Defining and Differentiating Vector-Valued Functions - AP Calculus BC

1. A swimmer's position is $\mathbf{r}(t)=\langle \cos(3t),\ \sin(3t),\ t^2\rangle$. What is $\mathbf{r}'(t)$?
   Answer: $\langle -3\sin(3t),\ 3\cos(3t),\ 2t\rangle$ This problem requires differentiating trigonometric functions with chain rule inside a vector function. For cos(3t), we get -sin(3t)·3 = -3sin(3t) by the chain rule. For sin(3t), we get cos(3t)·3 = 3cos(3t). For t², we simply get 2t. Therefore r'(t) = ⟨-3sin(3t), 3cos(3t), 2t⟩. Choice A forgets to apply the chain rule and misses the factor of 3 from the inner function's derivative, a common error when differentiating composite functions. Always remember to multiply by the derivative of the inner function when using the chain rule in vector differentiation.
2. A car's location is $\mathbf{r}(t)=\langle (t^2+1)^5,\ \tan t,\ 4t^{-1}\rangle$; determine $\mathbf{r}'(t)$.
   Answer: $\langle 10t(t^2+1)^4,\ \sec^2 t,\ -4t^{-2}\rangle$ 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 $(t² + 1)^5$ differentiates to 10t $(t² + 1)^4$ using the chain rule. The second component tan t becomes sec² t, and the third $4t^{-1}$ differentiates to -4 $t^{-2}$ by the power rule. A tempting distractor like choice A omits the inner derivative in the chain rule for the first component, missing the 10t factor. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
3. A kite's position is $r(t)=\langle \sec t,\ \csc t,\ \ln(\sin t)\rangle$. Find $r'(t)$.
   Answer: $\langle \sec t\tan t,\ -\csc t\cot t,\ \cot t\rangle$ This vector function tests knowledge of trigonometric derivatives. The derivative of sec t is sec t·tan t. The derivative of csc t is -csc t·cot t (note the negative sign). For ln(sin t), we use the chain rule: d/dt[ln(sin t)] = 1/sin t · cos t = cot t. Choice D incorrectly gives csc t·cot t for the third component, confusing it with a derivative of a different trigonometric function. To master vector differentiation with trigonometric components, memorize the derivatives of all six trig functions and their signs.
4. A drone's position is $\mathbf{r}(t)=\langle t^3-2t,\ \sin t,\ e^{2t}\rangle$ meters; which vector gives $\mathbf{r}'(t)$?
   Answer: $\langle 3t^2-2,\ \cos t,\ 2e^{2t}\rangle$ 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.
5. A puck moves with $\mathbf{r}(t)=\langle t\cos t,\ t\sin t,\ 5\rangle$. What is $\mathbf{r}'(t)$?
   Answer: $\langle \cos t-t\sin t,\ \sin t+t\cos t,\ 0\rangle$ This vector function requires the product rule for its first two components since each contains a product of functions. For $\mathbf{r}(t)=\langle t\cos t,\ t\sin t,\ 5\rangle$, applying the product rule: $(t\cos t)' = 1 \cdot \cos t + t \cdot (-\sin t) = \cos t - t\sin t$ and $(t\sin t)' = 1 \cdot \sin t + t \cdot \cos t = \sin t + t\cos t$, while the constant third component gives $5' = 0$. Choice A incorrectly shows just $\cos t$ and $\sin t$ without applying the product rule, essentially treating $t$ as a constant. Remember that vector differentiation follows all standard calculus rules including the product rule when needed.
6. A particle moves with $\mathbf{r}(t)=\langle \ln t,\ t^2\cos t,\ \sqrt{t}\rangle$ for $t>0$; find $\mathbf{r}'(t)$.
   Answer: $\langle \tfrac{1}{t},\ 2t\cos t-t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$ 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.
7. A satellite's path is $\mathbf{r}(t)=\langle e^t\cos t,\ e^t\sin t,\ t\rangle$; compute $\mathbf{r}'(t)$.
   Answer: $\langle e^t(\cos t-\sin t),\ e^t(\sin t+\cos t),\ 1\rangle$ This problem involves differentiating a vector-valued function for the satellite's velocity. The x-component is $e^t$ cos t, and using the product rule, its derivative is $e^t$ cos t - $e^t$ sin t or $e^t$ (cos t - sin t). The y-component is $e^t$ sin t, differentiating to $e^t$ sin t + $e^t$ cos t or $e^t$ (sin t + cos t). The z-component is t, with derivative 1. A tempting distractor is choice C, which omits the additional terms from the product rule, treating $e^t$ as constant. In general, to find the derivative of a vector-valued function, apply the appropriate differentiation rules to each component function independently.
