$\text { Find the velocity vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. }$
-$\mathrm { r } = \mathrm { a } ( 2 - \cos \theta )$ and $\frac { \mathrm { d } \theta } { \mathrm { dt } } = 7$
A) $\mathbf { v } = ( 7 a \sin \theta ) \mathbf { u } _ { \mathrm { r } } + 7 \mathrm { a } ( 2 - \cos \theta ) \mathbf { u } _ { \theta }$
B) $\mathbf { v } = ( 7 a \cos \theta ) \mathbf { u } _ { \mathrm { r } } + 7 \mathrm { a } ( 2 - \sin \theta ) \mathbf { u } _ { \theta }$
C) $\mathbf { v } = 7 a ( 2 - \cos \theta ) \mathbf { u } _ { \mathbf { r } } + ( 7 \mathrm { a } \sin \theta ) \mathbf { u } _ { \theta }$
D) $\mathbf { v } = ( 2 a \sin \theta ) \mathbf { u } _ { \mathrm { r } } + 2 \mathrm { a } ( 7 - \cos \theta ) \mathbf { u } _ { \theta }$

Question

Quizplus · Accepted Answer

The velocity vector in polar coordinates is given by $\mathbf{v} = \dot{r} \mathbf{u}_r + r\dot{\theta} \mathbf{u}_\theta$, where $\dot{r}$ is the derivative of $r$ with respect to time and $\dot{\theta}$ is the derivative of $\theta$ with respect to time. Given $r = a(2 - \cos\theta)$ and $\frac{d\theta}{dt} = 7$, we find $\dot{r} = 7a\sin\theta$ (by differentiating $r$ with respect to $\theta$ and then multiplying by $\frac{d\theta}{dt}$). Thus, the radial component of velocity is $7a\sin\theta \mathbf{u}_r$, and the angular component is $7a(2 - \cos\theta) \mathbf{u}_\theta$, making option A correct.

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​.

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