Problem 9 - Calculus an Applied Approach

Question

A spherical balloon is inflated with gas at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is (a) 1 foot and (b) 2 feet?

Step-by-Step Solution

Answer

At a radius of 1 foot, the rate of the change of the radius is approximately 0.796 ft/min, while at a radius of 2 feet, it is approximately 0.199 ft/min.

Step 1: Differentiating the Volume Equation

First, differentiate the volume equation with respect to time (t). Using the chain rule, the derivative of \(V = \frac{4}{3} \pi r^3\) is \( \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}\)

Step 2: Find the Radius Rate for Radius = 1 foot

Substitute \( \frac{dV}{dt} = 10\) ft³/min and \( r = 1\) into the equation to solve for \( \frac{dr}{dt}\). The calculation is as follows: \(10 = 4\pi(1)^2 \frac{dr}{dt}\). Solving for \(\frac{dr}{dt}\) gives approximately 0.796 ft/min

Step 3: Find the Radius Rate for Radius = 2 feet

Substitute \( \frac{dV}{dt} = 10\) ft³/min and \( r = 2\) into the equation to solve for \( \frac{dr}{dt}\). The calculation is as follows: \(10 = 4\pi(2)^2 \frac{dr}{dt}\). Solving for \(\frac{dr}{dt}\) gives approximately 0.199 ft/min

Key Concepts

Calculus

Calculus is a branch of mathematics focused on change. It helps us understand how things change at any given point in time. Calculus is used in a variety of fields, from physics to engineering and in everyday applications like calculating speeds and predicting outcomes.

In particular, calculus is the key to solving problems involving motion and growth, like the one in our exercise about an inflating spherical balloon. Two important concepts in calculus are differentiation and integration. In this exercise, we are primarily concerned with differentiation, which allows us to find the rate of change of a quantity. This is essential for finding how fast the radius of the balloon is changing as it inflates.

By applying calculus to this problem, we can calculate how quickly the radius grows for any given increase in the balloon's volume. This involves some understanding of other calculus tools, like the chain rule, which we'll discuss next.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It essentially allows you to take the derivative of a function that is nestled within another function.

In our balloon problem, the chain rule is used to differentiate the volume of a sphere with respect to time. The volume depends on the radius, which in turn depends on time. So the volume is a function of the radius, and the radius is a function of time. This requires using the chain rule given by the formula:

• \( \frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \)

This equation breaks down the problem into two parts:

• \( \frac{dV}{dr} \) – the rate at which volume changes with respect to the radius.
• \( \frac{dr}{dt} \) – the rate at which the radius changes with respect to time, which is what we want to find.

The chain rule helps us chain these derivatives together to express how the volume change relates to the change in radius over time.

Differentiation

Differentiation is a key concept in calculus used to find the rate at which a quantity changes. It involves finding the derivative of a function, which gives us this rate of change.

In the context of our exercise, differentiation is applied to the volume of the sphere to determine how fast the radius changes as the balloon is inflated. The volume of a sphere with radius \(r\) is given by the equation:

• \( V = \frac{4}{3}\pi r^3 \)

By differentiating this with respect to time using the chain rule, we find:

• \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)

Here, \( \frac{dV}{dt} \) is the rate at which the volume is changing, which is provided as 10 cubic feet per minute. Solving for \( \frac{dr}{dt} \) using this equation with specific radii gives us how fast the radius increases for each condition.

Through differentiation, we understand that although the volume increases steadily, the change in radius depends on the size of the radius itself. This demonstrates an essential aspect of calculus in analyzing rates of change.

Volume of a Sphere

The formula for the volume of a sphere is central to this exercise. This mathematical formula shows us how to compute the volume of a sphere when its radius is known. It is given by:

• \( V = \frac{4}{3} \pi r^3 \)

This equation highlights the relationship between the volume and the radius of a sphere. When the radius changes, the volume changes too, and understanding this relationship is crucial in solving real-world problems like this one.

The volume increases with the cube of the radius, which means small changes in the radius will lead to larger changes in volume. By relating the volume and the rate at which it changes to the rate of change of the radius, we can apply this concept in practical ways.

When a balloon is inflated, we aren't just interested in how much air it holds, but also in understanding how its dimensions change over time. By understanding the formula for the volume of a sphere, we gain insightful perspective into this process, linking size with capacity and the dynamic nature of an inflating balloon.