Question Video: Finding the Rate of Change of the Radius of an Expanding Cylinder given the Rate of Change of Its Surface Area Using Related Rates Mathematics • Third Year of Secondary School

Video Transcript

The height of a cylinder is equal to its base diameter. Maintaining this relationship between height and base diameter, the cylinder expands such that the rate of increase of its surface area is 32𝜋 square centimetres per second with respect to time. Calculate the rate of increase of its radius when its base has a radius of 18 centimetres.

The first thing we should do when dealing with related rates questions is identify what we’ve been given and what we’re looking to find. In this question, we’ve been given a cylinder whose height is equal to its base diameter. Letting 𝑟 be equal to the radius of the cross section of this cylinder, we find its base diameter and its height is equal to two 𝑟. We’re also told that the rate of increase of its surface area is 32𝜋 square centimetres per second with respect to time. We know that the rate of change of something is considered to be its derivative.

So, letting 𝑠 be equal to the surface area, then we know that d𝑠 by d𝑡 is equal to 32𝜋. We want to find the rate of increase of its radius. So, we’re looking to find d𝑟 by d𝑡. So, how do we link d𝑠 by d𝑡 with d𝑟 by d𝑡? Well, we’ll use the chain rule. We can see that d𝑠 by d𝑡 will be equal to d𝑠 by d𝑟 times d𝑟 by d𝑡. By dividing through by d𝑠 by d𝑟, we find that d𝑟 by d𝑡 will be calculated by dividing d𝑠 by d𝑡 by d𝑠 by d𝑟. We know d𝑠 by d𝑡, but how are we going to work out d𝑠 by d𝑟?

Well, we recall the formula for the surface area of a cylinder. It’s the area of the two circles, that’s two 𝜋𝑟 squared, plus the area of the rectangle in between, and that’s two 𝜋𝑟 times ℎ. We already said, though, that the height of our cylinder is two 𝑟, so let’s replace ℎ with two 𝑟 in our formula. Simplifying, we find that the surface area of our cylinder is six 𝜋𝑟 squared. And we see that we can find an expression for d𝑠 by d𝑟 by differentiating this expression with respect to 𝑟. The first derivative of six 𝜋𝑟 squared is two times six 𝜋𝑟, which is 12𝜋𝑟.

We’re looking to find the rate of increase when the radius is 18 centimetres, so we’ll evaluate d𝑠 by d𝑟 when 𝑟 is equal to 18. It’s 12𝜋 times 18, which is 216𝜋. So, we now have d𝑠 by d𝑡 and d𝑠 by d𝑟. We said the d𝑟 by d𝑡 was the quotient of these; it’s 32𝜋 over 216𝜋. Which simplifies to four twenty-sevenths. The rate of increase of the radius of our cylinder when its base has a radius of 18 centimetres is four twenty-sevenths centimetres per second.