Help required to differentiate a function.

1. Oct 12, 2012

CrazyNeutrino

I need help to take the time derivative of a physics equation.

The equation:

ME=1/2 mr^2 (dθ/dt)^2 +mgr θ^2/2

Where ME is the mechanical energy , m is the mass of the object, r is the radius of the path of the object and θ is the angle around the path.

Walter Lewin at MIT differentiates it and gets the result:

mr^2 (dθ/dt) (d^2 θ/dt^2)+mgr θ dθ/dt

Can someone please explain why this is so. I can't understand how that would be the derivative of the function. Please answer.

2. Oct 12, 2012

haruspex

It's standard chain rule. Which part don't you get? If y = y(t), the derivative of y2 is 2y dy/dt

3. Oct 12, 2012

CrazyNeutrino

The derivative of 1/2mr^2
Would be 2 1/2mr right? Why do we need to use the chain rule?

Last edited: Oct 12, 2012
4. Oct 12, 2012

CrazyNeutrino

Can you please just explain the entire derivative of the function.

5. Oct 12, 2012

CrazyNeutrino

I get that the derivative of 1/2mr^2=2 1/2 mr r = mr^2, what i don't get is the dtheta/dt part.

6. Oct 12, 2012

Wizlem

it appears as though r does not depend on t if Walter Lewin at MIT differentiates it correctly.

The chain rule is required because $\frac{d((f(t))^{2})}{dt} = 2 f \frac{d(f(t))}{dt}$

Thus, your derivative of 1/2mr^2 is wrong because you forgot the factor of dr/dt but that isn't really relavent

What you want to do is take the derivative of the θ stuff. What i mean is

$\frac{d}{dt} ME = \frac{1}{2}mr^{2} \frac{d}{dt} (\frac{dθ}{dt})^{2} + \frac{1}{2}mgr\frac{d}{dt}θ^{2}$

Last edited: Oct 12, 2012
7. Oct 12, 2012

CrazyNeutrino

R is a constant radius, it doesn't depend on t. But how would I do d/dt of (dtheta/dt)^2

8. Oct 12, 2012

CrazyNeutrino

Ok thanks a lot. I understand. Correct me if I'm wrong:
d/dt of 1/2 mr^2(dθ/dt)^2 is (1/2 mr^2) 2 dθ/dt d^2 θ/dt^2)=mr^2 dθ/dt d^2 θ/dt^2 and the 2 you bring down by the power rule is from dθ/dt squared and not from r square and then derivative of the inside is the second derivative of θ and then the second part d/dt of mgr θ^2/2 is mgr 2 θ/2 dθ/dt = mgr θ dθ/dt.

Is this correct?

9. Oct 12, 2012

haruspex

Yes.

10. Oct 12, 2012

CrazyNeutrino

Walter Lewin writes all the stuff he write on the board from his book so sometimes what he writes isn't what he says. Over here he said the 2 from the r square would cancel out the 1/2. That confused me because then the derivative of dtheta/dt square becomes 2dtheta/dt dtheta/dt square.

11. Oct 12, 2012

CrazyNeutrino

Thanks for the help.