# When can I treat derivative as a fraction? (Brachistochrone)

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?

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.
• Note that the final integral of the OP only works if $v(y)$ is injective on the interval from $y_1$ to $y_2$ May 18, 2020 at 21:13