The position vector of a particle is r(t). Find the requested vector.
-The acceleration at $t = \frac { \pi } { 8 }$ for $\mathbf { r } ( \mathrm { t } ) = \left( 10 \mathrm { t } - 3 \mathrm { t } ^ { 3 } \right) \mathbf { i } + 2 \tan ( 2 \mathrm { t } ) \mathbf { j } + \mathrm { e } ^ { 5 } \mathrm { t } \mathbf { k }$
A) $a \left( \frac { \pi } { 8 } \right) = - \frac { 9 } { 4 } \pi i \mathbf { i } - 32 \mathbf { j } + 25 \mathbf { k }$
B) $a \left( \frac { \pi } { 8 } \right) = \frac { 9 } { 4 } \pi \mathbf { i } + 32 \mathbf { j } + 25 \mathrm { e } ^ { 5 / 8 } \pi \mathbf { k }$
C) $a \left( \frac { \pi } { 8 } \right) = - \frac { 9 } { 4 } \pi \mathrm { i } + 32 \mathrm { j } + 25 \mathrm { e } ^ { 5 / 8 \pi } \mathbf { k }$
D) $a \left( \frac { \pi } { 8 } \right) = - \frac { 9 } { 4 } \pi i - 32 j + 25 e ^ { 5 / 8 } \pi k$

Question

Quizplus · Accepted Answer

The acceleration vector is the second derivative of the position vector. For $\mathbf{r}(t)$, the second derivative is $-6t \mathbf{i} + 8 \sec^2(2t) \mathbf{j} + 25 \mathbf{k}$. Evaluating at $t = \frac{\pi}{8}$, we get $-\frac{9}{4}\pi \mathbf{i} + 32 \mathbf{j} + 25 e^{5/8 \pi} \mathbf{k}$.

The Position Vector of a Particle Is R(t) t=π8t = \frac { \pi } { 8 }t=8π​

The Position Vector of a Particle Is R(t) $t = \frac { \pi } { 8 }$