Arectemieryf0

2022-07-21

Let $r=r\left(t\right)$ and $\theta =\theta \left(t\right)$ with $r\left(t\right)>0$. Let $x\left(t\right)=r\left(t\right)\mathrm{cos}\left(\theta \left(t\right)\right)$ and $y\left(t\right)=r\left(t\right)\mathrm{sin}\left(\theta \left(t\right)\right)$ . Prove that $\frac{d\theta }{dt}=\frac{1}{{x}^{2}+{y}^{2}}\left(x\frac{dy}{dt}-y\frac{dx}{dt}\right)$
The hint is to use $y\left(t\right)/x\left(t\right)$ and use implicit differentiation but I can't see how to use that hint to solve this problem.

Tristan Pittman

Expert

We have
$\frac{y\left(t\right)}{x\left(t\right)}=\mathrm{tan}\left(\theta \left(t\right)\right).$
Now differentiate both sides with respect to $t$. On the left-hand side, you can differentiate $y\left(t\right)/x\left(t\right)$ using the quotient rule. But on the other side we need to use the chain rule. We have
$\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{tan}\left(\theta \left(t\right)\right)=\frac{1}{{\mathrm{cos}}^{2}\left(\theta \left(t\right)\right)}×\frac{\mathrm{d}\theta }{\mathrm{d}t}.$
Now rearrange for $\frac{\mathrm{d}\theta }{\mathrm{d}t}$.

Do you have a similar question?