# Related rates with a cylinder

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.

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

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.

• Quick question: I often see the slope of a line described as follows: the slope of a line is the rate of change of the value of the function with respect to its argument. I have never thought about it before, but when we use the wording with respect to, does this really mean: [the slope of a line is] the rate of change of the value of the function as the argument changes? Feb 24 at 20:05
• @TaylorRendon: I would describe it as the amount by which the value of the function would change when the argument increases by $1$ unit if the rate of change remained constant. Feb 24 at 20:12
• I see, I like that description. However, regarding the wording "with respect to": when we say something like "[the slope of the line is] the rate of change of $y$ with respect to $x$", does this mean "the rate of change of $y$ as $x$ changes" (since $y$ changes relative to how $x$ changes)? Just wanted to make sure my way of thinking is correct :-). Feb 24 at 23:46
• @TaylorRendon: I won’t say that it’s wrong, but for me with respect to $x$ makes it clearer that we’re looking specifically at the changes in $y$ that are directly associated with changes in $x$, even if $y$ might be undergoing other changes as well, ones that are not being taken into account. This is even more important when we’re looking at partial derivatives. Possibly your wording has this connotation for you; for me it doesn’t say as clearly that the changes in $y$ are directly associated with the changes in $x$. Feb 25 at 0:34
• @TaylorRendon: You’re welcome. I’m all in favor of people wanting to be good teachers! Feb 25 at 1:04

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