Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (g)/Solution 1

The two variables that are changing in this problem are the volume of water and the height of the water. These two variables are related in the formula for volume of a cylinder:

$V=\pi r^{2}h$

For this problem, we know that the radius of the cylinder is 3, so we can put that into our formula to get:

$V=9\pi h$

Since the question is asking about rates of change, we will differentiate the above formula with respect to t.

${\frac {dV}{dt}}=9\pi {\frac {dh}{dt}}$

In this new formula, dh/dt stands for the rate at which the height of the water is changing, which is what we're solving for. We also know that dV/dt stands for the rate at which the volume of water is changing, which is 5 m³/min. So substituting in dV/dt = 5 we have:

$5=9\pi {\frac {dh}{dt}}$

And solving for dh/dt we get:

${\frac {dh}{dt}}={\frac {5}{9\pi }}m/min$