Problem 6 - Calculus Early Transcendentals

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 a rate of \(20\pi\, \text{cm}^2/\text{s}\).

Step 1: Understand the Circle

Draw a circle with the center labeled and a radius line extending to the edge of the circle. Label this radius as \(r\). The area \(A\) is the measure of the space contained within this circle.

Step 2: Establish Relationship Between Area and Radius

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

Step 3: Differentiate the Area with Respect to Time

Use implicit differentiation on the equation \(A = \pi r^2\) with respect to time \(t\). Since \(A\) and \(r\) both change over time, differentiate to get \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\), which relates the rate of change of area to the rate of change of radius.

Step 4: Calculate the Rate at the Given Instant

Substitute the given radius \(r = 5\, \text{cm}\) and the rate of change of the radius \(\frac{dr}{dt} = 2\, \text{cm/s}\) into the equation \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). This gives \(\frac{dA}{dt} = 2\pi \times 5 \times 2\), which simplifies to \(20\pi\, \text{cm}^2/\text{s}\).

Key Concepts

Differentiation

Differentiation is a core concept in calculus, and it captures the idea of how a quantity changes with respect to another. Imagine driving a car; differentiation helps us understand how fast the car is accelerating or decelerating at any point in time by examining the rate of change of speed.
When we relate differentiation to a circle's area, we think about how the area changes as the circle's radius increases. This concept is particularly useful in "related rates" problems, where two or more quantities change with time.
In the context of our circle, both the area of the circle and its radius are changing over time. We use the derivative to understand these changes. If we know how fast the radius is changing, differentiation allows us to calculate how fast the area is changing too. It’s like translating the change in one aspect of an object (radius) into a different aspect (area).

• We use implicit differentiation, which involves applying differentiation rules to both sides of an equation.
• It requires understanding the chain rule, which finds derivatives of composite functions.

The process is made simple if we practice identifying which quantities depend on each other and how to break down these dependencies using calculus.

Circle Area Formula

The circle area formula is fundamental in geometry and calculus. It's given as \[ A = \pi r^2 \]where \( A \) is the area of the circle, and \( r \) is its radius. This formula showcases the relationship between a circle's radius and its area.
To apply this formula, you simply need to know the radius. The area directly depends on the square of the radius, meaning as the radius changes, the area changes exponentially. If you double the radius, the area increases fourfold.

• \( \pi \) is a constant that approximates 3.14159, which links the circular dimensions in Euclidean geometry.

Knowing this formula is essential when dealing with related rates problems, as it provides a straightforward way to express the relationship that can be differentiated to find rates of change. Remember, it's not just a static formula; in calculus, it's dynamic and helps us explore changing scenarios.

Implicit Differentiation

Implicit differentiation is a method used when it's challenging to express one variable solely in terms of another. In many real-world problems, variables are interlinked, and solving for one explicitly isn't straightforward.
In our circle problem, we differentiate the formula \[ A = \pi r^2 \]with respect to time \( t \). As neither \( A \) nor \( r \) is independent of time, we apply implicit differentiation.
This means we differentiate each variable with respect to \( t \):

• The derivative of \( A \) becomes \( \frac{dA}{dt} \).
• The derivative of \( \pi r^2 \) becomes \( 2\pi r \frac{dr}{dt} \).

This technique allows us to find out how fast one quantity changes compared to another indirectly, without needing an explicit function. You'll encounter implicit differentiation often in calculus, especially when dealing with time-dependent processes.