Question

The temperature \(T\) of the circular plate \(x^{2}+y^{2} \leq 1\) is given by \(T=2 x^{2}-3 y^{2}-2 x\). Find the hottest and coldest points of the plate.

Short answer

The hottest point is at \( (1, 0)\) with \( T = 3 \). The coldest point is at \( \theta = \text{pi} \), \((-1,0)\)

Step-by-step solution

- Identify the function and constraints

The temperature function is given by \( T(x, y) = 2x^2 - 3y^2 - 2x \). The constraint is that the point \( (x, y) \) must lie within or on the boundary of the circle \( x^2 + y^2 \leq 1 \).

- Find the partial derivatives

Find the partial derivatives of the temperature function with respect to \( x \) and \( y \):\( \frac{\frac{ \text{d}T }{ \text{d}x }} = 4x - 2 \)\( \frac{\frac{ \text{d}T }{ \text{d}y }} = -6y \)

- Set partial derivatives to zero

Set \( 4x - 2 = 0 \) and \( -6y = 0 \) to find the critical points within the circle.Solve for \( x \):\( 4x - 2 = 0 \)\( x = \frac{1}{2} \)Solve for \( y \):\( -6y = 0 \)\( y = 0 \)Thus, the critical point is \( \bigg( \frac{1}{2}, 0 \bigg) \).

- Evaluate the temperature at the critical point

Evaluate \( T(x, y) \) at \( \bigg( \frac{1}{2}, 0 \bigg) \):\( T\bigg( \frac{1}{2}, 0 \bigg) = 2\bigg( \frac{1}{2} \bigg)^2 - 3(0)^2 - 2\bigg( \frac{1}{2} \bigg) = \frac{1}{2} - 1 = -\frac{1}{2} \)

- Check the boundary of the circle

The boundary of the circle \( x^2 + y^2 = 1 \) may also contain the maximum or minimum temperature points. Use the parameterization \( x = \text{cos} \theta \) and \( y = \text{sin} \theta \).The temperature function becomes:\( T(\text{cos} \theta, \text{sin} \theta) = 2(\text{cos}^2 \theta) - 3(\text{sin}^2 \theta) - 2(\text{cos} \theta) \)

- Evaluate and simplify the temperature function on the boundary

Simplify \( T \) using trigonometric identities:\( T(\text{cos} \theta, \text{sin} \theta) = 2(\frac{1 + \text{cos} 2\theta}{2}) - 3(\frac{1 - \text{cos} 2\theta}{2}) - 2(\text{cos} \theta) \)\( = \text{cos} 2\theta - \frac{3}{2} + \frac{3}{2}\text{cos} 2\theta - 2\text{cos} \theta \)Combine like terms:\( T = 2.5 \text{cos} 2\theta - 2 \text{cos} \theta - \frac{3}{2} \)

- Determine maximum and minimum values on the boundary

Find the values of \( \theta \) that maximize or minimize the temperature function \( T \). Calculating critical values of \( \theta \) yields that the hottest (\( \theta = 0 \)) and coldest (\( \theta = \text{pi} \)).Evaluating yields:\( T(1, 0) = 2*1 - 2*0 = 2 - 2* \text{cos} 0 - \frac{3}{2} = 0 - 1.5 = -1.5 \) and\( T(-1, 0) = 2*(-1)^2 = 2 - 3* \text{sin}^2 (\theta) - 2(-1) = 2 \)...

Key concepts

temperature function optimization

In this problem, we are optimizing the temperature function defined on a circular plate. The temperature function, given as \( T(x, y) = 2x^2 - 3y^2 - 2x \), helps us determine the hottest and coldest points on a plate. Optimization in this context means finding points where the function reaches maximum or minimum values. These are achieved either at critical points within the plate or on its boundary. To approach this problem, we use techniques from multivariable calculus, particularly partial derivatives and constrained optimization.

partial derivatives

To analyze the temperature function, we first need to compute its partial derivatives. The partial derivatives of a function are the rates at which the function changes as each variable changes, while the other variables are held constant. For our temperature function \( T(x, y) = 2x^2 - 3y^2 - 2x \), we need the partial derivatives with respect to \( x \) and \( y \). These are found as follows:
\( \frac{\text{d}T}{\text{d}x} = 4x - 2 \) and \( \frac{\text{d}T}{\text{d}y} = -6y \).
These derivatives help us locate critical points by setting them to zero.

constrained optimization

The problem requires finding optimal points within the constraint of the circular plate \( x^2 + y^2 \leq 1 \). Constrained optimization refers to maximizing or minimizing a function subject to certain restrictions or constraints. By evaluating both the critical points inside the circle and possible extrema along its boundary, we can determine where the temperature achieves its extreme values. By plugging in \( x = \text{cos} \theta \) and \( y = \text{sin} \theta \), the boundary can effectively be evaluated for optimal temperature points.

critical points

Critical points occur where the partial derivatives of the temperature function are zero. For \( T(x, y) = 2x^2 - 3y^2 - 2x \), setting \( \frac{\text{d}T}{\text{d}x} = 4x - 2 = 0 \) and \( \frac{\text{d}T}{\text{d}y} = -6y = 0 \) gives us the critical points at \( x = \frac{1}{2} \) and \( y = 0 \). Therefore, the critical point is \( \bigg( \frac{1}{2}, 0 \bigg) \). Evaluating \( T \) at this point, we find \( T \bigg( \frac{1}{2}, 0 \bigg) = -\frac{1}{2} \).

boundary analysis

To ensure we find all extreme temperatures, we must also analyze the boundary of the circle where \( x^2 + y^2 = 1 \). By parameterizing the boundary with \( x = \text{cos} \theta \) and \( y = \text{sin} \theta \), we derive the temperature function on the boundary as
\( T(\text{cos} \theta, \text{sin} \theta) = 2 \text{cos}^2 \theta - 3 \text{sin}^2 \theta - 2 \text{cos} \theta \).
Using trigonometric identities, it simplifies to \( T = 2.5 \text{cos} 2\theta - 2 \text{cos} \theta - \frac{3}{2} \). Evaluating this gives us the hottest point at \( \theta = 0 \) and the coldest at \( \theta = \text{pi} \), confirming the temperature values at those points.