Find derivative of ((rdot^2) + ((thetadot * r)^2))^(1/2)

1. Jun 2, 2009

toastie

1. The problem statement, all variables and given/known data
i have rdot which is equal to the dr/dt and I have thetadot which is equal to d(theta)/dt.
I need to find the derivative of ((rdot^2)+((thetadot*r)^2))^(1/2)

2. Relevant equations

3. The attempt at a solution
I believe that I am just making errors in my derivation, but I keep getting [2(dr^2/d^2t)+2thetadot*r+2((dtheta^2)/dt)*r]/((rdot^2)+((thetadot*r)^2))^(1/2)

2. Jun 2, 2009

Cyosis

Re: Derivative

Using the chainrule.

$$\frac{d \dot{r}^2}{dt}=\frac{d \dot{r}^2}{d \dot{r}} \frac{d \dot{r}}{dt}=2 \dot{r} \frac{d \dot{r}}{dt}=2\dot{r}\ddot{r}$$

3. Jun 2, 2009

toastie

Re: Derivative

okay. is there any other mistakes in my derivation now:

4. Jun 2, 2009

Cyosis

Re: Derivative

I did the first term (without the root) for you so you could see how to use the chain rule. All you have done now is use my answer and replace the first term of your answer with mine. This really won't help you a lot. You used the chain rule incorrectly for the other terms as well. Use my example to get the correct answers and post your steps here. Don't forget that theta and r are functions of t so the chain rule needs to be applied on every term.

Last edited: Jun 2, 2009
5. Jun 2, 2009

toastie

Re: Derivative

okay. I have used the chain rule and got:

6. Jun 2, 2009

Cyosis

Re: Derivative

It's almost correct. First off $(1\sqrt{x})'=1/(2\sqrt{x})$. Secondly you took the derivative of $$2\dot{\theta}r$$. You should take the derivative of $$(\dot{\theta}r)^2$$.

7. Jun 2, 2009

toastie

Re: Derivative

i thought the derivative of (theta*r)^2 was = 2*thetadot*dr/dt + 2*((dtheta^2)/d^2t)*r

thus i got:

8. Jun 2, 2009

Cyosis

Re: Derivative

That's basically saying dx^2/dx=2. Do you see where you went wrong now? You applied the product rule correctly after wards, the problem lies with the first step.

9. Jun 2, 2009

toastie

Re: Derivative

10. Jun 2, 2009

Cyosis

Re: Derivative

This part is wrong.

I am not sure which chain rule "method" you were taught. But try to go back to basics and use that method step by step also writer as r(t) and theta as theta(t). This should prevent you from making mistakes. Secondly show the steps instead of just showing the final answer and asking whether it is right or not.

11. Jun 2, 2009

toastie

Re: Derivative

okay I begain with theta^2 times r^2 because it is the samething as (theta*r)^2. So using the product rule I get 2*r^2*theta*d(theta)/dt + 2*(theta^2)*r*dr/dt.

12. Jun 2, 2009

Cyosis

Re: Derivative