#  ## Related Rates

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Erin K
1 year ago

What if you don’t have a value for dV/dt? I have a problem where I am given the height of the water, but not given any values for the speed at which it’s draining.

Matheno
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Reply to  Erin K
1 year ago

Thanks for asking, Erin. You need to be given _some_ rate, for either dV/dt or dh/dt, in order to find the other one’s related rate. If you provide some more detail about the question statement, we’ll do our best to help!

Nicky G
2 years ago

why does the height change from 20 to 10? I know you got the 10 from the smaller triangle when comparing it to the larger one, but why would you use that one instead of 20 if that makes sense?

Matheno
Editor
Reply to  Nicky G
2 years ago

Thanks for asking, Nicky! To answer it, look back at the figure at the top where we defined the different variables: “h” is the value of the water’s height in the cone at every instant, so as time passes and the water drains, the value of h changes. The problem asks us to find dh/dt *at the instant when h = 10 cm*,” so that’s the value of h we have to use here.

A key take-away is that that figure at the top matters a LOT: in a problem like this, it does a great deal of work in defining the variables for us and acts as a guide to everything that follows. When it comes to substitute values, it’s important to look back and see what each variable means so that you don’t use a wrong value.

Your physical sense of what’s going on here might help too: if you think about water draining out of a cone, you probably have the sense that at first (when the cone is full) the water level drops slowly, but then toward the end (when the cone is almost empty) the water level drops quickly. That is, dh/dt starts small and gets larger as the water level drops — it doesn’t stay the same throughout — which means that dh/dt _somehow_ depends on the value of h at that instant. So we have to calculate the value of dh/dt for the moment the question tells us, which here happens to be when h = 10 cm. (And it’s through our calculations that we find out exactly how dh/dt depends on h: as you see above, it turns out to depend on 1/h^2.)

Does that help? We hope so!