$\text { Find the velocity vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. }$
-$r = 2 \cos 2 t$ and $\theta = 4 t$
A) $\mathbf { v } = ( - 8 \sin 4 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } + 4 \cos 4 \mathrm { tu } _ { \theta }$
B) $\mathbf { v } = ( 4 \sin 2 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } + 8 \cos 2 \mathrm { tu } _ { \theta }$
C) $\mathbf { v } = ( - 4 \sin 2 \mathrm { t } ) \mathrm { u } _ { \mathrm { r } } + 8 \cos 2 \mathrm { tu } _ { \theta }$
D) $v = ( - 2 \sin 2 t ) u _ { r } + 8 \cos 2 t u _ { \theta }$

Question

Quizplus · Accepted Answer

Given, 
$r = 2 \cos 2t$ and $	heta = 4t$

To find the velocity vector, we differentiate both $r$ and $	heta$ with respect to time $t$.

$\dfrac {dr}{dt} = -4 \sin 2t$ and $\dfrac {d 	heta}{dt} = 4$

The velocity vector is given by

$\vec{v} = \left(\dfrac {dr}{dt}ight)\vec{u_r} + r\left(\dfrac {d 	heta}{dt}ight)\vec{u_	heta}$

Substituting the values, we get

$\vec{v} = (-4\sin2t) \vec{u_r} + 8 \cos2t \vec{u_	heta}$

Rewriting in terms of $u_r$ and $u_	heta$ we get,

$\vec{v} = (-4\sin2t) u_r + 8 \cos2t u_	heta$

Therefore, the answer is (C).

Find the velocity vector in terms of ur and uθ. \text { Find the velocity vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. } Find the velocity vector in terms of ur​ and uθ​.

$\text { Find the velocity vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. }$