If a cylindrical tank with radius 5 meters is being filled with water at a rate of 3 cubic meters per minute, how fast is the height of the water increasing?

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Aug 15, 2014

The answer is $\frac{\mathrm{dh}}{\mathrm{dt}} = \frac{3}{25 \pi} \frac{m}{\min}$.

With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is:
$V = \pi {r}^{2} h$

There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in:
$V = \pi {\left(5 m\right)}^{2} h$

Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:
$\frac{\mathrm{dV}}{\mathrm{dt}} = \left(25 {m}^{2}\right) \pi \frac{\mathrm{dh}}{\mathrm{dt}}$

In the problem, we are given $3 \frac{{m}^{3}}{\min}$ which is $\frac{\mathrm{dV}}{\mathrm{dt}}$. So we substitute this in:
$\frac{\mathrm{dh}}{\mathrm{dt}} = \frac{3 {m}^{3}}{\min \left(25 {m}^{2}\right) \pi} = \frac{3}{25 \pi} \frac{m}{\min}$

In general
- find a formula to relate the 2 variables
- substitute values to remove the constant variables
- implicitly differentiate wrt time (most often the case)
- substitute the given rate
- and solve for the desired rate.

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