$\ddot x$ vs. $\dot x^2$

Question

I'm working on a physics assignment and am having some trouble. I need to integrate $r\dot\theta^2$ with respect to $t$. However, my trouble lies in the definition of the upper-dot format.

Given: $$ \dot\theta=\frac{\mathrm{d}}{\mathrm{d}t}\theta $$ and $$ \ddot\theta=\frac{\mathrm{d}^2}{\mathrm{d}t^2}\theta $$

If I square $\dot\theta$, I get: $$\dot\theta^2 = \left(\frac{\mathrm{d}}{\mathrm{d}t}\theta\right)\left(\frac{\mathrm{d}}{\mathrm{d}t}\theta\right)=\frac{\mathrm{d}^2}{\mathrm{d}t^2}\theta^2$$

Does this mean that: $$\dot\theta^2 = \ddot\theta\theta$$

So if I go back to my original task: $$\int r\dot\theta^2\, \mathrm{d}t \rightarrow \int r\ddot\theta\theta \,\mathrm{d}t$$ Which I still don't really know how to integrate, my first thoughts are a combination of chain rule and integration by parts, any help is appreciated!

Answer 1

$$\dot\theta^2 = \left(\frac{d}{dt}\theta\right)\left(\frac{d}{dt}\theta\right)=\frac{d^2}{dt^2}\theta^2$$

This line assumes that you can interchange operators (the derivative) and functions ($\theta$). This is not the case. The most you can say is that $\dot \theta^2$ is the square of the derivative.

As an example, suppose $\theta(t) = t^2$. Then $$ \left( \frac{\mathrm{d}\theta}{\mathrm{d}t} \right)^2 = (2t)^2 = 4t^2 $$ but the result you claim would give $$ \frac{\mathrm{d}^2}{\mathrm{d}t^2} (t^2)^2 = \frac{\mathrm{d}^2}{\mathrm{d}t^2} t^4 = 12t^2 $$ Definitely not the same thing.