Problem 31 - Calculus Single Variable

Question

A spherical balloon is losing air at the rate of \(6 \mathrm{~cm}^{3} / \mathrm{min}\). At what rate is its radius \(r\) decreasing when \(r=24 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

The radius is decreasing at a rate of approximately \( -\frac{1}{384\pi} \) cm/min when \( r = 24 \) cm.

Step 1: Understand the Relationship between Volume and Radius of a Sphere

The volume of a sphere is calculated by the formula \( V = \frac{4}{3}\pi r^3 \). We have a problem involving the change in volume over time, \( \frac{dV}{dt} = -6 \, \text{cm}^3/\text{min} \) (negative because the balloon is losing air). This means we need to connect this with \( \frac{dr}{dt} \), the rate at which the radius is decreasing.

Step 2: Differentiate the Volume Formula with Respect to Time

Using the chain rule, differentiate both sides of the formula \( V = \frac{4}{3}\pi r^3 \) with respect to time \( t \):\[ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \]This equation links \( \frac{dV}{dt} \) to \( \frac{dr}{dt} \).

Step 3: Substitute Known Values into the Differentiated Volume Equation

We know that \( \frac{dV}{dt} = -6 \, \text{cm}^3/\text{min} \) and \( r = 24 \, \text{cm} \). Substitute these values into the equation from the previous step:\[ -6 = 4\pi (24)^2 \cdot \frac{dr}{dt} \]This will allow us to solve for \( \frac{dr}{dt} \).

Step 4: Solve for the Rate of Change of the Radius

Rearrange the equation to solve for \( \frac{dr}{dt} \):\[ \frac{dr}{dt} = \frac{-6}{4\pi (24)^2} \]Calculate the value:\[ \frac{dr}{dt} = \frac{-6}{2304\pi} \approx -\frac{1}{384\pi} \, \text{cm/min} \]This is the rate at which the radius is decreasing.

Key Concepts

Related Rates

Related rates problems involve understanding how different quantities change over time and how these changes are interconnected. They are very common in calculus, particularly in scenarios where different aspects of a mathematical model are evolving simultaneously.

In our balloon problem, we deal with two changing rates:

• The rate at which the volume of the balloon is decreasing ( \( \frac{dV}{dt} \)).
• The rate at which the radius of the balloon decreases ( \( \frac{dr}{dt} \)).

Both of these rates are connected through the geometry of the sphere. The task is to understand and calculate one rate by using the information given about the other rates and the geometry equations involved.

Differentiation

Differentiation is a fundamental concept in calculus used to find how a function is changing at any point. It involves computing the derivative of a function, which expresses the function's rate of change.

In this exercise, we use differentiation on the formula for the volume of a sphere:

• Volume formula: \( V = \frac{4}{3}\pi r^3 \).
• Differentiate with respect to time \( t \) to find \( \frac{dV}{dt} \).

By applying the chain rule, we connect the change in volume with the change in radius over time, resulting in: \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \). This step is crucial as it allows us to relate the rates and solve the problem.

Volume of a Sphere

The formula for the volume of a sphere, \( V = \frac{4}{3}\pi r^3 \), is fundamental to solving the problem. It relates the radius of the sphere to its volume, providing a basis for understanding how any change in radius impacts the volume.

This relationship is especially useful when considering how loss of volume affects the radius, as seen in the balloon losing air. As the volume shrinks, the radius decreases over time, and these dynamics are captured by differentiating the volume formula with respect to time.

Remember:

• The volume increases or decreases with the cube of the radius.
• Small changes in radius lead to significant changes in volume due to the cubic relationship.

Rate of Change

Rate of change refers to how a specific quantity varies with time. It plays a central role in this problem as we determine the rate at which the balloon's radius decreases when the volume changes at a specified rate.

In mathematical terms, this is expressed as \( \frac{dr}{dt} \), representing the instantaneous rate of change of the radius per unit of time. By setting up the relationship \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \) and substituting the known values, we solve for \( \frac{dr}{dt} \).

Understanding rates of change helps us predict and analyze dynamic systems—be it a deflating balloon or more complex real-world phenomena.