Problem 18 - Calculus Volume 1

Question

Draw and label diagrams to help solve the related-rates problems. The radius of a circle increases at a rate of \(2 \mathrm{~m} / \mathrm{sec}\). Find the rate at which the area of the circle increases when the radius is \(5 \mathrm{~m}\).

Step-by-Step Solution

Answer

The area increases at a rate of \( 20\pi \mathrm{~m^2/s} \).

Step 1: Understanding the Problem

We know the relationship between the radius and the area of a circle. The area \( A \) of a circle is given by the formula \( A = \pi r^2 \), where \( r \) is the radius.

Step 2: Identifying Given Values

The problem states that the radius is increasing at a rate of \( \frac{dr}{dt} = 2 \mathrm{~m/s} \). We are asked to find \( \frac{dA}{dt} \) when the radius \( r = 5 \mathrm{~m} \).

Step 3: Applying Derivatives

To find \( \frac{dA}{dt} \), we use the chain rule for derivatives. Differentiate \( A = \pi r^2 \) with respect to time \( t \):\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot 2r \cdot \frac{dr}{dt}. \]

Step 4: Substitute and Calculate

Substitute the given values, \( r = 5 \mathrm{~m} \) and \( \frac{dr}{dt} = 2 \mathrm{~m/s} \), into the derived formula:\[ \frac{dA}{dt} = \pi \cdot 2 \cdot 5 \cdot 2 = 20\pi \mathrm{~m^2/s}. \]

Step 5: Conclude the Solution

Thus, the rate at which the area of the circle increases when the radius is \( 5 \mathrm{~m} \) is \( 20\pi \mathrm{~m^2/s} \).

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate compositions of functions. When a function is composed of another function, such as the area of a circle given as a function of its radius, we use the chain rule to find its derivative with respect to an external variable like time.
The chain rule formula is written as:

• If you have a function composed as \( y = f(g(x)) \), the derivative using the chain rule is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In the related rates problem involving a circle, the area \( A = \pi r^2 \) depends on the radius \( r \), which changes with time \( t \). Therefore, to find the rate of change of the area with respect to time, we apply the chain rule:

• First, find the derivative of \( A \) with respect to \( r \), yielding \( 2\pi r \).
• Then multiply this by \( \frac{dr}{dt} \) (the rate of change of the radius with time) to get \( \frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} \).

This chain of derivation helps connect how a change in radius affects the area's change over time.

Circle Area Formula

The formula for calculating the area of a circle is a fundamental concept in geometry. Given as \( A = \pi r^2 \), it expresses how the area \( A \) of a circle is directly proportional to the square of its radius \( r \).
This relationship shows why a small change in the radius leads to a larger impact on the area due to squaring.

• \( \pi \) (pi) is a constant roughly equal to 3.14159...
• \( r \) is the radius of the circle, the distance from the center to the edge.

In problems involving changing rates, understanding that even small increases in \( r \) can significantly change \( A \) is crucial. In our specific task, as the radius increases, the area does as well, dynamically tied to \( r^2 \).
This understanding forms the basis of how we tackle related rates problems concerning circles.

Radius Rate Change

In related rates problems, how quickly a radius changes is pivotal. This is often denoted as \( \frac{dr}{dt} \), representing the change of the radius over time.
When given, this rate allows us to determine other linked rates, such as the change in area.

• Ensure you understand the unit of \( \frac{dr}{dt} \). Here, the radius rate change is \( 2 \text{ m/s} \).
• A positive rate indicates the radius is expanding, while a negative rate would imply contraction.

In our scenario, the steady rate at which the circle expands \( 2 \text{ m/s} \) dictates how other attributes—like area—adjust with time. Substituting this value into our derived formula for \( \frac{dA}{dt} \) showcases the interconnectedness of temporal changes.

Differentiation Techniques

Differentiation is the cornerstone of solving related rates problems. Here, you often employ various techniques to find derivatives of functions, making sense of how different variables dynamically interact.
For our circle area problem, we apply simple differentiation and the chain rule to get meaningful insights into rates of change.

• First, recognize the function to differentiate. Here, it's \( A = \pi r^2 \).
• Use basic differentiation: The derivative of \( r^2 \) with respect to \( r \) is \( 2r \).

Then apply the chain rule to extend this derivative to time-dependent changes:

• You find \( \frac{dA}{dr} = 2\pi r \) initially.
• To consider changes over time, complete it with \( \frac{dr}{dt} \), yielding \( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \).

Understanding these techniques enables us to seamlessly connect changes in radius to area, predicting how the circle grows over time.