AP Calculus BC Quiz: Defining And Differentiating Vector Valued Functions

What this quiz covers

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

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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. $\langle \tfrac{1}{t},\ 2t\cos t+t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$
2. $\langle \tfrac{1}{t},\ 2t\cos t-t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$ (correct answer)
3. $\langle \ln t,\ 2t\cos t-t^2\sin t,\ \tfrac{1}{2\sqrt{t}}\rangle$
4. $\langle \tfrac{1}{t},\ 2\cos t-t\sin t,\ \sqrt{t}\rangle$
5. $\langle -\tfrac{1}{t},\ 2t\cos t-t^2\sin t,\ \tfrac{1}{\sqrt{t}}\rangle$

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.

Question 2

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

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.

Question 3

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

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.

Question 4

1. $\langle \sec t\tan t,\ -\csc t\cot t,\ 2\cos(2t)\rangle$ (correct answer)
2. $\langle \sec^2 t,\ -\csc^2 t,\ 2\cos(2t)\rangle$
3. $\langle \sec t\tan t,\ \csc t\cot t,\ 2\cos(2t)\rangle$
4. $\langle \sec t\tan t,\ -\csc t\cot t,\ \cos(2t)\rangle$
5. $\langle \sec t\tan t,\ -\csc t\cot t,\ 2\sin(2t)\rangle$

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.

Question 5

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

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.

Question 6

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

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.

Question 7

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

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.

Question 8

1. $\langle -\tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$ (correct answer)
2. $\langle \tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$
3. $\langle -\tfrac{1}{t^2},\ \tfrac{1}{1+t^2},\ t^4\rangle$
4. $\langle -\tfrac{1}{t},\ \tfrac{1}{1+t^2},\ 4t^3\rangle$
5. $\langle -\tfrac{1}{t^2},\ \tfrac{1}{1-t^2},\ 4t^3\rangle$

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.

Question 9

1. $\langle e^{t}\cos t,\ e^{t}\sin t,\ 0\rangle$
2. $\langle e^{t}(\cos t-\sin t),\ e^{t}(\sin t+\cos t),\ 0\rangle$ (correct answer)
3. $\langle e^{t}(\cos t+\sin t),\ e^{t}(\sin t-\cos t),\ 0\rangle$
4. $\langle e^{t}(\sin t-\cos t),\ e^{t}(\cos t+\sin t),\ 0\rangle$
5. $\langle e^{t}(\cos t-\sin t),\ e^{t}(\sin t+\cos t),\ 7\rangle$

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.

Question 10

1. $\langle 12(3t-1)^3,\ -\sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$ (correct answer)
2. $\langle 4(3t-1)^3,\ -\sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$
3. $\langle 12(3t-1)^3,\ -\sin t-\cos t,\ \tfrac{1}{t^2+1}\rangle$
4. $\langle 12(3t-1)^3,\ \sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$
5. $\langle 12(3t-1)^4,\ -\sin t+\cos t,\ \tfrac{2t}{t^2+1}\rangle$

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.

Question 11

1. $\langle \sec t\tan t,\ -\csc t\cot t,\ 1\rangle$ (correct answer)
2. $\langle \sec^2 t,\ -\csc^2 t,\ 1\rangle$
3. $\langle \sec t\tan t,\ \csc t\cot t,\ 1\rangle$
4. $\langle \sec t\cot t,\ -\csc t\tan t,\ 1\rangle$
5. $\langle \sec t\tan t,\ -\csc t\cot t,\ t\rangle$

Explanation: For $\mathbf{r}(t)=\langle \sec t,\ \csc t,\ t\rangle$, we need to recall the derivatives of secant and cosecant. The derivative of $\sec t$ is $\sec t \tan t$. The derivative of $\csc t$ is $-\csc t \cot t$ (note the negative sign). The derivative of $t$ is 1. Therefore, $\mathbf{r}'(t)=\langle \sec t\tan t,\ -\csc t\cot t,\ 1\rangle$. Choice C incorrectly omits the negative sign in the derivative of cosecant, giving $\csc t \cot t$ instead of $-\csc t \cot t$. Always remember that the derivative of $\csc t$ includes a negative sign: $(\csc t)' = -\csc t \cot t$.

