Problem 56 - Calculus Early Transcendentals

Question

Temperature of an elliptical plate The temperature of points on an elliptical plate \(x^{2}+y^{2}+x y \leq 1\) is given by \(T(x,y)=25\left(x^{2}+y^{2}\right) .\) Find the hottest and coldest temperatures on the edge of the elliptical plate.

Step-by-Step Solution

Answer

Answer: The coldest temperature on the edge of the elliptical plate is 0°C, and the hottest temperature is 25°C.

Step 1: Find the equation of the edge of the elliptical plate

To find the equation of the ellipse that represents the edge of the plate, we need to set the inequality x² + y² + xy ≤ 1 to an equal sign, as follows: \(x^{2} + y^{2} + xy = 1\) This equation represents the boundary of the elliptical plate.

Step 2: Set up a function for T(x,y) along the boundary

Now, we can plug the equation that represents the boundary into T(x,y) to set up a function for the temperature on the edge of the plate: \(T(x,y) = 25(x^2 + y^2)\) We will substitute y from the edge equation into this temperature function. From the ellipse equation we have \(y = \frac{1-x^2}{1+x}\), now substitute it into the T(x,y) function. \(T(x) = 25\left( x^2 + \left(\frac{1-x^2}{1+x}\right)^2 \right)\) Now, we can find the maximum and minimum values of T(x) within the limits of -1 ≤ x ≤ 1.

Step 3: Differentiate T(x) with respect to x

To find the hottest and coldest temperatures on the edge of the elliptical plate, we need to find the critical points of the T(x) function, which can be obtained by finding the first derivative of T(x) with respect to x: \(T'(x) = \frac{d}{dx}\left(25\left(x^2 + \left(\frac{1-x^2}{1+x}\right)^2\right)\right)\) After finding the derivative, we get: \(T'(x) = 50x - 100x^3(1+x)^{-3}\)

Step 4: Find the critical points of T(x)

To find the critical points of the T(x) function, we need to set the first derivative equal to zero and solve for x: \(50x - 100x^3(1+x)^{-3} = 0\) After solving this equation, we get two critical points: x = 0 and x = 1.

Step 5: Evaluate T(x) at the critical points and endpoints

Now we need to evaluate T(x) at the critical points and the endpoints -1 and 1. 1. At x = -1: \(T(-1) = 25((-1)^2 + (\frac{1-(-1)^2}{1+(-1)})^2)\) = 0 2. At x = 0: \(T(0) = 25((0)^2 + (\frac{1-(0)^2}{1+(0)})^2)\) = 25 3. At x = 1: \(T(1) = 25((1)^2 + (\frac{1-(1)^2}{1+(1)})^2)\) = 0

Step 6: Compare the temperature values and find the hottest and coldest temperatures

Now, we can compare the temperature values obtained in the previous step: 1. T(-1) = 0 2. T(0) = 25 3. T(1) = 0 So, the coldest temperature on the edge of the elliptical plate is 0°C, and the hottest temperature is 25°C.

Key Concepts

Elliptical Geometry

In the context of this exercise, elliptical geometry is essential as it defines the shape of the plate on which we will analyze temperature distribution. An ellipse is a type of conic section that is characterized by a closed, symmetric curve. In our problem, the ellipse is described by the equation \(x^{2}+y^{2}+xy = 1\). This equation illustrates the boundary of the plate where we will evaluate the temperature.

• The ellipse is a geometrical shape that allows for smooth transitions between circles and lines.
• The presence of terms like \(xy\) indicates a more complex geometry than a simple circle.

This elliptical boundary confines all the points (x, y) under consideration in our analysis, and is key to understanding how variables are restricted within this domain.

Temperature Distribution

Temperature distribution across a surface describes how temperature varies at different points on that surface. For the elliptical plate in the problem, the temperature distribution is given by the function \(T(x,y) = 25(x^{2}+y^{2})\). This function provides a measure of the temperature at any point \((x, y)\) on the plate's surface.

• A basic square of the coordinates (\(x^{2} + y^{2}\)) is used to determine temperature.
• The constant coefficient (25 in this case) scales the temperature values.

To find the hottest and coldest points, we focus specifically on where temperature is defined on the boundary, using substitutions and simplifications from elliptical geometry. This allows us to evaluate changes in temperature around the edge of the ellipse.

Critical Points

Critical points are specific points on a graph where the derivative (or rate of change) of a function is zero or undefined. In the context of optimization, they help us identify maximum and minimum values of a function.
For this exercise, we are interested in finding critical points of the temperature function \(T(x)\) which is derived by substituting the boundary conditions into \(T(x,y)\). This involves:

• Differentiating the temperature function \(T(x)\) to find its critical points
• Identifying where the derivative equals zero or does not exist
• Evaluating the temperature at these points to find maximum (hottest) and minimum (coldest) temperatures

The critical points, alongside the endpoints of the domain \(x = -1\) and \(x = 1\), were evaluated, and temperatures were calculated to determine that the points x = 0 and x = ±1 offered the coldest and hottest temperatures at 0°C and 25°C, respectively.