8. A moving point has $\mathbf{r}(t)=\langle (3t-1)^5,\ e^{t^3},\ \sin(2t)\rangle$. What is $\mathbf{r}'(t)$?
   Answer: $\langle 15(3t-1)^4,\ 3t^2e^{t^3},\ 2\cos(2t)\rangle$ This final problem combines the chain rule with exponential and trigonometric differentiation in vector form. For $\mathbf{r}(t)=\langle (3t-1)^5,\ e^{t^3},\ \sin(2t)\rangle$, the first component uses the chain rule: $((3t-1)^5)' = 5(3t-1)^4 \cdot 3 = 15(3t-1)^4$, the second component also uses the chain rule: $(e^{t^3})' = e^{t^3} \cdot 3t^2 = 3t^2e^{t^3}$, and the third component gives $(\sin(2t))' = \cos(2t) \cdot 2 = 2\cos(2t)$. Choice E incorrectly shows $-2\sin(2t)$ for the third component, confusing the derivative of sine with the derivative of cosine. The universal strategy for vector derivatives is to differentiate each component as a scalar function, applying all necessary rules including the chain rule for composite functions.
9. An object's position is $r(t)=\langle t\cos t,\ t\sin t,\ t^2\rangle$. What is $r'(t)$?
   Answer: $\langle \cos t- t\sin t,\ \sin t+ t\cos t,\ 2t\rangle$ This vector function requires the product rule for each component since they all contain products. For t·cos t, we get d/dt[t·cos t] = 1·cos t + t·(-sin t) = cos t - t·sin t. Similarly, t·sin t differentiates to sin t + t·cos t. The third component t² simply becomes 2t. Choice B reverses the signs in the first two components, showing cos t + t·sin t and sin t - t·cos t, which results from sign errors in applying the product rule. To differentiate vector functions accurately, apply the product rule carefully to each component, keeping track of all signs.
10. A car's position is $r(t)=\langle (3t-1)^4,\ \cos t+ t,\ \sin(2t)\rangle$. What is $r'(t)$?
    Answer: $\langle 12(3t-1)^3,\ -\sin t+1,\ 2\cos(2t)\rangle$ This problem tests vector differentiation with various function types in each component. For (3t-1)⁴, the chain rule gives d/dt[(3t-1)⁴] = 4(3t-1)³·3 = 12(3t-1)³. The second component cos t + t differentiates to -sin t + 1. The third component sin(2t) becomes 2cos(2t) by the chain rule. Choice A incorrectly shows 4(3t-1)³ for the first component, forgetting to multiply by 3 (the derivative of 3t-1). When differentiating vector functions, ensure every application of the chain rule includes all necessary factors from inner function derivatives.
11. A puck moves with $\mathbf{r}(t)=\langle (t-1)^5,\ \cos t+\sin t,\ e^{-t}\rangle$; determine $\mathbf{r}'(t)$.
    Answer: $\langle 5(t-1)^4,\ -\sin t+\cos t,\ -e^{-t}\rangle$ For $\mathbf{r}(t)=\langle (t-1)^5,\ \cos t+\sin t,\ e^{-t}\rangle$, we differentiate component by component. Using the chain rule, the derivative of $(t-1)^5$ is $5(t-1)^4 \cdot 1 = 5(t-1)^4$. The derivative of $\cos t + \sin t$ is $-\sin t + \cos t$. The derivative of $e^{-t}$ is $e^{-t} \cdot (-1) = -e^{-t}$. Therefore, $\mathbf{r}'(t)=\langle 5(t-1)^4,\ -\sin t+\cos t,\ -e^{-t}\rangle$. Choice E incorrectly gives $e^{-t}$ instead of $-e^{-t}$ for the third component, missing the negative sign from the chain rule. When differentiating $e^{-t}$, remember to multiply by the derivative of $-t$, which is $-1$.
12. A drone's position is $\mathbf{r}(t)=\langle t^2-3t,\ \sin t,\ e^{2t}\rangle$ meters; what is $\mathbf{r}'(t)$?
    Answer: $\langle 2t-3,\ \cos t,\ 2e^{2t}\rangle$ 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.
13. A camera dolly follows $\mathbf{r}(t)=\langle (\sin t)^2,\ (\cos t)^2,\ t\sin t\rangle$; determine $\mathbf{r}'(t)$.
    Answer: $\langle 2\sin t\cos t,\ -2\sin t\cos t,\ t\cos t+\sin t\rangle$ 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.
