My teacher was solving the Brachistochrone problem in class. She parametrized the required path with $x(y)$, then said $T=\int_0^Tdt=\int_{y_1}^{y_2}\frac{dt}{dy}dy=\int_{y_1}^{y_2}\frac{dy}{dy/dt}$. Why is this conversion from $dt/dy$ to $dy/dt$ allowed? There is a similar question on math stack https://math.stackexchange.com/q/1784671/ which has a green answer saying that this can be found from the chain rule wherever it is legal. How does the chain rule explain what my teacher did? Or is there a different explanation?
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2$\begingroup$ Have you taken a look at this: Inverse functions and differentiation? If so, does it answer your question? $\endgroup$– Alfred CentauriMay 18, 2020 at 20:52
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In mathematics people discuss a lot about this things but in physics one is usually happy with differential formulas as if they were algebraic simple formulas, thus $$\frac{\mathrm{d}y}{\mathrm{d}y} \equiv 1,$$ which is what your teacher probably intended. Then $$ \mathrm{d}t = \frac{\mathrm{d}y}{\mathrm{d}y} \mathrm{d}t = \frac{\mathrm{d}t}{\mathrm{d}y} \mathrm{d}y = \frac{ \mathrm{d}y}{v_y}.$$
One can add that in a more general way $$v = \sqrt{ \Big( \frac{\mathrm{d}x}{\mathrm{d}t} \Big)^2 + \Big( \frac{\mathrm{d}y}{\mathrm{d}t} \Big)^2}$$ $$ \mathrm{d}t = \frac{ \sqrt{ \Big( \mathrm{d}x \Big)^2 + \Big( \mathrm{d}y\Big)^2}}{\sqrt{ \Big( \frac{\mathrm{d}x}{\mathrm{d}t} \Big)^2 + \Big( \frac{\mathrm{d}y}{\mathrm{d}t} \Big)^2}} = \frac{\mathrm{d}s}{v}.$$ Which is not exactly the chain rule but works better for a trajectory in a plane.
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1$\begingroup$ Note that the final integral of the OP only works if $v(y)$ is injective on the interval from $y_1$ to $y_2$ $\endgroup$ May 18, 2020 at 21:13
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$\begingroup$ true; to the point of the question, in physics we seem contended with differentials as fractions although, as @BioPhysicist said, one needs to be careful. However, it works pretty well given that one can see experiments with this stuff working, for example youtube.com/watch?v=1BdO8J0iynY then physicist are mostly happy with the final result even if mathematicians might not be with the steps in between. $\endgroup$ May 18, 2020 at 22:01