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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.


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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 $x$.

  • $\begingroup$ What division are you talking about? Because I can't tell what you are talking about. $\endgroup$ – Alexandra Mar 28 at 6:40
  • $\begingroup$ In the previous step, both sides of the equation were divided by 2y. $\endgroup$ – Diatche Mar 28 at 6:51

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