Problem 133 - Calculus 3

Question

Express the volume of a right circular cylinder as a function of two variables: a. its radius \(r\) and its height \(h .\) b. Show that the rate of change of the volume of the cylinder with respect to its radius is the product of its circumference multiplied by its height. c. Show that the rate of change of the volume of the cylinder with respect to its height is equal to the area of the circular base.

Step-by-Step Solution

Answer

The volume is \( V = \pi r^2 h \); \( \frac{\partial V}{\partial r} = 2\pi r h \); \( \frac{\partial V}{\partial h} = \pi r^2 \).

Step 1: Volume Formula of the Cylinder

The volume \( V \) of a right circular cylinder can be expressed in terms of its radius \( r \) and height \( h \) using the formula: \[ V = \pi r^2 h \]This formula comes from the area of the base circle (\( \pi r^2 \)) multiplied by the height \( h \) of the cylinder.

Step 2: Derivative with Respect to Radius

To find the rate of change of the volume with respect to the radius \( r \), we differentiate the volume \( V = \pi r^2 h \) with respect to \( r \):\[ \frac{\partial V}{\partial r} = \frac{\partial}{\partial r}(\pi r^2 h) = 2\pi r h \]This result means that the rate of change of the volume with respect to the radius is the product of the circumference of the base \( 2\pi r \) and the height \( h \).

Step 3: Derivative with Respect to Height

To find the rate of change of the volume with respect to the height \( h \), we differentiate the volume \( V = \pi r^2 h \) with respect to \( h \):\[ \frac{\partial V}{\partial h} = \frac{\partial}{\partial h}(\pi r^2 h) = \pi r^2 \]This result shows that the rate of change of the volume with respect to the height is equal to the area of the circular base \( \pi r^2 \).

Key Concepts

Rate of Change

Understanding the rate of change is essential when analyzing how a cylinder's volume varies with changes in its dimensions. The rate of change essentially looks at how one quantity changes in relation to another. When dealing with a cylinder, key factors include changes in its radius or its height. For instance, adjusting the radius while keeping the height constant leads to a different rate of change compared to the scenario where the height changes and the radius remains fixed.
This concept ties deeply into how we approach calculating derivatives, allowing us to understand not just the 'how,' but also the 'why' behind these changes. Thus, the rate of change gives us insight into the behavior of the volume as its dimensions vary.

Partial Derivatives

Partial derivatives play a crucial role in finding the rate of change of a cylinder's volume with respect to one variable at a time, treating other variables as constants. When considering a function of more than one variable, such as the volume of a cylinder dependent on both radius and height, partial derivatives show how the function changes relative to one variable.
In the formula for the cylinder volume, \[ V = \pi r^2 h \]we can focus on the effect of changing the radius or the height:

• For the radius, the partial derivative is \[ \frac{\partial V}{\partial r} = 2\pi r h \]
• For the height, it's \[ \frac{\partial V}{\partial h} = \pi r^2 \]

This calculation highlights the separate impact of each variable on the volume, providing a clearer application of how these derivatives function in applied problems.

Volume Formula

The volume formula for a cylinder is a foundational concept in geometry and calculus. The standard formula for the volume of a right circular cylinder is \[ V = \pi r^2 h \]This represents the area of the base circle, \[ \pi r^2 \]multiplied by the height, \[ h \]. This formula neatly ties together the three-dimensional shape with its two-dimensional base and one-dimensional height.
Understanding this formula allows us to see how changing either the radius or the height will affect the volume. It takes its roots from basic principles of geometry, expanding into more complex problem-solving through differentiation to study how dimensions influence volume. This formula not only simplifies calculating the volume for practical problems but also serves as a gateway to exploring more advanced mathematical concepts such as rate of change and partial derivatives.