Problem 51 - Calculus Single Variable

Question

The radius of a certain cone is increasing at a rate of 6 centimeters per minute while the height is decreasing at a rate of 4 centimeters per minute. At the instant when the radius is 9 centimeters and the height is 12 centimeters, how is the volume changing?

Step-by-Step Solution

Answer

The volume is increasing at a rate of \( 324\pi \text{ cm}^3/\text{min} \).

Step 1: Understanding the Formula for the Volume of a Cone

The volume \( V \) of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \) where \( r \) is the radius and \( h \) is the height of the cone. We need to find how this volume changes over time.

Step 2: Defining Known Rates and Variables

We know \( \frac{dr}{dt} = 6 \) cm/min (radius increasing) and \( \frac{dh}{dt} = -4 \) cm/min (height decreasing). We're given \( r = 9 \) cm and \( h = 12 \) cm at the specific instant. Our goal is to determine \( \frac{dV}{dt} \), the rate at which the volume is changing.

Step 3: Applying the Chain Rule

To find \( \frac{dV}{dt} \), we need to differentiate the volume formula with respect to time \( t \). This requires using the chain rule: \( \frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right) = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right) \).

Step 4: Substituting Known Values into the Derivative

Substitute \( r = 9 \) cm, \( \frac{dr}{dt} = 6 \) cm/min, \( h = 12 \) cm, and \( \frac{dh}{dt} = -4 \) cm/min into the differentiated equation: \[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2 \times 9 \times 6 \times 12 + 9^2 \times (-4) \right) \].

Step 5: Calculating the Rate of Change of Volume

Simplify the expression to find \( \frac{dV}{dt} \): first, compute \( 2 \times 9 \times 6 \times 12 = 1296 \), and \( 9^2 = 81 \) giving \( 81 \times (-4) = -324 \). Therefore, \[ \frac{dV}{dt} = \frac{1}{3} \pi (1296 - 324) = \frac{1}{3} \pi \times 972 = 324\pi \text{ cm}^3/\text{min} \].

Key Concepts

Volume of a Cone

When dealing with problems involving a cone, understanding the basic volume formula is crucial. The volume, denoted as \( V \), of a cone is determined using the equation \( V = \frac{1}{3} \pi r^2 h \). In this formula, \( r \) stands for the radius of the base, and \( h \) represents the height of the cone.
This relationship shows how both the square of the radius and the height linearly affect the volume. Essentially, if either the radius or the height increases, the volume of the cone changes significantly. Our task in related rates problems is to determine how alterations in \( r \) and \( h \) manifest as changes in the volume over time.

Chain Rule

The chain rule is an essential tool in calculus. It helps us differentiate composite functions. In the context of related rates, it assists us in connecting changes in one variable to changes in another where they are interdependent.
To apply the chain rule, we break down the derivative of the volume formula with respect to time \( t \):

• Differentiate \( r^2 h \) using the product rule and chain rule simultaneously.
• Calculate \( \frac{d}{dt}(r^2) \) and \( \frac{d}{dt}(h) \).

The chain rule allows us to connect these derivatives back to our known rates \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \), facilitating a connection to \( \frac{dV}{dt} \), the desired rate of change of volume.

Differentiation

Differentiation is the process of finding a derivative, which indicates how a function changes as its inputs change. In this exercise, we are interested in how the volume of the cone changes when its dimensions change.
We begin by differentiating the volume formula relative to time:

• This involves applying the product and chain rule on \( r^2 h \).
• We obtain that: \( \frac{dV}{dt} = \frac{1}{3} \pi (2r \frac{dr}{dt} h + r^2 \frac{dh}{dt}) \).

Understanding differentiation in this form allows us to substitute known values directly into the derivative to assess real-world changes in volume.

Rate of Change

The rate of change provides insights into how quickly a quantity, like volume, changes over time with respect to the alterations in the related variables: the radius and the height of the cone.
In this specific scenario:

• Radius increases at \( 6 \, \text{cm/min} \).
• Height decreases at \( 4 \, \text{cm/min} \).
• These rates are substituted into our differentiated volume formula.

Substituting our known values, we compute to find the volume changing at a rate of \( 324\pi \, \text{cm}^3/\text{min} \). This solution not only shows how fast the volume is changing, but also illustrates the dynamic nature of related quantities in calculus.