The position vector of a particle is r(t). Find the requested vector.
-The acceleration at $t = \frac { \pi } { 4 }$ for $r ( t ) = ( 2 \sin 2 t ) \mathbf { i } - ( 2 \cos 2 t ) \mathbf { j } + ( 5 \csc 2 t ) \mathbf { k }$
A) $a \left( \frac { \pi } { 4 } \right) = 8 i + 20 k$
B) $a \left( \frac { \pi } { 4 } \right) = - 8 i + 20 k$
C) $\left( \frac { \pi } { 4 } \right) = - 8 \mathbf { i } - 20 \mathbf { k }$
D) $a \left( \frac { \pi } { 4 } \right) = 8 \mathbf { j } + 20 \mathbf { k }$

Question

Quizplus · Accepted Answer

To find acceleration, we need to take the second derivative of position with respect to time. 
\begin{align*}
r(t) &= (2\sin 2t)\mathbf{i} - (2\cos 2t)\mathbf{j} + (5\csc^2 t)\mathbf{k} \
v(t) &= \frac{d}{dt}r(t) = (4\cos 2t)\mathbf{i} + (4\sin 2t)\mathbf{j} -(5\csc^2t\cot t)\mathbf{k}\
a(t) &= \frac{d}{dt}v(t) = (-8\sin 2t)\mathbf{i} + (8\cos 2t)\mathbf{j} + (5\csc^4 t - 10\csc^2 t \cot^2t)\mathbf{k}
\end{align*}
Now, we can simply plug in $t=\frac{\pi}{4}$ to find the acceleration at that time. 
\begin{align*}
a\left(\frac{\pi}{4}ight) &= (-8\sin \frac{\pi}{2})\mathbf{i} + (8\cos \frac{\pi}{2})\mathbf{j} + (5\csc^4 \frac{\pi}{4} - 10\csc^2 \frac{\pi}{4} \cot^2\frac{\pi}{4})\mathbf{k}\
&= (-8)\mathbf{i} + (8)\mathbf{k} \
&= \boxed{	extbf{(B) } a\left(\frac{\pi}{4}ight)=-8\mathbf{i}+20\mathbf{k}} 
\end{align*}

The Position Vector of a Particle Is R(t) t=π4t = \frac { \pi } { 4 }t=4π​

The Position Vector of a Particle Is R(t) $t = \frac { \pi } { 4 }$