Derivative related to equal areas proof

[mishima]

Kleppner and Kolenkow's Introduction to Mechanics text on page 241 (1st edition) has:

[itex]\frac{d}{dt}[/itex](r²[itex]\dot{\theta}[/itex])=r(2[itex]\dot{r}[/itex][itex]\theta[/itex]+r[itex]\ddot{\theta}[/itex])

Is this wrong or am I missing something? I get:

r(2[itex]\dot{\theta}[/itex]+r[itex]\ddot{\theta}[/itex])

...by product rule. It seems at the least the book should have a theta dot in the first term. I don't see where the r dot comes from though. Thank you.

 

[PAllen]

Kleppner and Kolenkow's Introduction to Mechanics text on page 241 (1st edition) has:

[itex]\frac{d}{dt}[/itex](r²[itex]\dot{\theta}[/itex])=r(2[itex]\dot{r}[/itex][itex]\theta[/itex]+r[itex]\ddot{\theta}[/itex])

Is this wrong or am I missing something? I get:

r(2[itex]\dot{\theta}[/itex]+r[itex]\ddot{\theta}[/itex])

...by product rule. It seems at the least the book should have a theta dot in the first term. I don't see where the r dot comes from though. Thank you.

It does seem the book is missing a dot over the first theta on the right hand side. Otherwise the book answer is correct, and your is not. As for the r dot, what do you think is the result of:

d/dt (r^2)

 

[mishima]

[itex]\frac{d}{dx} (f(x))^2 = 2f(x)\frac{d}{dx}f(x)[/itex]

Thanks, I was doing d/dr instead of d/dt. And thanks for confirming the theta dot.