Let $\mathbf { r } ( t ) = e ^ { \cos t } \mathbf { i } + \tan t \mathbf { j } + \cos ( 5 - t ) \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is
A) $\sin t e ^ { \cos t } \mathbf { i } + \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
B) $- \sin t e ^ { \cos t } \mathbf { i } + \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
C) $- \sin t e ^ { \cos t } \mathbf { i } - \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
D) $\sin t e ^ { \cos t } \mathbf { i } - \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
E) $\sin t e ^ { \cos t } \mathbf { i } + \sec ^ { 2 } t \mathbf { j } - \sin ( 5 - t ) \mathbf { k }$

Question

Quizplus · Accepted Answer

To find $\mathbf { r } ^ { \prime } ( t )$, we differentiate each component of $\mathbf { r } ( t )$ with respect to $t$:- For the $\mathbf{i}$ component, the derivative of $e^{\cos t}$ is $-\sin t e^{\cos t}$ using the chain rule.- For the $\mathbf{j}$ component, the derivative of $\tan t$ is $\sec^2 t$.- For the $\mathbf{k}$ component, the derivative of $\cos(5 - t)$ is $\sin(5 - t)$ but with a negative sign due to the chain rule, making it $\sin(t - 5)$, which is equivalent to $\sin(5 - t)$ in magnitude but opposite in sign. However, the correct sign is positive as in the original statement, indicating a mistake in my explanation regarding the sign. The correct derivative, considering the chain rule correctly, should indeed result in a positive sign for the $\mathbf{k}$ component, aligning with the options provided. Thus, the correct derivative is $\sin(5 - t)$, maintaining the positive sign as per the chain rule applied correctly to the derivative of $\cos(5 - t)$, which involves multiplying by the derivative of the inside function, $-1$, resulting in $-\sin(5 - t)$, but this contradicts the correct interpretation of the derivative as presented in the options. The correct interpretation should acknowledge the derivative of $\cos(5 - t)$ as $\sin(5 - t)$ with a negative sign due to the derivative of $-t$ being $-1$, which correctly results in $-\sin(5 - t)$, aligning with the principles of differentiation and the chain rule.

Let r(t)=ecos⁡ti+tan⁡tj+cos⁡(5−t)k\mathbf { r } ( t ) = e ^ { \cos t } \mathbf { i } + \tan t \mathbf { j } + \cos ( 5 - t ) \mathbf { k }r(t)=ecosti+tantj+cos(5−t)k

Let $\mathbf { r } ( t ) = e ^ { \cos t } \mathbf { i } + \tan t \mathbf { j } + \cos ( 5 - t ) \mathbf { k }$