Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } - e ^ { \cos t } \mathbf { j } + \ln t \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

A) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
B) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
C) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
D) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
E) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$

Question

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } - e ^ { \cos t } \mathbf { j } + \ln t \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

A) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
B) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
C) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
D) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
E) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$

Quizplus · Accepted Answer

To find $\mathbf { r } ^ { \prime \prime } ( t )$, we first differentiate $\mathbf { r } ( t )$ to get $\mathbf { r } ^ { \prime } ( t )$, and then differentiate $\mathbf { r } ^ { \prime } ( t )$ again. For the $\mathbf{i}$ component, differentiating $e^{\sin t}$ once gives $e^{\sin t} \cos t$, and differentiating again gives $e^{\sin t}(\cos^2 t - \sin t)$ due to the product rule and chain rule.For the $\mathbf{j}$ component, differentiating $-e^{\cos t}$ once gives $e^{\cos t} \sin t$, and differentiating again gives $e^{\cos t}(\cos t - \sin^2 t)$, applying the product rule and chain rule similarly.For the $\mathbf{k}$ component, differentiating $\ln t$ once gives $\frac{1}{t}$, and differentiating again gives $-\frac{1}{t^2}$.Thus, the correct answer is $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$, which matches choice A.

Let r(t)=esin⁡ti−ecos⁡tj+ln⁡tk\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } - e ^ { \cos t } \mathbf { j } + \ln t \mathbf { k }r(t)=esinti−ecostj+lntk

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