Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

The volume \(V\) of a right circular cylinder is given by \(V=\pi r^{2} h\), where \(r\) is the radius and \(h\) is the height. If \(h\) is held fixed at \(h=10\) inches, find the rate of change of \(V\) with respect to \(r\) when \(r=6\) inches.

Step-by-Step Solution

Answer

The rate of change of the volume is \( 120\pi \) cubic inches per inch.

Step 1: Understanding the Given Formula

We are given the volume of a right circular cylinder as \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. We need to find the rate of change of the volume \( V \) with respect to the radius \( r \), specifically when \( r = 6 \) and \( h = 10 \).

Step 2: Substitute the Fixed Height

Since the height \( h \) is fixed at 10 inches, substitute \( h = 10 \) into the volume formula: \[ V = \pi r^2 \times 10 = 10 \pi r^2. \] This simplifies the problem to a function of just \( r \).

Step 3: Differentiate with Respect to r

Find the derivative of \( V \) with respect to \( r \) to determine how the volume changes as the radius changes. Using the formula from the previous step, we have:\[ \frac{dV}{dr} = \frac{d}{dr}(10 \pi r^2) = 20 \pi r. \] This is the rate of change of the volume with respect to the radius.

Step 4: Evaluate the Derivative at r = 6

Substitute \( r = 6 \) into the derivative to find the rate of change of the volume when the radius is 6 inches:\[ \frac{dV}{dr} = 20 \pi \times 6 = 120 \pi. \] This means the volume is changing at \( 120 \pi \) cubic inches per inch when \( r = 6 \).

Key Concepts

Derivative

In calculus, a derivative represents how a quantity changes as another variable changes. It is the mathematical tool used to calculate the "rate of change" of a function. You can think of the derivative as a way to find the slope of a function at any given point, indicating how steep the curve is at that point.

When dealing with the volume of a cylinder, using differentiation helps us understand how the volume changes as we change the radius. Calculating the derivative involves a process called differentiation. In this exercise, we differentiate the volume formula of a cylinder with respect to the radius. This shows how sensitive the volume is to changes in the radius.

• Differentiate the formula to get the rate of change.
• Use the power rule: if you have a term like \( r^n \), its derivative is \( n \cdot r^{n-1} \).

By applying this rule, we simplified the problem to finding \( \frac{dV}{dr} \), which gives us the rate at which the volume changes as the radius increments.

Rate of Change

The rate of change tells us how one variable changes in response to another variable. It is extremely useful in calculus to understand how different quantities interact and influence each other.

In our exercise, the rate of change of the cylinder's volume with respect to its radius indicates how quickly the volume grows when the radius increases. When you calculate the derivative, the resulting expression represents this rate of change.

• Specifically, if the radius increases slightly, the rate of change tells you how much the volume is expected to increase.
• In our step-by-step solution, we found that the rate of change is \( 20 \pi r \).

Understanding the rate of change is crucial, especially when solving problems involving optimization or predicting the effects of changing one quantity in a formula.

Cylinder Volume

The volume of a cylinder is a fundamental concept in geometry and calculus. It is given by the formula \( V = \pi r^2 h \). This formula tells us the space a right circular cylinder occupies based on its radius \( r \) and height \( h \).

To derive deeper insights, it's crucial to know how each dimension affects the volume. In our problem, the height is constant at 10 inches, so the only variable affecting the volume is the radius.

• When the radius changes, it affects the volume quadratically. Since the radius is squared, small changes in radius produce larger changes in volume.
• The formula shows a direct relationship between the radius and the volume, but this relationship can change if we alter the radius.

Understanding how volume varies with respect to your variable of interest, which is the radius in this case, allows one to predict or control the outcome of real-world scenarios and design problems.