Problem 256 - Calculus Volume 1

Question

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 \(t=4 \mathrm{~s}\).

Step-by-Step Solution

Answer

The area is expanding at approximately \(0.647\pi \) square inches per second when \(t = 4\) seconds.

Step 1: Differentiate the area with respect to the radius

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

Step 2: Differentiate the radius with respect to time

Given the radius as a function of time \(r = 2 - \frac{100}{(t+7)^2}\), differentiate it with respect to \(t\) to find \(\frac{dr}{dt}\). Use the chain rule for differentiation. \[\frac{dr}{dt} = 0 - \left(-100 \cdot \frac{-2}{(t+7)^3}\right) = \frac{200}{(t+7)^3}\]

Step 3: Apply the chain rule to find \(\frac{d A}{d t}\)

Use the chain rule \(\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}\). Substitute the results from Steps 1 and 2. \[\frac{dA}{dt} = 2\pi r \cdot \frac{200}{(t+7)^3}\]

Step 4: Evaluate \(\frac{d A}{d t}\) at \(t=4\)

First, find the radius \(r\) when \(t = 4\). \[r = 2 - \frac{100}{(4+7)^2} = 2 - \frac{100}{121} = 2 - \frac{100}{121} \approx 1.174\]Then substitute \(r\) and \(t = 4\) into \(\frac{dA}{dt}\). \[\frac{dA}{dt} = 2\pi (1.174) \cdot \frac{200}{(4+7)^3} = 2\pi (1.174) \cdot \frac{200}{1331}\]Simplify the expression to find the rate at which area is expanding.

Key Concepts

Differentiation Techniques

The chain rule is a key technique in calculus that helps us differentiate composite functions. It becomes especially important when dealing with complex differentiation problems, such as those involving multiple variables. In our example, the area of a circle depends on its radius, which in turn depends on time. This is where the chain rule shines. The formula for the chain rule is \( \frac{d A}{d t} = \frac{d A}{d r} \cdot \frac{d r}{d t} \), showing us how to obtain the derivative of \( A \) with respect to \( t \) by multiplying two derivatives: \( \frac{d A}{d r} \) and \( \frac{d r}{d t} \).

When differentiating functions, like the area \( A = \pi r^2 \) with respect to \( r \), straightforward rules such as the power rule come in handy. We simplify this by taking the derivative, which becomes \( 2\pi r \). For differentiating \( r \) with respect to \( t \), where \( r = 2 - \frac{100}{(t+7)^2} \), the chain rule guides us through the complexity, ultimately giving us \( \frac{200}{(t+7)^3} \). This is due to working with a composite function, emphasizing how powerful differentiation techniques can simplify our calculations.

Rate of Change

Understanding the rate of change is crucial in assessing how one quantity varies concerning another. In calculus, it often comes down to using derivatives. The rate of change of a circle's area, as a function of time, tells us how quickly the area changes as the circle expands. In our problem, once we have \( \frac{d A}{d r} = 2\pi r \) and \( \frac{d r}{d t} = \frac{200}{(t+7)^3} \), we multiply them to find \( \frac{d A}{d t} \), providing the rate at which the area grows over time. It illustrates how small changes in time result in changes in both the radius and the area. Rates of change thus provide essential insights into environments where change is constant, like physics or engineering, where they allow us to predict and understand behaviors of systems over time.

Applied Calculus Problems

Calculus plays a pivotal role in solving real-world problems, such as modeling how objects change over time. Such applied calculus problems take theoretical concepts and use them in practical scenarios. In our study, we explored a circle with a changing radius over time.

The solution involved understanding both the mathematical concept of differentiation and its real-life application to an expanding circle. By calculating derivatives, we determined how the area of the circle adjusts over time, an important factor in numerous scientific fields, from astronomy to engineering, where materials or spaces change dynamically. In applied calculus problems, setting up the correct equations, understanding relationships between variables, and solving them efficiently is key to deriving meaningful conclusions. It’s this bridge from theory to practice that enables us to innovate and optimize in everyday applications.