Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose we toss a rock into a pond causing a circular ripple. If the radius is increasing at a rate of 3 feet per second when the diameter is 4 , how fast is the area increasing?

Step-by-Step Solution

Answer

\(\frac{dA}{dt} = 12 \pi\) square feet per second. That is, the area of the ripple is increasing at a rate of \(12 \pi\) square feet per second when the diameter is 4.

Step 1: Express Given Information in Terms of Radius

The diameter is given as 4 feet, so the radius at this moment is half of this: \(r = 2\) feet. We know that radius is increasing at a rate of 3 feet per second, so \(\frac{dr}{dt} = 3\) feet per second.

Step 2: Use the Formula for Area to Differentiate Implicitly with Respect to Time

Since \(A = \pi r^2\), differentiating with respect to \(t\) gives us \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). This formula shows the relationship between the rate of change of the area and the other given quantities.

Step 3: Substituting Known Values to Find \(\frac{dA}{dt}\)

Substituting the values charted in step 1 into the formula obtained in step 2, \(\frac{dA}{dt} = 2\pi(2)(3)\). Calculate the value to get \(\frac{dA}{dt}\), which is the rate at which the area is increasing.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique in calculus used when dealing with equations that define one variable implicitly in terms of another. Instead of explicitly isolating one variable and differentiating, both sides of an equation are differentiated as they are.

Consider the equation for the area of a circle, which is not explicitly solved for one variable: \(A = \(pi\) r^2\).To find the rate of change of the area (\(\frac{dA}{dt}\)) with respect to time (\(t\)) when we only know the rate of change of the radius (\(\frac{dr}{dt}\)), implicit differentiation is employed.We differentiate both sides of the area equation with respect to time, applying the chain rule to the right-hand side. This gives us \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\), which tells us how fast the area is changing.

Related Rates

Related rates are a concept in calculus that deals with finding the rate at which one quantity is changing in relation to another. The connection between two or more variables that are changing over time is frequently given or implied in word problems.

In the case of the circular ripple in the pond, we have two related rates:

• The rate of change of the radius (\(\frac{dr}{dt}\))
The rate of change of the area (\(\frac{dA}{dt}\))

To solve for the latter, we need to connect the two using the known relationships between radius and area, as seen in the solution to the exercise. This typical application of the related rates concept allows us to see how one changing quantity, like radius, can affect another, such as the area.

Calculus Word Problems

Calculus word problems, such as the one involving the expanding ripple in a pond, challenge students to translate a real-world scenario into mathematical terms. Key steps to tackle these problems include:

• Identifying the quantities that are changing over time.
Expressing the relationships between these quantities with equations.Using calculus, particularly differentiation, to find the rate at which one quantity is changing in relation to another.

Clarity in representing known rates (like \(\frac{dr}{dt} = 3\ \text{feet per second}\)) and the relationship between variables (such as \(A = \(pi\) r^2\)) is crucial for solving these problems efficiently and accurately.

Area of a Circle

The area of a circle is a fundamental concept in geometry defined by the formula \(A = \(pi\) r^2\), where \(A\) is the area, \(\pi\) is a constant (approximately 3.14159), and \(r\) is the radius of the circle.

Understanding how to work with the area formula is essential in calculus when dealing with rates of change, as it is the basis for setting up the related rates equation. In the rock and ripple problem, knowing that the diameter is 4 feet allowed us to establish the radius as \(r = 2\) feet. Then, clearly understanding this geometric relationship helped us to correctly apply calculus principles and find the rate at which the area of the circle is increasing.