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

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Homework Help Overview

The original poster is attempting to find the derivative of the expression \(((\dot{r}^2) + (\dot{\theta} r)^2)^{1/2}\), where \(\dot{r}\) and \(\dot{\theta}\) represent the derivatives of \(r\) and \(\theta\) with respect to time. The problem involves applying the chain rule and product rule in the context of derivatives in polar coordinates.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the chain rule and product rule, with some questioning the correctness of previous derivations. There are attempts to clarify the differentiation of terms involving products of functions of time.

Discussion Status

Several participants have provided feedback on the original poster's attempts, suggesting corrections and clarifications regarding the application of differentiation rules. There is ongoing exploration of the correct approach to differentiate the terms involved, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of treating \(r\) and \(\theta\) as functions of time, which affects the differentiation process. There is also mention of potential errors in applying the chain rule and product rule correctly.

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Homework Statement


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)

Homework Equations





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)
 
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Using the chainrule.

[tex] \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}[/tex]
 


okay. is there any other mistakes in my derivation now:
[2(dr/dt)*(dr^2/d^2t)+2thetadot*r+2((dtheta^2)/dt)*r]/((rdot^2)+((thetadot*r)^2))^(1/2)
 


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:


okay. I have used the chain rule and got:
[2*rdot*((dr^2)/d^2t) + 2*rdot*thetadot + 2*((dtheta^2)/d^2t)*r]/((rdot^2)+((thetadot*r)^2))^(1/2)
 


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


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

thus i got:
[rdot*((dr^2)/d^2t) + rdot*thetadot + ((dthetadot^2)/d^2t)*r]/((rdot^2)+((thetadot*r)^2))^(1/2)
 


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.
 


okay so should (theta*r)^2 be 2*thetadot*((dthetadot^2)/dt)*r^2 + 2*(thetadot^2)*dr/dt ?
 
  • #10


toastie said:
2*(thetadot^2)*dr/dt ?

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


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


If you replace theta by thetadot and thetadot by thetadotdot you have the correct answer. I guess that was a typo?
 
  • #13


thank you for all your help.
 

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