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Question Video: Finding the Rate of Change of the Surface Area of a Shrinking Sphere Given the Rate of Change of Its Volume Using Related Rates Mathematics • Higher Education

A spherical balloon leaks helium at a rate of 48 cmΒ³/s. What is the rate of change of its surface area when its radius is 41 cm?

06:59

Video Transcript

A spherical balloon leaks helium at a rate of 48 centimeters cubed per second. What is the rate of change of its surface area when its radius is 41 centimeters?

The first thing to identify about this question is that it’s a related rates problem. This means a problem in which two related quantities are changing over time. We have been given a spherical balloon. Recall that one of the defining characteristics of a sphere is its radius, that is, the distance from the center of the sphere to its circumference. We can mark this on our balloon diagram. Let’s now examine our question in more detail. The first piece of information we are given is the rate at which helium is being lost in centimeters cubed per second. Here, we are being told the volume of helium that is being lost per unit of time.

Next, we are asked for the rate of change of the surface area when the radius takes a given value. More specifically, this means the rate of change of area per unit of time. The first tools that we will need for this question are the formulas for the volume and surface area of a sphere with respect to its radius. These should be familiar to you. The volume of a sphere which we’ll represent as 𝑉 is equal to four-thirds πœ‹π‘Ÿ cubed. The surface area of a sphere which we’ll represent as 𝐴 is equal to four πœ‹π‘Ÿ squared.

Okay, so our question has asked us to find the rate of change of the surface area with respect to time. We can express this as d𝐴 by d𝑑. Unfortunately, the formula that we have for volume is in terms of the radius of the sphere instead of time. This means we cannot directly differentiate our formula and reach the desired result. Instead, to make progress, we’ll use an application of the chain rule to write an equivalent statement. That is, d𝐴 by d𝑑 is equal to d𝐴 by dπ‘Ÿ multiplied by dπ‘Ÿ by d𝑑. Here, we can see the first of these terms is the derivative of the surface area with respect to the radius. Since we have a formula for surface area in terms of radius, we can use differentiation to find this term.

Differentiating four πœ‹π‘Ÿ squared with respect to π‘Ÿ, we get eight πœ‹π‘Ÿ. This expression can be substituted back into our equation for d𝐴 by d𝑑. After substituting this back into our equation, we are still stuck with this dπ‘Ÿ by d𝑑 term. This represents the rate of change of the radius with respect to time, but we currently don’t have an expression for that. Right now, we have formulas for the volume and surface area in terms of the radius. In order to move forward, we’ll need to use these to express dπ‘Ÿ by d𝑑 in a more useful way. To do so, let’s go back to the first piece of information given to us in the question.

The balloon is leaking helium at a rate of 48 centimeters cubed per second. Here, we are being given the rate of change of the volume with respect to time, that is, d𝑉 by d𝑑. The question tells us that the balloon is losing helium. We therefore say that our rate of change is negative 48 centimeters cubed per second. It is also worth noting here that as time increases, our balloon is always decreasing in volume. In other words, it is shrinking. Logically, since the shape remains a sphere, we should expect that the surface area of our shrinking balloon will also be decreasing over time. When we finally get to the answer our question, we should therefore see a negative number to represent that the surface area is decreasing over time when the radius of the balloon is 41 centimeters.

Another quick side note, let’s simply remember that consistently throughout the problem, the units of length in the system are centimeters and the units of time are seconds. This allows us to set aside the units during our calculations and add them back at the end of the question. We now have that d𝑉 by d𝑑 is equal to negative 48, but this does not yet help us with our goal of reexpressing dπ‘Ÿ by d𝑑. Instead, we can return to the previous trick that we used involving our friend, the chain rule. This allows us to construct an equivalent statement that d𝑉 by d𝑑 is equal to d𝑉 by dπ‘Ÿ multiplied by dπ‘Ÿ by d𝑑. Since these two statements are equivalent, we can also say this is equal to negative 48.

Looking back at our original formulas, we have an equation for 𝑉 in terms of π‘Ÿ. This means that we can make progress by finding d𝑉 by dπ‘Ÿ. Using differentiation and a little simplification, we find d𝑉 by dπ‘Ÿ is equal to four πœ‹π‘Ÿ squared. Let’s substitute this result back into the equation. Now dividing both sides of the equation by four πœ‹π‘Ÿ squared allows us to isolate dπ‘Ÿ by d𝑑 on the left-hand side. This is now an equation expressing dπ‘Ÿ by d𝑑 in terms of the radius of the sphere.

It’s worth mentioning that we have to be careful since our equation now involves a division by the variable π‘Ÿ. In this case, our equation requires us to divide by zero when π‘Ÿ is equal to zero. And hence, dπ‘Ÿ by d𝑑 is undefined at this point. Fortunately, we know the balloon is shrinking, and we’ll be evaluating the system when the radius is 41 centimeters. This means the undefined point when the radius is zero does not affect us. Now, we can simplify the right-hand side of the equation to find that dπ‘Ÿ by d𝑑 is equal to negative 12 over πœ‹π‘Ÿ squared.

Great! Let’s substitute this back into the equation for d𝐴 by d𝑑. On the right-hand side of this equation, we now have eight πœ‹π‘Ÿ multiplied by negative 12 over πœ‹π‘Ÿ squared. We can cancel the πœ‹ terms and one of the π‘Ÿ terms in the denominator. We also multiply eight by negative 12. This leaves us with the result that d𝐴 by d𝑑 is equal to negative 96 over π‘Ÿ. We now have a general expression that allows us to find the rate of change of the surface area of our balloon with respect to time for any given radius π‘Ÿ. Remember this is not valid for the case of π‘Ÿ equals zero as we mentioned earlier.

For our final step, we recall that the question asked us for d𝐴 by d𝑑 when the radius is 41 centimeters. We are now in a position to substitute the value of 41 directly into our equation. Since there are no useful simplifications for the fraction negative 96 over 41, we can leave our answer as it is. However, we must remember to add the units for surface area over time, that is, centimeters squared per second.

Great! We have now found that the rate of change of the surface area of the balloon is negative 96 over 41 centimeters squared per second when the radius of the balloon is 41 centimeters. This is the answer to our question. As a final note, the negative result we see matches our earlier observation that the rate of change of the surface area should be negative, since the balloon is shrinking.

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