How Does Velocity Change with Time in a Spiral Path?

[roam]

Homework Statement

http://img534.imageshack.us/img534/6164/questionv.jpg

Homework Equations

ω = dθ/dt

v = dS/dt

The Attempt at a Solution

I used the second expression for the tangential speed:

$v = \frac{dS}{dt} = r \frac{d \dot{\theta}}{d t} = (b-ct) \frac{d(kt)}{dt}$

$\therefore \ v(t) = (b-ct) k$

So is this a correct expression for speed as a function of time?

So when r=0, the velocity would also be 0?

 

[pcm]

in polar coordinates v²=(dr/dt)²+(rd$\theta$/dt)²

 

[pcm]

$v = \frac{dS}{dt} = r \frac{d \dot{\theta}}{d t} = (b-ct) \frac{d(kt)}{dt}$

is it dimensionally correct?

 

[roam]

I just substituted the rate of change of radial distance into the equation v = r.dθ/dt, and then I differentiated it. What's wrong with that?

 

[roam]

I think it's dimensionally correct. Why? Did I use the wrong equations?

 

[Vipho]

I just substituted the rate of change of radial distance into the equation v = r.dθ/dt, and then I differentiated it. What's wrong with that?

In this case, I think this formula isn't correct.
So, $\vec{r}=r\cos(\theta)\hat{x}+r\sin(\theta)\hat{y}<br /> \\<br /> \vec{v}=\frac{d\vec{r}}{dt}<br /> \\<br /> v_x=\frac{d(r\cos(\theta))}{dt}=\cos(\theta)\frac{dr}{dt}+r\frac{d(\cos\theta)}{dt}=\cos(\theta) \frac{dr}{dt}-r\sin\theta \frac{d\theta}{dt}<br /> \\<br /> v_y=\frac{d(r\sin(\theta))}{dt}=\sin(\theta)\frac{dr}{dt}+r\cos\theta\frac{d\theta}{dt}<br /> \\<br /> v^2=(v_x)^2+(v_y)^2=(\frac{dr}{dt})^2+(r\frac{d \theta}{dt})^2<br />$