Question

The temperature of a rectangular plate bounded by the lines \(x=\pm 1, y=\pm 1,\) is given by \(T=2 x^{2}-3 y^{2}-2 x+10 .\) Find the hottest and coldest points of the plate.

Short answer

The hottest points are (-1,1) and (-1,-1) with a temperature of 11. The coldest points are (1,1) and (1,-1) with a temperature of 7.

Step-by-step solution

- Set Up the Problem

Given the temperature function of the plate: \[ T = 2x^2 - 3y^2 - 2x + 10 \]Identify the constraints based on the plate's boundaries: \[ x = \pm 1, \quad y = \pm 1 \]

- Evaluate the Function at Boundary Points

Calculate the temperature at each boundary point: \[ (1,1), (1,-1), (-1,1), (-1,-1) \]Let's find each:For (1,1): \[ T = 2(1)^2 - 3(1)^2 - 2(1) + 10 = 2 - 3 - 2 + 10 = 7 \]For (1,-1): \[ T = 2(1)^2 - 3(-1)^2 - 2(1) + 10 = 2 - 3 - 2 + 10 = 7 \]For (-1,1): \[ T = 2(-1)^2 - 3(1)^2 - 2(-1) + 10 = 2 - 3 + 2 + 10 = 11 \]For (-1,-1): \[ T = 2(-1)^2 - 3(-1)^2 - 2(-1) + 10 = 2 - 3 + 2 + 10 = 11 \]

- Determine the Hottest and Coldest Points

The temperatures at the boundary points are:\[T(1,1) = 7, \quad T(1,-1) = 7, \quad T(-1,1) = 11, \quad T(-1,-1) = 11\]The hottest points are at \[ (-1,1) \text{ and } (-1,-1) \text{ with a temperature of } 11 \]The coldest points are at \[ (1,1) \text{ and } (1,-1) \text{ with a temperature of } 7 \]

Key concepts

boundary conditions

In many real-world problems, boundary conditions play a crucial role in determining the solution. For this exercise, the boundaries are defined by the rectangular plate’s edges, represented by the lines: \(x = \pm 1\) and \(y = \pm 1\). Boundary conditions help to limit the domain where we need to examine the function.
By restricting the domain, we simplify our calculations and focus only on the points that lie on these boundaries. Understanding and applying boundary conditions correctly ensures that the desired constraints are well-integrated into the solution.

extreme values

Extreme values, or extremum, refer to the highest and lowest points on a function within a given domain. In our exercise, extreme values help us determine the hottest and coldest points on the plate. Calculating extreme values involves evaluating the function at specific points.
The function's given boundary points include \((1, 1), (1, -1), (-1, 1), (-1, -1)\). These points need to be tested to figure out where the function reaches its maximum and minimum values. In this case, the maximum temperature is 11 and the minimum is 7, indicating the extreme values of the function within the bounded area.

optimization in calculus

Optimization in calculus is all about finding the best possible solution under given constraints. In the context of the temperature distribution, we are looking to optimize or find the extreme points. This involves evaluating the function and ensuring we have considered all potential points within the boundaries.
Optimization often involves taking derivatives to find critical points. However, in this problem, simple substitution at boundary points suffices to identify the extreme temperatures. This method of optimization ensures the result fully adheres to the conditions set by the plate's physical boundaries.

evaluating functions at specific points

Evaluating functions at specific points is a fundamental step in solving calculus problems. Here, we evaluate the temperature function \(T = 2x^2 - 3y^2 - 2x + 10\) at the boundary points given by \((1,1), (1,-1), (-1,1), (-1,-1)\).
This involves plugging in these coordinates one at a time into the temperature function to calculate the temperature at each point: For example, at \((1,1)\), \( T = 2(1)^2 - 3(1)^2 - 2(1) + 10 = 7 \).
This simple yet crucial step helps us pinpoint the hottest and coldest regions on the plate accurately.

temperature function analysis

Understanding the temperature distribution on a surface involves analyzing the given temperature function. For this exercise, the function is \(T = 2x^2 - 3y^2 - 2x + 10\). To comprehend its behavior:

• Identify how each term affects the temperature at various points.
• The quadratic terms \(2x^2\) and \(-3y^2\) influence how temperature changes with the position.
• The linear term \(-2x\) shifts the temperature in the x-direction.
• The constant term +10 provides a baseline temperature level.
  

By thoroughly analyzing the function, one can predict not just the extreme values, but also understand the broader trend of temperature variation over the entire plate.