t)\rangle$
$\langle \sec t\tan t,\ -\csc t\cot t,\ 2\cos(2t)\rangle$
$\langle \sec t\tan t,\ \csc t\cot t,\ 2\cos(2t)\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find r'(t), differentiate each component separately. The first component sec t differentiates to sec t tan t using the known derivative. The second component csc t becomes -csc t cot t, and the third sin(2t) differentiates to 2 cos(2t) via the chain rule. A tempting distractor like choice B confuses the derivatives with squares like sec² t, which is actually for tan t. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
A camera dolly follows $\mathbf{r}(t)=\langle(\sin t)^2,\ (\cos t)^2,\ t\sin t\rangle$; determine $\mathbf{r}'(t)$.
$\langle 2\sin t\cos t,\ -2\sin t\cos t,\ t\cos t+\sin t\rangle$
$\langle 2\sin t\cos t,\ -2\sin t\cos t,\ t\cos t-\sin t\rangle$
$\langle 2\sin t\cos t,\ 2\sin t\cos t,\ t\cos t+\sin t\rangle$
$\langle 2\sin t\cos t,\ -2\sin t\cos t,\ t\sin t\rangle$
$\langle \cos t,\ -\sin t,\ t\cos t+\sin t\rangle$
Explanation
This problem involves differentiating a vector-valued function for the camera dolly's velocity. The x-component is (sin $t)^2$, and by the chain rule, its derivative is 2 sin t * cos t. The y-component is (cos $t)^2$, differentiating to 2 cos t * (-sin t) or -2 sin t cos t. The z-component is t sin t, with product rule derivative t cos t + sin t. A tempting distractor is choice E, which has -sin t for the z-component, likely from reversing the sign in the product rule. In general, to find the derivative of a vector-valued function, apply the appropriate differentiation rules to each component function independently.
A swimmer’s position is $\mathbf{r}(t)=\langle t\sin t,\ t\cos t,\ t^2\rangle$; what is $\mathbf{r}'(t)$?
$\langle \sin t+t\cos t,\ -\sin t-t\cos t,\ 2t\rangle$
$\langle \sin t,\ \cos t,\ 2t\rangle$
$\langle \sin t+t\cos t,\ \cos t-t\sin t,\ 2t\rangle$
$\langle t\cos t,\ -t\sin t,\ 2t\rangle$
$\langle \sin t+t\cos t,\ \cos t+t\sin t,\ 2t\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find $r'(t)$, differentiate each component separately. The first component $t \sin t$ differentiates to $\sin t + t \cos t$ using the product rule. The second component $t \cos t$ becomes $\cos t - t \sin t$ also via the product rule, and the third $t^2$ differentiates to $2t$ by the power rule. A tempting distractor like choice B forgets to include the non-product terms in the derivatives of the first two components. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
Given $\mathbf{r}(t)=\langle \cos(3t),\ \sin(3t),\ t\rangle$, representing a helix, what is $\mathbf{r}'(t)$?
$\langle -3\cos(3t),\ 3\sin(3t),\ 1\rangle$
$\langle -3\sin(3t),\ 3\cos(3t),\ t\rangle$
$\langle -\sin(3t),\ \cos(3t),\ 1\rangle$
$\langle -3\sin(3t),\ 3\cos(3t),\ 1\rangle$
$\langle 3\sin(3t),\ 3\cos(3t),\ 1\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find r'(t), differentiate each component separately. The first component cos(3t) differentiates to -3 sin(3t) using the chain rule. The second component sin(3t) becomes 3 cos(3t) also via the chain rule, and the third t differentiates to 1. A tempting distractor like choice A forgets the chain rule multiplier of 3, leading to incorrect coefficients. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
A puck moves as $\mathbf{r}(t)=\langle \frac{1}{t},\ \arctan t,\ t^4\rangle$ for $t\ne0$; compute $\mathbf{r}'(t)$.
$\langle -\tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ t^4\rangle$
$\langle -\tfrac{1}{t^2},\ \tfrac{1}{1-t^2},\ 4t^3\rangle$
$\langle -\tfrac{1}{t},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$
$\langle \tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$
$\langle -\tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find r'(t), differentiate each component separately. The first component 1/t differentiates to -1/t² using the power rule. The second component arctan t becomes 1/(1 + t²), and the third $t^4$ differentiates to 4t³ by the power rule. A tempting distractor like choice B reverses the sign of the first component's derivative, confusing it with a positive term. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
A satellite follows $\mathbf{r}(t)=\langle e^{t}\cos t,\ e^{t}\sin t,\ 7\rangle$; find $\mathbf{r}'(t)$.
$\langle e^{t}(\sin t-\cos t),\ e^{t}(\cos t+\sin t),\ 0\rangle$
$\langle e^{t}(\cos t+\sin t),\ e^{t}(\sin t-\cos t),\ 0\rangle$
$\langle e^{t}(\cos t-\sin t),\ e^{t}(\sin t+\cos t),\ 7\rangle$
$\langle e^{t}(\cos t-\sin t),\ e^{t}(\sin t+\cos t),\ 0\rangle$
$\langle e^{t}\cos t,\ e^{t}\sin t,\ 0\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find r'(t), differentiate each component separately. The first component $e^t$ cos t differentiates to $e^t$ (cos t - sin t) using the product rule. The second component $e^t$ sin t becomes $e^t$ (sin t + cos t) also via the product rule, and the third constant 7 differentiates to 0. A tempting distractor like choice A neglects the product rule, treating it as if only one part is differentiated. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
A robot arm tip is $\mathbf{r}(t)=\langle(3t-1)^4,\ \cos t+\sin t,\ \ln(t^2+1)\rangle$; find $\mathbf{r}'(t)$.
$\langle 12(3t-1)^4,\ -\sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$
$\langle 12(3t-1)^3,\ -\sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$
$\langle 4(3t-1)^3,\ -\sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$
$\langle 12(3t-1)^3,\ -\sin t-\cos t,\ \tfrac{1}{t^2+1}\rangle$
$\langle 12(3t-1)^3,\ \sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$
Explanation
This problem tests the skill of defining and differentiating vector-valued functions by finding the derivative of a position vector. To find r'(t), differentiate each component separately. The first component (3t - $1)^4$ differentiates to 12 (3t - $1)^3$ using the chain rule. The second component cos t + sin t becomes -sin t + cos t, and the third ln(t² + 1) differentiates to 2t/(t² + 1) via the chain rule. A tempting distractor like choice B underestimates the chain rule multiplier, using 4 instead of 12 for the first component. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.