# Related rates calculus: how is $\frac{x}{y} \frac{dx}{dt} = \frac{dy}{dt}$?

Hello, in this problem the person wrote dy/dt the same as x/y times dx/dt. I don't get how this works. Wouldn't you get (xdx)/(ydt)?

Also in this question I posted: Related Rates Calculus - Confused About What dx/dt, dy/dt and dx/dy mean

I was confused as to how dx/dy = y? One of the commenters said that when you do dx/dy you get x(y) = 1/2(y^2 - 1) but I have no idea how they got to that point or how that equals to y in the first place.

Any help is greatly appreciated. Thank you.

Each quantity there is a function of time. By the Pythagorean theorem we have: $$y^2=x^2+1$$ Differentiate both sides with respect to time ($$t$$) (this is called implicit differentiation, by the way) and don't forget that when you differentiate a composition of functions, you use the chain rule (for example, $$\left(x^2(t)\right)'=2x(t)x'(t)$$): $$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0\implies \frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}$$
$$\frac{dx}{dt}$$ and $$y$$ are given. Finding $$x$$ is easy. Again, use the fact that what we have there geometrically is a right triangle (since $$x$$ is a distance, it should be a positive quantity): $$y^2=x^2+1^2\implies 2^2=x^2+1\implies x=\sqrt{3}\ mi$$ Thus: $$\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}=\frac{\sqrt{3}}{2}\cdot 500\ mph$$
In this example, $$\frac{dy}{dt} = \frac{x}{y} * \frac{dx}{dt}$$ because of the division that was performed in the previous step.
It appears you are getting hung up on a misconception about how you think the notation should work. It may be useful to consider an analogy: when presented with $$\frac{f(x)}{x}$$ you cannot cancel out of the $$x$$.