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.

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.

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.

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.

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.

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.

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.

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.

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.

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.