Determining the velocity function

[doktorwho]

Homework Statement

Given the $r(t) = ae^{kt}$ , $θ(t)=kt$ find the velocity function that is dependent on $r$.
$v(r)=?$

Homework Equations

3. The Attempt at a Solution [/B]
My attempt:
1)$r(t) = ae^{kt}$
2)${\dot r(t)} = ake^{kt}$
From the first equation:
$\ln {\frac{r(t)}{a}}=\ln e^{kt}$
$\ln {\frac{r(t)}{a}}=kt$
$t=\frac{\ln {\frac{r(t)}{a}}}{k}$
Replacing the $t$ in the second equation i get:
${\dot r}=akr$
Shouldn't this be the answer? In the answers it says ${\dot r}=\sqrt2r$?

 

[Chestermiller]

Your original equations are incorrect. $\theta$ is not a vectorl where are your unit vetors i these equations?

 

[doktorwho]

Your original equations are incorrect. $\theta$ is not a vectorl where are your unit vetors i these equations?

Yeah, no vectors, just the parametric equations of motion given. So what's wrong now?

 

[Chestermiller]

Yeah, no vectors, just the parametric equations of motion given. So what's wrong now?

If you are going to determine the velocity vector, you need to start out by expressing the position vector as $\vec{r}=r\vec{i}_r(\theta)$ and taking into account the fact that $\vec{i}_r$ is a function of $\theta$, that $\theta$ is a function of time, and that derivative of $\vec{i}_r$ with respect to $\theta$ can be expressed in terms of $\vec{i}_{\theta}$.

 

[doktorwho]

If you are going to determine the velocity vector, you need to start out by expressing the position vector as $\vec{r}=r\vec{i}_r(\theta)$ and taking into account the fact that $\vec{i}_r$ is a function of $\theta$, that $\theta$ is a function of time, and that derivative of $\vec{i}_r$ with respect to $\theta$ can be expressed in terms of $\vec{i}_{\theta}$.

So the polar coordinate,
$\vec r(t)=ae^{kt}\vec e_r$
$θ=kt$
$\vec v(t)=\dot r\vec e_r + r\dot θ\vec e_θ$
$\vec v(t)=ake^{kt}\vec e_r + ae^{kt}k\vec e_θ$
$v(r)=\sqrt2r$
This should be it.