Problem 20 - Calculus for Biology and Medicine

Question

A circular sector with radius \(r\) and angle \(\theta\) has area \(A .\) Find \(r\) and \(\theta\) so that the perimeter is smallest for a given area \(A .(\) Note: \(A=\frac{1}{2} r^{2} \theta\), and the length of the arc \(s=r \theta\), when \(\theta\) is measured in radians; see Figure \(5.59 .\) )

Step-by-Step Solution

Answer

For smallest perimeter, set \( r = \sqrt{A} \) and \( \theta = 2 \).

Step 1: Express Perimeter in terms of r and θ

The perimeter of the sector is the sum of two radii and the arc length. Therefore, \( P = 2r + r\theta \).

Step 2: Use given area expression

We know from the problem statement that the area \( A \) of the sector is \( \frac{1}{2} r^2 \theta \). We need to find \( \theta \) in terms of \( r \) and \( A \). By rearranging, \( \theta = \frac{2A}{r^2} \).

Step 3: Substitute θ in Perimeter Expression

Substitute \( \theta = \frac{2A}{r^2} \) into the perimeter expression: \( P = 2r + r(\frac{2A}{r^2}) = 2r + \frac{2A}{r} \).

Step 4: Differentiate Perimeter with respect to r

For minimizing the perimeter, we need to differentiate \( P \) with respect to \( r \) and set the derivative to zero. Compute \( \frac{dP}{dr} = 2 - \frac{2A}{r^2} \).

Step 5: Solve for r when derivative is zero

Set \( 2 - \frac{2A}{r^2} = 0 \) and solve for \( r \). This gives \( r^2 = A \), so \( r = \sqrt{A} \).

Step 6: Solve for θ using r

Now, substitute \( r = \sqrt{A} \) back into \( \theta = \frac{2A}{r^2} \) to find \( \theta \). This gives \( \theta = 2 \).

Step 7: Verify the critical point is a minimum

Second derivative \( \frac{d^2P}{dr^2} = \frac{4A}{r^3} \), which is positive for \( r = \sqrt{A} \), confirming it's a minimum.

Key Concepts

Perimeter Optimization

In calculus, perimeter optimization is a common problem where you aim to find the best dimensions that minimize or maximize a quantity. Here, we focus on minimizing the perimeter of a circular sector with a given area. The perimeter, in this case, includes the length of two radii and the arc. Thus, the formula for the perimeter is given by:

• Perimeter (\( P \)) = 2\( r \) + \( r\theta \)

For optimization, we also have an area constraint. The area of a sector is:

• Area (\( A \)) = \( \frac{1}{2} r^2 \theta \)

To achieve optimization, we express the perimeter in terms of only one variable by substituting expressions, simplifying the equation, and using calculus techniques.

Differentiation Techniques

Differentiation is a cornerstone of calculus used to find the rate of change of one quantity with respect to another. In the context of perimeter optimization, we use differentiation to find the minimum value of the perimeter function. Here's how it works:First, we substitute \( \theta \) from the area formula into the perimeter formula:

• \( \theta = \frac{2A}{r^2} \)
• \( P = 2r + \frac{2A}{r} \)

The next step is to differentiate this new perimeter function with respect to \( r \):

• \( \frac{dP}{dr} = 2 - \frac{2A}{r^2} \)

Setting the derivative equal to zero helps find the critical points which indicate potential minimum or maximum perimeter values.

Critical Points Analysis

After differentiating the perimeter function, critical points analysis is essential to determine whether they are minima, maxima, or saddle points. These points occur when the derivative equals zero or is undefined. In this exercise, once we found the derivative:- \( \frac{dP}{dr} = 2 - \frac{2A}{r^2} \)We set it to zero and solve for \( r \):- \( r^2 = A \), leading to \( r = \sqrt{A} \)The critical points need verification. This comes through the second derivative test:

• \( \frac{d^2P}{dr^2} = \frac{4A}{r^3} \)

Since this second derivative is positive for our calculated \( r \), this confirms we have a minimum perimeter, successfully optimizing with respect to our constraints.