Problem 18 - Calculus Volume 1

Question

For the following exercises, draw and label diagrams to help solve the related-rates problems. The radius of a circle increases at a rate of 2 m/sec. Find the rate at which the area of the circle increases when the radius is 5 m.

Step-by-Step Solution

Answer

The area increases at a rate of \(20\pi\) m²/sec when the radius is 5 m.

Step 1: Understand the Problem

We are given a circle whose radius is increasing at a constant rate of 2 m/sec. We need to find how fast the area of the circle increases when the radius is 5 m.

Step 2: Write Down the Known Rates

The rate of change of the radius with respect to time is given by \( \frac{dr}{dt} = 2 \text{ m/sec} \).

Step 3: Express the Area of the Circle

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

Step 4: Relate the Area Change to Radius Change

To find how fast the area is changing, differentiate the area equation with respect to time \( t \):\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} \].

Step 5: Substitute Known Values

Substitute \( r = 5 \) m and \( \frac{dr}{dt} = 2 \text{ m/sec} \) into the derivative:\[ \frac{dA}{dt} = 2\pi (5)(2) = 20\pi \text{ m}^2/\text{sec} \].

Step 6: Solve for the Rate of Change of Area

The rate at which the area of the circle is increasing when the radius is 5 m is \( 20\pi \text{ m}^2/\text{sec} \).

Key Concepts

Rate of Change

In mathematical problems, especially those involving motion or growth, understanding the rate of change is crucial. A rate of change tells us how much a quantity changes over a given period. In our exercise, we are dealing with the rate of change of the radius of a circle and the resulting rate of change in its area. The radius increases at a rate of 2 meters per second, which means that every second, the radius gets 2 meters longer.

When solving related rates problems, we often relate two or more rates of change to find an unknown rate. In this case, we are looking at how the continuous increase of the radius affects the area of the circle. By finding how fast the area changes, we apply what we've learned about the relationship between different variables and their respective rates.

Differentiation

Differentiation is a powerful tool in calculus that allows us to find the rate at which one quantity changes concerning another. It's essential in solving related rates problems because it helps us express how quantities, like area, are affected by changing elements like the radius.

To find how fast the circle's area is increasing, we differentiated the area formula with respect to time. The area of a circle is given by the formula \( A = \pi r^2 \). By applying differentiation, \( \frac{d}{dt} (\pi r^2) \), we find the rate of change of the area in terms of the rate of change of the radius. This results in the expression \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \), which links how fast the area changes to how fast the radius grows.

Circle Area

A circle's area is one of the simplest yet fundamental concepts in geometry. It provides insight into the physical space enclosed by a circle. The formula for the area of a circle is \( A = \pi r^2 \), where \( r \) is the radius. This tells us that the area depends on the square of the radius, illustrating that any change in the radius significantly impacts the circle's area.

• The larger the radius, the exponentially larger the area.
• Changes in the radius affect the area more than linearly due to the square in the formula.

In the given problem, as the radius increases over time, so does the area. Understanding this dependent relationship is vital, as it helps us realize why changes in one dimension (like the radius) can lead to more significant changes in outcomes like area. Consequently, by calculating \( \frac{dA}{dt} \) when \( r = 5 \) meters with \( \frac{dr}{dt} = 2 \) meters per second, we see how a small increase in radius results in a substantial area change, emphasizing the square relationship.