Problem 253 - Calculus 3

Question

The radius of a right circular cone is increasing at 3 \(\mathrm{cm} / \mathrm{min}\) whereas the height of the cone is decreasing at 2 \(\mathrm{cm} / \mathrm{min}\) . Find the rate of change of the volume of the cone when the radius is 13 \(\mathrm{cm}\) and the height is 18 \(\mathrm{cm} .\)

Step-by-Step Solution

Answer

The rate is approximately 1115.38 cm³/min.

Step 1: Understand the Cone Volume Formula

The volume \( V \) of a right circular cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius and \( h \) is the height.

Step 2: Differentiate the Volume with Respect to Time

We need to find the rate of change of the volume \( \frac{dV}{dt} \). We apply the chain rule to differentiate \( V = \frac{1}{3} \pi r^2 h \):\[ \frac{dV}{dt} = \frac{1}{3} \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right). \]

Step 3: Substitute Given Rates of Change

We know \( \frac{dr}{dt} = 3 \, \text{cm/min} \) and \( \frac{dh}{dt} = -2 \, \text{cm/min} \). Substitute these into the differentiated volume formula:\[ \frac{dV}{dt} = \frac{1}{3} \pi (2rh \cdot 3 + r^2 \cdot (-2)). \]

Step 4: Substitute Specific Values of Radius and Height

Substitute the given values \( r = 13 \, \text{cm} \) and \( h = 18 \, \text{cm} \) into the equation:\[ \frac{dV}{dt} = \frac{1}{3} \pi (2 \cdot 13 \cdot 18 \cdot 3 + 13^2 \cdot (-2)). \]

Step 5: Calculate the Result

First, calculate each part separately:\(2 \cdot 13 \cdot 18 \cdot 3 = 1404\) and \(13^2 \cdot (-2) = -338\).Then substitute these back to find \( \frac{dV}{dt} \):\[ \frac{dV}{dt} = \frac{1}{3} \pi (1404 - 338) = \frac{1}{3} \pi (1066). \]

Step 6: Solve for the Rate of Change

Now calculate the numeric value:\[ \frac{dV}{dt} = \frac{1}{3} \times 1066 \times \pi = 355.33\pi \, \text{cm}^3/\text{min}. \]Thus, the rate of change of the cone's volume is approximately \( 1115.38 \, \text{cm}^3/\text{min} \).

Key Concepts

Right Circular Cone

A right circular cone is a three-dimensional geometric shape that has a flat circular base and a single curved surface that tapers to a point known as the apex. This type of cone is symmetrical around its vertical axis, and it is important in many mathematical calculations. Understanding the properties of a right circular cone can help us solve problems involving its surface area and volume.

Let's focus on its formula for volume calculation:

• The volume of a right circular cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius of the base, \( h \) is the height of the cone from the base to the apex, and \( \pi \) is a mathematical constant approximately equal to 3.14159.
• This formula is derived from the volume of a cylinder, \( V = \pi r^2 h \), considering a cone as one-third of it with the same base and height.

To tackle related rates problems involving cones, one needs to be comfortable with this volumetric formula and how changes in \( r \) and \( h \) impact the cone's volume.

Volume Derivative

In related rates problems, we often need to find the derivative of the volume with respect to time. For a right circular cone, this involves differentiating the volume formula \( V = \frac{1}{3} \pi r^2 h \) with respect to time \( t \).

Remember, both \( r \) (radius) and \( h \) (height) can be functions of time, which means their rates of change, \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \), influence how quickly the volume changes. To find \( \frac{dV}{dt} \), the derivative of volume with respect to time, you apply the formula:

• The derivative of the volume is: \( \frac{dV}{dt} = \frac{1}{3} \pi \left(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}\right) \). This is obtained using the product rule and the chain rule of calculus as we differentiate with respect to time.
• It reflects how both the increase in radius and the decrease in height simultaneously affect the overall volume of the cone.

Understanding this derivative is crucial to solving related rates problems involving the volume of cones.

Chain Rule Differentiation

The chain rule is a fundamental concept in calculus used to differentiate composite functions. When solving related rates problems that involve multiple variables changing over time, the chain rule allows us to find how one quantity changes in relation to another.

In our cone problem, the chain rule helps articulate the relationship between the rates of radius and height changes, and how they combine to affect the volume:

• Applying the chain rule, \( \frac{dV}{dt} \) involves differentiating \( r^2 \) and \( h \), both functions of time. This means incorporating both \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \) into the derivative of the volume.
• The derivative \( 2rh \frac{dr}{dt} \) concerns the part of the volume relating to \( r^2h \), while \( r^2 \frac{dh}{dt} \) corresponds to the height \( h \).

Mastering the chain rule is essential as it turns complex relationships into manageable calculus equations—allowing us to compute exact rates of change in geometrical problems like these.