Question

The radius \(r\) of a circle is increasing at a rate of 3 inches per minute. Find the rates of change of the area when (a) \(r=6\) inches and (b) \(r=24\) inches.

Short answer

The rate of change of the area when (a) \(r = 6\) inches is \(36\pi\) square inches per minute, and when (b) \(r = 24\) inches is \(144\pi\) square inches per minute.

Step-by-step solution

Write the Formula for the Area of the Circle

We know that the area \(A\) of a circle with radius \(r\) is given by \(A = \pi r^2\). This will be our starting point.

Implicitly Differentiate the Area formula with respect to Time

When we differentiate the area \(A = \pi r^2\) with respect to time \(t\), we get \(\frac{{dA}}{{dt}} = 2\pi r \cdot \frac{{dr}}{{dt}}\). Here \(\frac{{dA}}{{dt}}\) is the rate of change of the area and \(\frac{{dr}}{{dt}}\) is the rate of change of the radius.

Find the Rate of Change of Area When \(r = 6\) inches

Substitute \(r = 6\) inches and \( \frac{{dr}}{{dt}} = 3\) inches/minute into the equation \(\frac{{dA}}{{dt}} = 2\pi r \cdot \frac{{dr}}{{dt}}\). Thus, \(\frac{{dA}}{{dt}} = 2\pi (6) \cdot 3 = 36\pi\) square inches/minute.

Find the Rate of Change of Area When \(r = 24\) inches

Similarly, substitute \(r = 24\) inches and \( \frac{{dr}}{{dt}} = 3\) inches/minute into the equation \(\frac{{dA}}{{dt}} = 2\pi r \cdot \frac{{dr}}{{dt}}\). Thus, \(\frac{{dA}}{{dt}} = 2\pi (24) \cdot 3 = 144\pi\) square inches/minute.

Key concepts

Implicit Differentiation

Implicit differentiation is a technique in calculus used to find the derivative of a function that is not explicitly solved for one variable in terms of another. It's quite common in situations where you have an equation that defines one variable implicitly in terms of others.

In the context of our problem, we need to differentiate the area of a circle with respect to time, but the area formula, \( A = \pi r^2 \), does not directly depend on time. So, we treat the radius \( r \) as a function of time \( t \) and apply the chain rule to differentiate \( A \) with respect to \( t \). This gives us \( \frac{{dA}}{{dt}} = 2\pi r \cdot \frac{{dr}}{{dt}} \), with \( \frac{{dr}}{{dt}} \) representing the rate of change of the radius.

Related Rates

Related rates are a category of problems in calculus that determine how different rates are connected when they occur simultaneously. These problems typically involve equations where multiple variables change with respect to time.

When solving the given exercise, we assume the relationship between the area of the circle and its radius. Since the radius changes over time, the area also changes over time, and the rates at which they change are 'related'. By figuring out one rate, like the rate of change of the radius (\( \frac{{dr}}{{dt}} \), which is given as 3 inches per minute), we can determine the rate of change of the area (\( \frac{{dA}}{{dt}} \) ) using implicit differentiation.

Area of a Circle

An essential concept in geometry is the area of a circle, which is the amount of space enclosed within its circumference. The formula for the area is \( A = \pi r^2 \), where \( A \) represents the area, and \( r \) is the radius of the circle.

This simple yet pivotal relationship plays a direct role in related rates problems involving circles, as seen in our exercise. Understanding how the area changes as the radius grows or shrinks is critical for solving these types of problems. In the case study provided, knowing this formula allows us to connect the radius and the area through differentiation.

Derivative with Respect to Time

Derivatives are a foundational concept in calculus that represent the rate at which a quantity changes. When we take the derivative with respect to time, denoted as \( \frac{d}{dt} \), we're looking at how a particular variable changes as time progresses.

In our exercise, \( \frac{{dA}}{{dt}} \) represents the derivative of the area with respect to time, or simply put, it shows how the area of the circle changes for each minute that passes. By understanding the rate at which the radius changes (\( \frac{{dr}}{{dt}} \)), we can use the derivative to find the rate of change for the area.