A cylindrical tank with a radius of 90 centimeters is being filled with water so that the volume increases at a rate of 180 centimeters cubed per second. How fast is the height of the water increasing? The volume of the cylinder of height ℎ is given by the formula 𝑉 equals 𝜋𝑟 squared ℎ.
Let’s draw a picture using the information we’ve been given. We have a cylindrical tank with a radius of 90 centimeters. We also know that the volume increases at a rate of 180 centimeters cubed per second. The question is how fast is the height of the water increasing. But what we’re really being asked is to find the rate of change of the height of the water. Let’s write down what we know already.
We know that the volume of a cylinder is 𝜋𝑟 squared ℎ. We know that the rate of change of volume d𝑉 by d𝑡 is equal to 180 centimeters cubed per second, which is a constant. And we know that the radius of the cylinder is 90 centimeters. So 𝑟 squared is 8100 centimeters. We can substitute this into the formula for volume which gives us 𝑉 equals 8100𝜋ℎ.
Now, remember, we want to find the rate of change of the height ℎ. And we know that both volume and height are functions of time. They both change over time as water is poured into the cylinder. So if we differentiate volume with respect to time, as volume is a function of height, we also need to differentiate height with respect to time. So d𝑉 by d𝑡 is 8100𝜋 times dℎ by d𝑡.
We know that the volume changes at a constant rate of 180 centimeters cubed per second. So d𝑉 by d𝑡 is 180. And that’s equal to 8100𝜋 times dℎ by d𝑡. So now to find dℎ by d𝑡 which is actually what we’re looking for, we make dℎ by d𝑡 the subject of this equation. Dividing both sides by 8100𝜋 leaves dℎ by d𝑡 on the right-hand side. Cancelling this down, we find dℎ by d𝑡 is one over 45𝜋. Remember, volume is measured in cubic centimeters, but height is measured in centimeters. So dℎ by d𝑡 is one over 45𝜋 centimeters per second.