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

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œ‹ cmΒ²/s with respect to time. Calculate the rate of increase of its radius when its base has a radius of 18 cm.

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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œ‹ 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 π‘Ÿ be equal to the radius of the cross section of this cylinder, we find its base diameter and its height is equal to two π‘Ÿ. We’re also told that the rate of increase of its surface area is 32πœ‹ 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 d𝑑 is equal to 32πœ‹. We want to find the rate of increase of its radius. So, we’re looking to find dπ‘Ÿ by d𝑑. So, how do we link d𝑠 by d𝑑 with dπ‘Ÿ by d𝑑? Well, we’ll use the chain rule. We can see that d𝑠 by d𝑑 will be equal to d𝑠 by dπ‘Ÿ times dπ‘Ÿ by d𝑑. By dividing through by d𝑠 by dπ‘Ÿ, we find that dπ‘Ÿ by d𝑑 will be calculated by dividing d𝑠 by d𝑑 by d𝑠 by dπ‘Ÿ. We know d𝑠 by d𝑑, but how are we going to work out d𝑠 by dπ‘Ÿ?

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

We’re looking to find the rate of increase when the radius is 18 centimetres, so we’ll evaluate d𝑠 by dπ‘Ÿ when π‘Ÿ is equal to 18. It’s 12πœ‹ times 18, which is 216πœ‹. So, we now have d𝑠 by d𝑑 and d𝑠 by dπ‘Ÿ. We said the dπ‘Ÿ by d𝑑 was the quotient of these; it’s 32πœ‹ over 216πœ‹. 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.

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