Question 12

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

Explanation: This problem involves differentiating a vector function where each component requires the product rule. For $\mathbf{r}(t)=\langle (t^2+1)\cos t,\ (t^2+1)\sin t,\ t\rangle$, the first component's derivative is $(2t)\cos t + (t^2+1)(-\sin t) = 2t\cos t - (t^2+1)\sin t$. The second component's derivative is $(2t)\sin t + (t^2+1)(\cos t) = 2t\sin t + (t^2+1)\cos t$. The third component's derivative is simply 1. Therefore, $\mathbf{r}'(t)=\langle 2t\cos t-(t^2+1)\sin t,\ 2t\sin t+(t^2+1)\cos t,\ 1\rangle$. Choice A has incorrect signs in both the first and second components, likely from sign errors when applying the product rule. When using the product rule $(uv)' = u'v + uv'$, carefully track the signs of each term.

Question 13

1. $\langle e^t+t e^t,\ \sec^2 t,\ 0\rangle$ (correct answer)
2. $\langle t e^t,\ \sec^2 t,\ 0\rangle$
3. $\langle e^t+t e^t,\ \sec t,\ 0\rangle$
4. $\langle e^t+t e^t,\ \sec^2 t,\ 5\rangle$
5. $\langle e^{t-1}+t e^t,\ \sec^2 t,\ 0\rangle$

Explanation: For $\mathbf{r}(t)=\langle t\,e^t,\ \tan t,\ 5\rangle$, we differentiate each component separately. The first component requires the product rule: $(t\,e^t)' = 1 \cdot e^t + t \cdot e^t = e^t + te^t = e^t(1+t)$. The derivative of $\tan t$ is $\sec^2 t$. The derivative of the constant 5 is 0. Therefore, $\mathbf{r}'(t)=\langle e^t+t e^t,\ \sec^2 t,\ 0\rangle$. Choice D incorrectly gives 5 as the third component, forgetting that the derivative of a constant is zero. Remember that constant components in vector functions always have zero derivatives.

Question 14

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

Explanation: This problem requires differentiating a vector-valued function to find the puck's velocity. The x-component is t sin t, and by the product rule, its derivative is sin t + t cos t. The y-component is t cos t, differentiating to cos t - t sin t. The z-component is arctan t, with derivative 1/(1 + $t^2$). A tempting distractor is choice D, which has +t sin t for the y-component, possibly from missing 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.

Question 15

1. $\langle e^t(\cos t-\sin t),\ e^t(\sin t+\cos t),\ 1\rangle$ (correct answer)
2. $\langle e^t(\cos t+\sin t),\ e^t(\sin t-\cos t),\ 1\rangle$
3. $\langle e^t\cos t,\ e^t\sin t,\ 1\rangle$
4. $\langle e^t(\cos t-\sin t),\ e^t(\sin t+\cos t),\ t\rangle$
5. $\langle e^t(\cos t-\sin t),\ e^t(\sin t-\cos t),\ 0\rangle$

Explanation: 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.

Question 16

1. $\langle 6t(t^2+1)^2,\ \sec^2 t,\ -10t^{-3}\rangle$ (correct answer)
2. $\langle 3(t^2+1)^2,\ \sec^2 t,\ -10t^{-3}\rangle$
3. $\langle 6t(t^2+1)^2,\ \sec t,\ -10t^{-3}\rangle$
4. $\langle 6t(t^2+1)^3,\ \sec^2 t,\ -10t^{-3}\rangle$
5. $\langle 6t(t^2+1)^2,\ \sec^2 t,\ 10t^{-3}\rangle$

Explanation: This problem involves differentiating a vector-valued function for the robot arm's velocity. The x-component is (t^2 + 1)^3, and by the chain rule, its derivative is 3(t^2 + 1)^2 * 2t or 6t(t^2 + 1)^2. The y-component is tan t, with derivative sec^2 t. The z-component is 5 t^{-2}, differentiating to -10 t^{-3}. A tempting distractor is choice B, which lacks the 2t factor in the x-component, likely from forgetting the inner derivative in the chain rule. In general, to find the derivative of a vector-valued function, apply the appropriate differentiation rules to each component function independently.

