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 volume with respect to radius is \(120\pi\) cubic inches per inch.

Step 1: Identify Known Values

We are given the formula for the volume of a cylinder: \( V = \pi r^2 h \). We are also given that the height \( h = 10 \) inches and radius \( r = 6 \) inches in this scenario. We need to find the rate of change of volume with respect to radius \( r \).

Step 2: Formulate the Problem

Since we are looking for how the volume \( V \) changes with respect to the radius \( r \), we will compute the derivative of the volume \( V \) with respect to \( r \). This derivative is \( \frac{dV}{dr} \).

Step 3: Compute the Derivative

The formula used for the volume is \( V = \pi r^2 h \). Since \( h \) is fixed at 10, substitute to get \( V = \pi r^2 \times 10 = 10\pi r^2 \). Differentiate \( V \) with respect to \( r \):\[ \frac{dV}{dr} = \frac{d}{dr}(10\pi r^2) = 20\pi r \]

Step 4: Substitute Known Radius

Now, substitute \( r = 6 \) into the derivative \( \frac{dV}{dr} = 20\pi r \) to find the rate of change of volume:\[ \frac{dV}{dr} = 20\pi \times 6 = 120\pi \]

Key Concepts

Derivative

Calculus revolves around the idea of change and how variables interact and affect one another. The derivative is a core concept in calculus that helps us understand this notion of change. Essentially, the derivative of a function measures how the function's output changes as we change its input. For instance, if we think of the volume of a cylinder, knowing the derivative allows us to see how sensitive the volume is to changes in the radius, assuming other factors such as the height remain constant.

When we compute a derivative, we are looking for the rate at which one quantity changes with respect to another. In our exercise, we have the function for the volume of a cylinder, given by the formula:

• \( V = \, \pi r^2 h \)

With \( h \) being constant, the derivative \( \frac{dV}{dr} \) will tell us how much the volume changes for a tiny change in radius \( r \). Calculating this derivative involves using basic differentiation rules, where you multiply powers and reduce them by one. This helps in predicting and understanding the behavior of physical systems like the cylinder in various scenarios.

Rate of Change

The rate of change is a concept in mathematics that describes how one quantity varies in relation to another. It's essentially a comparison of how much one variable changes concerning the changes in another variable. When we talk about the rate of change in calculus, we're often referring to the derivative.

In the context of the volume of a cylinder, we are interested in how the volume (\(V\)) changes as the radius (\(r\)) changes. By finding the derivative \( \frac{dV}{dr} \), we discover the rate of change of the volume with respect to the radius. This gives us a precise measure of how sensitive the volume is to small variations in radius. To calculate this rate of change, consider the derivative calculation:

• Differentiate the volume formula \(V = 10\pi r^2\) with respect to \(r\).
• This leads to \( \frac{dV}{dr} = 20\pi r \).
• By substituting \( r = 6 \), we find \( \frac{dV}{dr} = 120\pi \).

This result tells us that for every unit increase in the radius when \(r=6\), the volume increases by \(120\pi\) cubic inches.

Volume of a Cylinder

The volume of a cylinder is a measure of how much space the cylinder occupies, and it can be easily computed using its dimensions: the radius and the height. The formula to calculate the volume \(V\) of a cylinder is given by:

• \( V = \pi r^2 h \)

where \( r \) is the radius of the base and \( h \) is the height of the cylinder. This formula comes from the idea that the base of the cylinder is a circle (with area \( \pi r^2 \)) and this circular area is extended along the height to form the entire volume.

In our exercise, we kept the height \(h\) constant at 10 inches, simplifying the volume formula to \(V = 10\pi r^2\). This tells us that, in this scenario, the volume of our cylinder directly depends on the square of its radius. Understanding this formula is crucial in solving problems involving cylinders, as it allows us to compute exactly how changes in dimensions affect the amount of space the cylinder occupies. Knowing the rate of change further informs us about how sensitive this volume is to the changes in radius, providing a deeper insight into the problem at hand.