Problem 77 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

OIL SPILLS In calm waters, the oil spilling from the ruptured hull of a grounded tanker spreads in all directions. Assuming that the area polluted is a circle and that its radius is increasing at a rate of \(2 \mathrm{ft} / \mathrm{sec}\), determine how fast the area is increasing when the radius of the circle is \(40 \mathrm{ft}\).

Step-by-Step Solution

Answer

When the radius of the polluted area is 40 ft, the rate at which the area is increasing is \(160π\) square feet per second.

Step 1: Write down the given information

We are given: - The area of a circle formula: \(A = πr^2\) - Rate of change of radius: \(\frac{dr}{dt} = 2\) ft/sec - The radius of the circle when we want to find the rate of change of the area: \(r = 40\) ft

Step 2: Differentiate the area expression with respect to time

Since we want to find how fast the area is increasing with respect to time, we need to differentiate the area expression concerning time. Using implicit differentiation, we get: \(\frac{dA}{dt} = \frac{d(πr^2)}{dt}\) Now apply the chain rule which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, our outer function is the area, \(A\), and the inner function is the radius, \(r\). Using the chain rule, we get: \(\frac{dA}{dt} = 2πr \cdot \frac{dr}{dt}\)

Step 3: Plug in the given values and solve for dA/dt

Now, plug in the given values of the radius, \(r = 40\) ft, and the rate of change of the radius, \(\frac{dr}{dt} = 2\) ft/sec, into the formula obtained in Step 2: \(\frac{dA}{dt} = 2π(40) \cdot (2)\) Calculate \(2π(40) \cdot (2)\) to find the value of \(\frac{dA}{dt}\): \(\frac{dA}{dt} = 160π\)

Step 4: Interpret the result

The rate of change of the area of the polluted water, \(\frac{dA}{dt}\), is equal to \(160π\) square feet per second when the radius of the circle is 40 feet. This means that the polluted area is increasing at a rate of \(160π\) square feet for every second at that moment.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique in calculus used when a function is not given in the form of a single explicit equation, like \( y = f(x) \). In problems where you have to differentiate an equation involving two variables that are not easily separable, implicit differentiation comes in handy.
Here's how it works:

• First, identify the relationship given, like \( A = \pi r^2 \). In our problem, \( A \) is a function of \( r \), and since \( r \) is changing with time, so is \( A \).
• Then, you differentiate each term with respect to the chosen variable you're interested in, usually time (\( t \) in related rates problems).
• Remember to use the chain rule where necessary to account for the rate of change in each variable involved.

This method is particularly useful in related rates problems, as it allows us to find out how fast one quantity changes with respect to another.

Area of a Circle

The area of a circle is an essential concept in geometry and is calculated using the formula \( A = \pi r^2 \), where \( A \) is the area and \( r \) is the radius. The radius is the distance from the center of the circle to any point on its boundary.

• As the radius changes, the area of the circle changes. This is particularly relevant in problems involving expanding or contracting circles, like an oil spill growing in size.
• If the radius is known, calculating the area is straightforward as you only need to square the radius and multiply it by \( \pi \).

Understanding how the area changes as the radius changes helps in solving related rates problems, where the goal is to determine how fast the area is increasing at a particular moment.

Chain Rule

The chain rule is a fundamental rule in calculus for finding the derivative of a composite function. When you have a function nested inside another, this rule becomes indispensable.
The basic idea is: if you have a function \( y = f(u) \) and \( u = g(x) \), then the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).

• In our exercise, the area \( A \) is a function of the radius \( r \), which itself changes over time. To find \( \frac{dA}{dt} \), we derive \( \frac{d}{dt}(\pi r^2) \) using the chain rule.
• This involves first multiplying by the derivative of the outer function \( (2\pi r) \) and then by the derivative of the inner function \( \frac{dr}{dt} \).

By applying the chain rule in related rates problems, you can determine the rate at which one variable is changing by also knowing how another variable it depends on is changing.