# Related rates and "given equation" problems

We received this great question on our blog post " 4 Steps to Solve Any Related Rates Problem – Part 2:
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We can only take the derivative with respect to one variable, so we need to eliminate one of those two.

This statement, if I understand it correctly, means that “related rate” problems are only relating two functions; i.e. Volume to Height or Area to Radius, etc…

How does that apply to related rate problems that use x^2+y^2=z^2? We don’t eliminate any variable here, typically we have the rate of change for two of the three and are solving for the third rate at a specific time.

I understand the reasoning of this cone problem, however I am trying to generalize the process of solving related rates for all scenarios and the quoted statement contradicts a pythagorean related rate.

P.S. You can solve for dr/dt and r using similar triangles as well.
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First, we LOVE this question. Thanks so much for asking!

We created our “4 Steps” the way we did because students tend to have the most trouble with problems that, as you cite, relate volume to height, or area to radius, or similar. It’s tough to know where to start, or how to relate the quantities you need to! Hence when we think “related rates,” those are the scenarios that typically come to mind. (And also that tend to appear on exams.)

You are completely correct, however, that a “given equation” can be used as a related rates problem. (We have just such an example here: “Given an equation, find the rate.”) The difference is that, using the example you wrote, x^2+y^2=z^2, each quantity is immediately assumed to be a function of time, which we can emphasize by writing the equation as

Finding how the rates are related is then simply a matter of taking the derivative with respect to time, \dfrac{d}{dt} :

2\,x(t) \, \dfrac{dx}{dt} + 2\,y(t) \, \dfrac{dy}{dt} = 2\,z(t) \, \dfrac{dz}{dt}

As you say, the problem will have provided two of the rates, and ask for the third at a particular moment t.

Thinking about our 4-step strategy: Because in this case the equation [*] immediately relates the time-dependent quantities for which we’re given the rates, you get to skip our Steps 1 and 2, and move directly to Step 3: “3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.” These “given equation” problems are thus less interesting (and challenging), but certainly no less valid.

We hope that all makes sense, and trust you’ll let us know if not – or if any of this leads to other thoughts you’d like to discuss.

P.S. You can solve for dr/dt and r using similar triangles as well.

Yes indeed - nice!

For now, thanks again for writing in, and please let us know what other questions or thoughts you have. : )