Transform $dx/dt$ to $dr/dt$ polar coordinates
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2 Answers
$\begingroup$The transformations can be written as
$$\tag 1 r^2 = x^2 + y^2~~ \text{and}~~ \theta = \tan^{-1}\left(\frac{y}{x}\right)$$
Differentiating $(1)$
$$r^2 = x^2 + y^2 \implies 2 r r' = 2x x' + 2y y' \implies r ' = \dfrac{x x' + y y'}{r}$$
Substituting $x = r \cos \theta, y = r \sin \theta$, we have
$$\begin{align} r' &= \dfrac{x x' + y y'}{r} \\&= \dfrac{(r \cos \theta \left(-r \cos \theta \left(r^2 \right)-r \sin \theta +r \cos \theta\right)+r \sin \theta \left(-r \sin \theta \left(r^2 \right)+r \sin \theta+r \cos \theta)\right)}{r}\\ &= r(1-r^2)\end{align}$$
We can also derive
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) \implies \theta' = \dfrac{xy'-yx'}{r^2}$$
Make the appropriate substitutions, do some trig simplifications and cleanup to derive $\theta'$.
$\endgroup$ $\begingroup$Your error is in writing $$\frac{dr}{dt}=\frac{dr}{dx}\frac{dx}{dt}\\\frac{dr}{dt}=\frac{dr}{dy}\frac{dy}{dt}$$ This is not true. Use the multivariable chain rule here to get $$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r }{\partial y}\frac{dy}{dt}$$Then sub in your two equations, as well as $$\frac{\partial r}{\partial x}={\cos \theta}\\\frac{\partial r}{\partial y}={\sin \theta}$$ to get the correct final result
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