Question 17

1. $\langle -4\sin t,\ 4\cos t,\ 2t\rangle$ (correct answer)
2. $\langle 4\sin t,\ -4\cos t,\ 2t\rangle$
3. $\langle -4\sin t,\ 4\cos t,\ t^2\rangle$
4. $\langle -4\cos t,\ 4\sin t,\ 2t\rangle$
5. $\langle -4\sin t,\ 4\cos t,\ 2\rangle$

Explanation: This problem requires differentiating a vector-valued function to find the boat's velocity vector. The x-component is 4 cos t, with derivative -4 sin t. The y-component is 4 sin t, differentiating to 4 cos t. The z-component is t^2, and its derivative is 2t using the power rule. A tempting distractor is choice E, which has a constant 2 for the z-component, perhaps from forgetting to differentiate t^2 entirely. In general, to find the derivative of a vector-valued function, apply the appropriate differentiation rules to each component function independently.

Question 18

1. $\langle \tfrac{1}{t},\ -2t^{-3},\ \tfrac{1}{2\sqrt{t}}\rangle$ (correct answer)
2. $\langle \tfrac{1}{\ln t},\ -2t^{-1},\ \tfrac{1}{2\sqrt{t}}\rangle$
3. $\langle \tfrac{1}{t},\ 2t^{-3},\ \tfrac{1}{\sqrt{t}}\rangle$
4. $\langle \tfrac{1}{t},\ -2t^{-2},\ \tfrac{1}{2\sqrt{t}}\rangle$
5. $\langle \tfrac{1}{t},\ -2t^{-3},\ \sqrt{t}\rangle$

Explanation: To find the derivative of this vector-valued function, we differentiate each component independently. For $\mathbf{r}(t)=\langle \ln t,\ t^{-2},\ \sqrt{t}\rangle$, the first component gives $(\ln t)' = \frac{1}{t}$, the second component gives $(t^{-2})' = -2t^{-3}$ using the power rule, and the third component gives $(\sqrt{t})' = (t^{1/2})' = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}}$. Choice D incorrectly has $-2t^{-2}$ for the second component, missing the correct application of the power rule where we multiply by the original exponent. Remember that when differentiating vector functions, apply standard calculus rules to each component separately.

Question 19

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

Explanation: To differentiate this vector function, we apply standard rules to each component separately. For $\mathbf{r}(t)=\langle e^{-t},\ t^5-2t,\ \arctan t\rangle$, the first component uses the chain rule: $(e^{-t})' = e^{-t} \cdot (-1) = -e^{-t}$, the second component gives $(t^5-2t)' = 5t^4-2$, and the third component gives $(\arctan t)' = \frac{1}{1+t^2}$. Choice B incorrectly has $e^{-t}$ instead of $-e^{-t}$ for the first component, missing the negative sign from the chain rule. The strategy for vector derivatives is to methodically apply differentiation rules to each component, being especially careful with chain rule applications.

Question 20

1. $\langle -\sin(3t),\ \cos(3t),\ 2t\rangle$
2. $\langle -3\sin(3t),\ 3\cos(3t),\ 2t\rangle$ (correct answer)
3. $\langle -3\sin(3t),\ 3\cos(3t),\ t^2\rangle$
4. $\langle 3\sin(3t),\ 3\cos(3t),\ 2t\rangle$
5. $\langle -3\cos(3t),\ 3\sin(3t),\ 2t\rangle$

Explanation: This problem involves differentiating a vector function with trigonometric and polynomial components. For $\mathbf{r}(t)=\langle \cos(3t),\ \sin(3t),\ t^2\rangle$, we apply the chain rule to the first two components: $(\cos(3t))' = -\sin(3t) \cdot 3 = -3\sin(3t)$ and $(\sin(3t))' = \cos(3t) \cdot 3 = 3\cos(3t)$, while the third component simply gives $(t^2)' = 2t$. Choice A incorrectly omits the factor of 3 from the chain rule in the trigonometric derivatives, showing $-\sin(3t)$ instead of $-3\sin(3t)$. When differentiating composite functions in vector components, always remember to multiply by the derivative of the inner function.