Question

The part of a tree normally sawed into lumber is the trunk, a solid shaped approximately like a right circular cylinder. If the radius of the trunk of a certain tree is growing \(\frac{1}{2}\) inch per year and the height is increasing 8 inches per year, how fast is the volume increasing when the radius is 20 inches and the height is 400 inches? Express your answer in board feet per year ( 1 board foot \(=1\) inch by 12 inches by 12 inches).

Short answer

The volume is increasing at \( 100\pi \) board feet per year.

Step-by-step solution

Define the Volume Formula

The volume of a cylinder is given by the formula \( V = \pi r^2 h \), where \( V \) is the volume, \( r \) is the radius, and \( h \) is the height. We will use this formula to find the rate at which the volume is increasing with respect to time.

Differentiate the Volume Formula

To determine how fast the volume is increasing, we need to find the derivative \( \frac{dV}{dt} \). Using the product rule, we have:\[ \frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]This accounts for the rate of change of both the radius and the height.

Substitute Known Values

We know the current radius \( r = 20 \) inches, the height \( h = 400 \) inches, the rate of change of radius \( \frac{dr}{dt} = \frac{1}{2} \) inch/year, and the rate of change of height \( \frac{dh}{dt} = 8 \) inches/year. Substitute these values into the differentiated volume equation:\[ \frac{dV}{dt} = \pi \left( 2(20)(400)\left(\frac{1}{2}\right) + (20)^2(8) \right) \]

Calculate the Rate of Change of Volume

Now, compute \( \frac{dV}{dt} \) using the substituted values:\[ \frac{dV}{dt} = \pi \left( 2 \times 20 \times 400 \times \frac{1}{2} + 400 \times 8 \right) \]Simplifying this gives:\[ \frac{dV}{dt} = \pi \left( 8000 + 6400 \right) = \pi \times 14400 \]Thus, \( \frac{dV}{dt} = 14400\pi \) cubic inches per year.

Convert to Board Feet per Year

To convert the rate from cubic inches per year to board feet per year, use the conversion factor: 1 board foot = 144 cubic inches.\[ \text{Rate in Board Feet} = \frac{14400\pi}{144} = 100\pi \]Therefore, the volume is increasing at \( 100\pi \) board feet per year.

Key concepts

Cylinder Volume Differentiation

When it comes to solving problems involving the volume of a cylinder, a key focus is understanding how changes in dimensions affect volume. The formula for the volume of a cylinder is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cylinder. This formula comes from multiplying the area of the circular base \( \pi r^2 \) by the height \( h \).

To find how the volume changes over time, we differentiate the volume formula with respect to time \( t \). This requires applying the product rule for differentiation since two variables, \( r \) and \( h \), are involved. The derivative \( \frac{dV}{dt} = \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) \) gives us the rate at which the volume increases, accounting for changes in both radius and height over time.

This differentiation process is fundamental because it helps determine how sensitive the volume is to the changes in either the radius or the height, which is crucial in many real-world applications like manufacturing or environment studies where such changes are common.

Related Rates

Related rates problems like these occur often in calculus whenever you have multiple variables that change over time. They are called 'related rates' because the rate of change of one variable is connected or related to the rate of change of another variable through an equation. In this scenario, the problem involves the rates of change of the radius and height of the cylinder, both contributing to the change in the cylinder's volume.

To solve related rates problems, follow these steps:

• Identify all variables and their rates of change.
• Express the relationship between the variables using a formula (e.g., volume of a cylinder).
• Differentiate the formula with respect to time, using the chain rule and the product rule as necessary.
• Substitute the known values and solve for the unknown rate.

Related rates questions typically involve real-life applications, making it important to grasp how variables affect each other. Their usefulness spans across fields such as physics, engineering, and economics—all areas where understanding how one thing changes concerning another is crucial.

Rate of Change in Geometry

In geometry, understanding how different parameters of shapes and objects change is critical. Rate of change refers to how fast a quantity, like volume or area, increases or decreases over time as some of its dimensions change. This concept is fundamental in solving many calculus problems, providing insights into the dynamics of geometric figures over time.

In the cylinder example, rate of change in geometry is illustrated by the changing radius and height, directly influencing the volume. This problem becomes a study of how rates can be manipulated and understood using calculus, recognizing that changes in physical dimensions often have a large impact on other properties, like the amount of material or space contained.

By examining rates of change, we can predict future dimensions, optimize designs for maximum volume or efficiency, and better understand natural processes like growth in biological systems. The ability to model these changes is a major advantage in numerous areas such as architecture, agriculture, and environmental science.