Calculating r(dot): Dynamics Time Derivative of Radius

[Spectre5]

This is from Dynamics where r(dot) is the time derivative of the radius.


$$\dot{r} = \frac{d(r)}{dt}$$
$$\dot{r} = \frac{d}{dt}(2R\sin(\frac{\theta}{2})) \ \ \ \ \ \mbox{note: this is given}$$
$$\dot{r} = 2R\frac{d}{dt}(\sin(\frac{\theta}{2}))$$
$$\dot{r} = 2R\cos(\frac{\theta}{2})\frac{d}{dt}(\frac{\theta}{2})$$
$$\dot{r} = 2R\cos(\frac{\theta}{2})(\frac{1}{2})\dot{\theta}$$
$$\dot{r} = R\dot{\theta}\cos(\frac{\theta}{2})$$

I am confused on one part...in this step:

$$\dot{r} = 2R\frac{d}{dt}(\sin(\frac{\theta}{2}))$$
$$\dot{r} = 2R\cos(\frac{\theta}{2})\frac{d}{dt}(\frac{\theta}{2})$$

Why do you have to differentiate the sin(theta/2)?? Since you are differentiating with respect to t, don't you count the theta as a constant and therefore not differentiate it? But my profressor said that the above is correct.

PS: Sorry for my bad tex.

 

[maverick280857]

Hello Spectre5

The derivative of a function $$f(\phi)$$ with respect to a variable it does not explicitly depend on is given by the chain rule,

$$\frac{df}{du} = \frac{df}{d\phi}\frac{d\phi}{du} = f'(\phi)\frac{d\phi}{du}$$

In your case therefore

$$\dot{r} = 2R\frac{d}{dt}(\sin(\frac{\theta}{2})) = 2R\cos(\frac{\theta}{2})\frac{d}{dt}(\frac{\theta}{2})$$

You must understand that you are differentiating $$\dot{r}$$ with respect to time so you must first differentiate with respect to theta and then multiply by the derivative of theta with respect to time in accordance with the chain rule for total derivatives as shown above.

You are probably thinking that when you differentiate with respect to time, theta will remain constant (with time) but that is not so since theta too can (and in a general situation, will) change with time. Unless you are explicitly told that $$\dot{\theta} = 0$$ you cannot say so.

Hope that helps...

Cheers
Vivek

 

[Spectre5]

yea, I was not paying attention to the fact that it uses the chain rule :-/

Thanks for the help, I understand it now.