Problem 256 - Calculus Volume 1

Question

[T] The formula for the area of a circle is \(A=\pi r^{2}\) where \(r\) is the radius of the circle. Suppose a circle is expanding, meaning that both the area \(A\) and the radius \(r\) (in inches) are expanding. a. Suppose \(r=2-\frac{100}{(t+7)^{2}}\) where \(t\) is time in seconds. Use the chain rule \(\frac{d A}{d t}=\frac{d A}{d r} \cdot \frac{d r}{d t}\) to find the rate at which the area is expanding. b. Use a. to find the rate at which the area is expanding at \(t=4\) s.

Step-by-Step Solution

Answer

At \( t = 4 \) seconds, the area is expanding at approximately 3.53 square inches per second.

Step 1: Differentiate the Area Formula with Respect to Radius

The formula for the area of a circle is given by \( A = \pi r^2 \). To find \( \frac{dA}{dr} \), differentiate the area \( A \) with respect to the radius \( r \): \( \frac{dA}{dr} = \pi \cdot 2 \cdot r = 2\pi r \).

Step 2: Differentiate the Radius Formula with Respect to Time

The given radius as a function of time is \( r = 2 - \frac{100}{(t+7)^2} \). Differentiate this function with respect to \( t \) to find \( \frac{dr}{dt} \). You can use the chain rule: \( \frac{d}{dt} \left( 2 - \frac{100}{(t+7)^2} \right) = 0 + \frac{200}{(t+7)^3} \).

Step 3: Apply the Chain Rule to Find \(\frac{dA}{dt}\)

According to the chain rule, \( \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} \). Substituting in the derivatives we found, \( \frac{dA}{dt} = (2\pi r) \cdot \frac{200}{(t+7)^3} = \frac{400\pi r}{(t+7)^3} \).

Step 4: Substitute Values to Find Rate at \( t = 4 \)

First, find \( r \) at \( t = 4 \) using \( r = 2 - \frac{100}{(t+7)^2} \): \( r = 2 - \frac{100}{(11)^2} = 2 - \frac{100}{121} = 2 - \frac{100}{121} \approx 1.17 \). Now, substitute this \( r \) and \( t = 4 \) into the rate equation: \( \frac{dA}{dt} = \frac{400\pi \times 1.17}{(11)^3} \approx \frac{400\pi \times 1.17}{1331} \). Calculate the final result.

Step 5: Calculate the Numerical Values

Compute \( \frac{dA}{dt} \approx \frac{400 \times 3.1416 \times 1.17}{1331} \approx \frac{1468.8 \pi}{1331} \approx 3.53 \) square inches per second.

Key Concepts

Area of a Circle

Understanding the area of a circle is a fundamental concept in geometry. The formula for the area of a circle is given by \( A = \pi r^2 \), where \( A \) represents the area and \( r \) is the radius. This formula originates from the relationship between a circle's diameter and its circumference, defined by the constant \( \pi \) (pi), approximately 3.1416.
To visualize this, think of \( \pi r^2 \) as the number of unit squares that can fit inside the circle. The radius \( r \) is crucial because it determines how large or small the circle is. Changes in the radius have a direct impact on the area due to the squared term, meaning that as the radius increases, the area grows rapidly.

Differentiation

Differentiation is a core concept in calculus, offering a way to understand how a function changes at any point. When we talk about differentiating the area of a circle with respect to its radius, we want to know how the area changes as the radius changes.
Using calculus, you differentiate \( A = \pi r^2 \) to get \( \frac{dA}{dr} = 2\pi r \). Here, \( \frac{dA}{dr} \) represents the rate of change of the area concerning the radius. This derivative tells us how small increments in the radius affect the area. It's like stretching a rubber band to make a larger circle, where the change in length impacts the size of the area.

Rate of Change

The rate of change provides insights into how quickly something is happening. In the context of our circle, we're interested in how fast the area is expanding over time when the radius is also changing. This is particularly useful in applications where the size of circles is modified dynamically, such as inflating balloons.
The chain rule in calculus is our tool here. We use \( \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} \) to combine the rates of change of area with respect to radius and radius with respect to time. This formula allows us to calculate how quickly the area changes as both the radius and time vary simultaneously.

Derivatives

Derivatives are a mathematical tool used to determine how a quantity changes over time or another variable. They form the backbone of calculus and provide the means to solve problems involving rates of change.
In our problem, derivatives help us find how the area of an expanding circle changes over time. We first find the derivative of the area with respect to the radius, \( \frac{dA}{dr} \), and the derivative of the radius with respect to time, \( \frac{dr}{dt} \), and then use the chain rule to find \( \frac{dA}{dt} \). Computing these derivatives allows us to understand and quantify dynamic processes, like the growth of the circle's area as its radius increases over time.