14. A puck moves as $\mathbf{r}(t)=\langle \frac{1}{t},\ \arctan t,\ t^4\rangle$ for $t\ne0$; compute $\mathbf{r}'(t)$.
    Answer: $\langle -\tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$ 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.
15. A particle moves with $r(t)=\langle t^5,\ e^t\sin t,\ \frac{t}{t+1}\rangle$. Determine $r'(t)$.
    Answer: $\langle 5t^4,\ e^t(\sin t+\cos t),\ \frac{1}{(t+1)^2}\rangle$ Vector differentiation here involves power, product, and quotient rules across components. The first component t⁵ simply becomes 5t⁴. For $e^t$·sin t, we need the product rule: d/dt[$e^t$·sin t] = $e^t$·sin t + $e^t$·cos t = $e^t$(sin t + cos t). The third component t/(t+1) requires the quotient rule: d/dt[t/(t+1)] = [(t+1)·1 - t·1]/(t+1)² = 1/(t+1)². Choice E incorrectly computes the product rule as $e^t$(sin t - cos t), using a minus sign instead of plus. Vector differentiation demands careful attention to signs when applying product and quotient rules.
16. A ball's position is $\mathbf{r}(t)=\langle (\sin t)^2,\ (\cos t)^2,\ t\sin t\rangle$. What is $\mathbf{r}'(t)$?
    Answer: $\langle 2\sin t\cos t,\ -2\sin t\cos t,\ \sin t+t\cos t\rangle$ This problem involves differentiating squared trigonometric functions and a product. For (sin t)² = sin²t, we use the chain rule: 2sin t·cos t. For (cos t)² = cos²t, we get 2cos t·(-sin t) = -2sin t cos t. For t sin t, we apply the product rule: 1·sin t + t·cos t = sin t + t cos t. Thus r'(t) = ⟨2sin t cos t, -2sin t cos t, sin t + t cos t⟩. Choice A incorrectly simplifies 2sin t cos t as just 2sin t, missing the cosine factor from the chain rule. Remember that differentiating sin²t or cos²t always produces a product of sine and cosine terms.
17. A particle has $\mathbf{r}(t)=\langle t^2\cos t,\ t^2\sin t,\ 5t\rangle$; what is $\mathbf{r}'(t)$?
    Answer: $\langle 2t\cos t-t^2\sin t,\ 2t\sin t+t^2\cos t,\ 5\rangle$ 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 t² cos t differentiates to 2t cos t - t² sin t using the product rule. The second component t² sin t becomes 2t sin t + t² cos t also via the product rule, and the third 5t differentiates to 5. A tempting distractor like choice B flips the signs in the product rule terms, leading to incorrect expressions. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
18. A robot arm tip is at $r(t)=\langle (t^2+1)^5,\ \sqrt{t},\ \tan t\rangle$. What is $r'(t)$?
    Answer: $\langle 10t(t^2+1)^4,\ \frac{1}{2\sqrt{t}},\ \sec^2 t\rangle$ This problem tests vector differentiation with more complex component functions. For the first component (t²+1)⁵, we use the chain rule: d/dt[(t²+1)⁵] = 5(t²+1)⁴·2t = 10t(t²+1)⁴. The second component √t = $t^{1/2}$ differentiates to (1/2)$t^{-1/2}$ = 1/(2√t). The third component tan t has derivative sec²t. Choice B incorrectly differentiates the first component as 5(t²+1)⁴, forgetting to multiply by the derivative of the inner function (2t). Remember that vector differentiation combines component-wise differentiation with careful application of all relevant calculus rules.
19. A particle moves with $\mathbf{r}(t)=\langle \ln t,\ t^3,\ \sqrt{t}\rangle$ for $t>0$; find $\mathbf{r}'(t)$.
    Answer: $\langle \tfrac{1}{t},\ 3t^2,\ \tfrac{1}{2\sqrt{t}}\rangle$ 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 $\ln t$ differentiates to $1/t$ using the derivative of the natural logarithm. The second component $t^3$ becomes $3t^2$ by the power rule, and the third $\sqrt{t}$ differentiates to $1/(2\sqrt{t})$ applying the chain rule. A tempting distractor like choice D confuses the derivative of $t^3$ with its antiderivative, resulting in $t^3$ instead of $3t^2$. Always remember that the derivative of a vector-valued function is obtained by differentiating each of its scalar component functions independently.
20. A swimmer's position is $\mathbf{r}(t)=\langle t\sin t,\ t\cos t,\ t^2\rangle$; what is $\mathbf{r}'(t)$?
    Answer: $\langle \sin t+t\cos t,\ \cos t-t\sin t,\ 2t\rangle$ 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.

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