Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

A humidifier uses a rotating disk of radius \(r\), which is partially submerged in water. The most evaporation occurs when the exposed wetted region (shown as the upper shaded region in Figure 27 ) is maximized. Show that this happens when \(h\) (the distance from the center to the water) is equal to \(r / \sqrt{1+\pi^{2}}\).

Step-by-Step Solution

Answer

The most evaporation occurs when \(h = \frac{r}{\sqrt{1+\pi^2}}\).

Step 1: Understand the Geometry Involved

Visualize the problem by identifying the circular disk with radius \(r\) that is partially submerged in water. The goal is to maximize the area of the wetted region exposed to air, which is the upper shaded segment of the circle.

Step 2: Formula for the Area of a Circle Segment

Recall that the area \(A\) of a circle segment can be calculated using the central angle \(\theta\) in radians as:\[ A = \frac{1}{2} r^2 (\theta - \sin \theta) \]where \(\theta\) is the angle in radians subtended by the wetted arc at the center.

Step 3: Relationship between Height and Angle

The angle \(\theta\) can be derived from the height \(h\), the distance from the circle's center to the water level. Using geometry, \(\theta\) can be expressed in terms of \((h/r)\). This is given by:\[ \theta = 2 \cos^{-1}(h/r) \]

Step 4: Differentiate the Area with Respect to Height

The area function is \( A(h) = \frac{1}{2} r^2 (2 \cos^{-1}(h/r) - (h/r) \sqrt{1-(h/r)^2}) \). Differentiate this function with respect to \(h\) and set it to zero to find the maximum area. Utilize the chain rule and implicit differentiation as needed.

Step 5: Solve for Maximum Height

Upon differentiating, solve the equation obtained by setting the derivative equal to zero to find the value of \(h\) that maximizes \(A\). The calculated derivative will involve trigonometric and algebraic terms, simplifying to the condition:\[ h = \frac{r}{\sqrt{1+\pi^2}} \]

Step 6: Verify Maximum Conditions

Ensure that the second derivative test confirms there is a maximum by checking the sign of the second derivative at \(h = \frac{r}{\sqrt{1+\pi^2}}\). It should be negative, indicating a local maximum.

Key Concepts

Geometry of Circles

Understanding the geometry of circles is crucial to solving problems involving circular disks and circular segments. In our humidifier example, the circle's geometry helps identify the region of interest, which is the portion of the disk that is not submerged. Here’s how to break it down:

• The radius of the circle, denoted by \(r\), represents the distance from the center of the circle to any point on its edge.
• Since the disk is partially submerged, the visible "wetted" region forms a circular segment, a part of the circle's area.
• The task is to maximize the area of this exposed wetted region to achieve maximum evaporation.

Through geometric considerations, the angle and height relationship helps establish the key variables that will be differentiated to find the optimal evaporation height.

Differentiation Techniques

Differentiation is a fundamental calculus technique used to find the rate of change of a function. In our problem, we use differentiation to discover how the area changes with respect to height \(h\). Here’s a closer look at this process:

• The area function, \( A(h) = \frac{1}{2} r^2 (2 \cos^{-1}(h/r) - (h/r) \sqrt{1-(h/r)^2}) \), needs differentiation with respect to \(h\).
• Applying the chain rule is essential because \(A\) is a composite function involving both trigonometric and algebraic components.
• By setting the derivative of \(A(h)\) to zero, we find the critical points where local maxima, minima, or saddle points could exist.

Differentiation helps identify the height \(h\) at which the wetted area is maximized, leading to optimal performance of the humidifier.

Trigonometric Functions

Trigonometric functions play a pivotal role in relating angles and sides of triangles. In the context of this problem, they help link the geometry of the circle to optimization tasks. Here’s how trigonometry is applied:

• The central angle \(\theta\), which the circle's arc subtends at the center, can connect to the height \(h\) using the inverse cosine function: \(\theta = 2 \cos^{-1}(h/r)\).
• The \(\cos^{-1}\) function helps translate the blunt angle into a mathematically manageable form to work with during differentiation.
• Trigonometric identities simplify expressions during differentiation and integration, especially with functions involving multiple terms.

Understanding how to manipulate these functions is crucial for solving such optimization problems.

Implicit Differentiation

Implicit differentiation is a technique needed when dealing with functions not easily solved for a single variable. Here, it helps to differentiate expressions involving \(h\) and \(\theta\) effectively. This is how it works in this problem:

• Implicit differentiation is essential for the area function \( A(h)\), which includes terms like \( \cos^{-1}(h/r) \) and \(\sqrt{1-(h/r)^2}\).
• It allows us to obtain the derivative with respect to \(h\) without explicitly solving for \(\theta\) in terms of \(h\).
• Applying this method efficiently deals with differentiating complex trigonometric functions combined with algebraic terms.

Using implicit differentiation simplifies finding critical points required to ensure the function reaches its maximum under the given conditions.