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A cylindrical tank with radius 5m is being filled with water at a rate of $3m^3/min$ How fast is the height of the water increasing?

I know that $$dv/dt = 3$$

And that $$v= \pi r^2 h$$

So I want to find the derivative of h

$$h= \frac {v}{ \pi r^2}$$

$$h' = \frac{3 \pi * 25}{ (\pi * 25)^2 }$$

$h' * v'$

This gives an incorrect answer and I am not sure why.

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We have as you said $h=\frac{v}{\pi r^2}$. Note that $r$ is a constant. Differentiate with respect to time. We get $\frac{dh}{dt}=\frac{1}{\pi r^2}\frac{dv}{dt}$. – André Nicolas May 10 '12 at 18:04
up vote 5 down vote accepted

You want $\dfrac{dh}{dt}$; by the chain rule this is $\dfrac{dh}{dv}\dfrac{dv}{dt}$.

You have $h=\dfrac{v}{\pi r^2}=\dfrac1{\pi r^2}v$, where $\dfrac1{\pi r^2}$ is a constant, so $\dfrac{dh}{dv}=\dfrac1{\pi r^2}$; you don't need the quotient rule for this differentiation. Finally, you have $\dfrac{dv}{dt}=3$, so $$\frac{dh}{dt}=\frac{dh}{dv}\frac{dv}{dt}=\frac3{\pi r^2}\text{ m/min}\;.$$ Since $r=5$ m, the actual rate is $\dfrac3{25\pi}$ m/min.

In a problem like this it's a good idea to use the $\dfrac{dv}{dt}$ notation instead of the $v'$ notation, because you're taking derivatives with respect to more than one variable, and you need to keep track of what that variable is in each calculation.

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Hint: $\displaystyle h'(t) = \frac{v'(t)}{\pi r^2}$, since $r$ is constant.

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