Problem 53 - Applied Calculus

Question

(a) Use the formula for the area of a circle of radius \(r\), \(A=\pi r^{2}\), to find \(d A / d r\) (b) The result from part (a) should look familiar. What does \(d A / d r\) represent geometrically? (c) Use the difference quotient to explain the observation you made in part (b).

Step-by-Step Solution

Answer

\(dA/dr = 2\pi r\), representing the circumference.

Step 1: Differentiate the Area Formula

Given the formula for the area of a circle \(A = \pi r^2\), let's differentiate with respect to \(r\) to find \(\frac{dA}{dr}\). Differentiating, we get: \[ \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r \]

Step 2: Interpret the Derivative

The derivative \(\frac{dA}{dr} = 2\pi r\) represents the rate of change of the area with respect to the radius. Geometrically, it gives the circumference of the circle, since \(2\pi r\) is the formula for the circumference.

Step 3: Use the Difference Quotient

The difference quotient for the area with respect to the radius is:\[\frac{A(r+h) - A(r)}{h} = \frac{\pi (r+h)^2 - \pi r^2}{h}\]Expanding the numerator, we get:\[\frac{\pi r^2 + 2\pi rh + \pi h^2 - \pi r^2}{h} = \frac{2\pi rh + \pi h^2}{h}\]Simplifying, we have:\[2\pi r + \pi h\]As \(h \rightarrow 0\), this approaches \(2\pi r\), confirming that \(\frac{dA}{dr} = 2\pi r\). This illustrates that the rate of change of the area with respect to the radius is indeed the circumference.

Key Concepts

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. A derivative represents how a function changes as its input changes. One way to visualize this is to think about the slope of a tangent line to the curve of the function on a graph.

In the case of finding the derivative of the area of a circle with respect to its radius, we start with the area formula:

• Given: \( A = \pi r^2 \)
• To find: \( \frac{dA}{dr} \)

By differentiating \( \pi r^2 \) with respect to \( r \), we use the power rule of differentiation, bringing down the power and reducing it by one:
\( \frac{d}{dr}(\pi r^2) = 2\pi r \).

This derivative, \( 2\pi r \), gives us an instantaneous rate of change of the area as the radius changes, crucial for understanding how small changes in radius alter the area of a circle.

Rate of Change

The rate of change is a way to understand how one quantity changes in relation to another. For the area of a circle, the rate of change with respect to the radius is represented by the derivative \( \frac{dA}{dr} = 2\pi r \). This value tells us how fast the area increases as the radius increases.

To break it down:

• When the radius of a circle increases, the circle does not just grow in a straightforward, linear way; the area increases more rapidly.
• The derivative \( 2\pi r \) specifically shows how the small increase in radius results in a larger change in area.

In practical terms, this provides valuable insight into how altering one dimension (the radius) affects another key property (the area) of a circle. Understanding rates of change is essential in fields ranging from physics to economics, where dynamic systems are under study.

Geometric Interpretation

Geometric interpretation provides a visual perspective to mathematical concepts, making them easier to grasp. The derivative \( \frac{dA}{dr} = 2\pi r \) has an interesting geometric meaning.

In the context of a circle:

• The derivative corresponds to the circumference of the circle, \( 2\pi r \).
• This means that the rate at which the area of the circle increases with the radius is the same as the circle's perimeter.

This geometric interpretation reveals that as you imagine the circle's radius slightly extending, the new perimeter is what directly correlates with the area change. This visual understanding helps in connecting the abstract idea of derivatives to a tangible geometric reality.

By visualizing the circle swelling with an increase in radius, it becomes clear that the circumference, the boundary of change, represents the changing area. This concept is pivotal in learning how derivatives link algebraic expressions to geometric shapes.