The position vector of a particle is r(t). Find the requested vector.
-The velocity at $t = \frac { \pi } { 4 }$ for $\mathbf { r } ( \mathrm { t } ) = 4 \sec ^ { 2 } ( \mathrm { t } ) \mathbf { i } - 5 \tan ( \mathrm { t } ) \mathbf { j } + 10 \mathrm { t } ^ { 2 } \mathbf { k }$
A) $\mathbf { v } \left( \frac { \pi } { 4 } \right) = - 10 \mathbf { j } - 5 \pi \mathbf { k }$
B) $\mathbf { v } \left( \frac { \pi } { 4 } \right) = - 10 \mathbf { j } + 5 \pi \mathbf { k }$
C) $\mathbf { v } \left( \frac { \pi } { 4 } \right) = 16 \mathbf { i } - 10 \mathbf { j } + 5 \pi \mathbf { k }$
D) $\mathbf { v } \left( \frac { \pi } { 4 } \right) = 16 \mathbf { i } + 10 \mathbf { j } - 5 \pi \mathbf { k }$

Question

Quizplus · Accepted Answer

To find the velocity vector, we differentiate the position vector.

$$\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt}(4\sec^2(t)\mathbf{i}-5	an(t)\mathbf{j}+10t^2\mathbf{k})$$$$=8\sec(t)	an(t)\mathbf{i}-5\sec^2(t)\mathbf{j}+20t\mathbf{k}$$

Then, we substitute $t=\frac{\pi}{4}$ to get the velocity at that time.

$$\mathbf{v}\left(\frac{\pi}{4}ight)=8\sec\left(\frac{\pi}{4}ight)	an\left(\frac{\pi}{4}ight)\mathbf{i}-5\sec^2\left(\frac{\pi}{4}ight)\mathbf{j}+20\left(\frac{\pi}{4}ight)\mathbf{k}$$$$=8(1)(1)\mathbf{i}-5(2)\mathbf{j}+5\pi\mathbf{k}=16\mathbf{i}-10\mathbf{j}+5\pi\mathbf{k}$$

Therefore, the correct choice is (C).

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) $t = \frac { \pi } { 4 }$