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

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!

• No, $\dot \theta^2$ is the square of the first derivative, not the second derivative of $\theta ^2.$ Commented May 10 at 5:53

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