Problem 6 - Calculus - AP Edition

Question

In parts (a)-(d), let \(A\) be the area of a circle of radius \(r\) and assume that \(r\) increases with the time \(t\) (a) Draw a picture of the circle with the labels \(A\) and \(r\) placed appropriately. (b) Write an equation that relates \(A\) and \(r\) (c) Use the equation in part (b) to find an equation that relates \(d A / d t\) and \(d r / d t\) (d) At a certain instant the radius is \(5 \mathrm{cm}\) and increasing at the rate of \(2 \mathrm{cm} / \mathrm{s}\). How fast is the area increasing at that instant?

Step-by-Step Solution

Answer

The area is increasing at \( 20\pi \) cm²/s.

Step 1: Draw the circle

Draw a circle and label its radius as \( r \) and its area as \( A \). The radius extends from the center of the circle to any point on its circumference. Place \( r \) as a line segment inside the circle and \( A \) as a reference to the entire area inside the circle.

Step 2: Relationship between Area and Radius

Recall that the area \( A \) of a circle is given by the formula \( A = \pi r^2 \). This equation relates the area of the circle to its radius.

Step 3: Differentiate the Area Equation with Respect to Time

To find how the area changes with time, differentiate the equation \( A = \pi r^2 \) with respect to time \( t \). The derivative is \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \). This equation relates the rate of change of the area \( \frac{dA}{dt} \) to the rate of change of the radius \( \frac{dr}{dt} \).

Step 4: Calculate How Fast the Area is Increasing

At the given instant, the radius is \( r = 5 \) cm and is increasing at \( \frac{dr}{dt} = 2 \) cm/s. Substitute these values into the differentiated equation: \( \frac{dA}{dt} = 2\pi (5)(2) \). Simplifying gives \( \frac{dA}{dt} = 20\pi \) cm²/s. Therefore, the area is increasing at a rate of \( 20\pi \) cm²/s.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate of change of one quantity concerning another. When we talk about differentiation with related rates, it's all about understanding how two or more related variables evolve together as time progresses.

In the context of the circle, the radius and the area of the circle are linked. As the radius of the circle changes over time, so does its area. By differentiating the area with respect to time, we're essentially finding out how the area changes as the radius changes.

Consider the formula for the area of a circle:

• \( A = \pi r^2 \)

Differentiating both sides with respect to time \( t \) gives us:

• \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)

This equation tells us the instantaneous rate at which the area \( A \) changes, based on the rate at which the radius \( r \) changes over time. Differentiation in related rates provides a way to link those dynamic changes.

Circle Area Formula

The area formula of a circle is one of the most fundamental formulas in geometry. It's given by

• \(A = \pi r^2 \)

, where \( A \) is the area, \( r \) is the radius, and \( \pi \) is a constant approximately equal to 3.14159. This formula arises from the relationship between the radius and the space enclosed within the circle.

The circle area formula helps in determining how much space is inside the circle. When solving problems involving circles, grasping this formula allows students to calculate areas easily.

The formula is crucial later on when finding the rate of increase of the area as its radius changes over time. The connection between \( A \) and \( r \) forms the basis for understanding many real-world applications, where precise area measurements are critical.

Radius and Area Relationship

The radius and area relationship in a circle is a key connection that defines how changes in the radius affect the area. When computing related rates, you need to use this very relationship to understand how one quantity influences another.

The circle's area formula,

• \( A = \pi r^2 \)

, directly ties the area \( A \) to the square of the radius \( r \). This means that any increase or decrease in the radius results in a quadratic effect on the area.

For example, when the radius increases by a small amount, the area grows even more significantly because it's linked by the square of the radius.

• For instance, if the radius is 5 cm and increases at a rate of 2 cm/s, we can see the area increase drastically based on the previous calculations to \( \frac{dA}{dt} = 20\pi \) cm²/s.

This pronounced effect highlights the substantial impact that changes in the radius have on the area. In related rates problems, focusing on this dependency helps unravel how intertwined these measures are over time.