The derivative of rotational kinetic energy in terms of period gives me the wrong answer. Why should I use the product rule? [closed]

Question

This is my first question here so I hope I do it correctly. I've tried to solve this, and google it, but I can't find the answer to this particular question. This equation comes from Carroll and Ostille, "An introduction to astrophysics"(chapter on Neutron stars) and it is not a homework problem. I'm reading the book on my own and I got stuck. I will try to show my work.

We begin by knowing $K = {{1} \over {2}}I\omega $ for an object with rotational energy. Using $\omega = {{2\pi} \over {P}}$ we can show that $K = {{1} \over {2}}I\omega = {{2\pi^2 I} \over {P^2}}$. So far so good, this is the formula we want.

Now the text says we want the ${{dK} \over {dt}}$ of ${{2\pi^2 I} \over {P^2}}$. And this is where I'm stuck, I think it's because I don't understand how $dt$ is applied when there's no t.

I've taken calculus and assumed that I just applied it like a polynomial getting $ = {{-4\pi^2 I} \over {P^3}}$. But the book claims the real answer should be $ = {{-4\pi^2 I\dot{P}} \over {P^3}}$. The only way I could get that answer is via the product rule(I assume) but I have no clue where that $\dot{P}$ comes from, or why it's a reasonable answer. I do know what the symbols mean though. Please help me?

Answer 1

$ KE = \frac{I \omega ^2}{2} $

From the chain rule,

$ \frac{dK}{dt}= I \omega \frac{d\omega}{dt}$

Using $\omega= 2 \pi /P $ we get:

$ \frac{dK}{dt}= I(\frac{(2\pi)^2}{P})(\frac{d(\frac{1}{P})}{dt})$

Again applying chain rule on P:

$ \frac{dK}{dt}= I(\frac{(2\pi)^2}{P})(\frac{d(\frac{1}{P})}{dP})(\frac{dP}{dt}) $

Which gives us $ \frac{dK}{dt}= \frac{I \cdot 4\pi^2}{P} \cdot (\frac{-1}{P^2})(\frac{dP}{dt})$

The term $ \dot{P} $ is nothing but the first time derivative of P

Hope it helps