Find the temperatures at the hottest and coldest points on the plate.

Steps to Solve

1. Identify the Temperature Function

   The temperature at any point ((x, y)) on the plate is given by the function: $$ T(x, y) = x^2 + 2y^2 - x $$

2. Find Critical Points Inside the Plate

   First, we need to calculate the partial derivatives of (T) with respect to (x) and (y) and set them to zero to find critical points inside the plate.

   • The partial derivative with respect to (x): $$ \frac{\partial T}{\partial x} = 2x - 1 $$
   • The partial derivative with respect to (y): $$ \frac{\partial T}{\partial y} = 4y $$

   Setting the derivatives to zero gives: $$ 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2} $$ $$ 4y = 0 \quad \Rightarrow \quad y = 0 $$

   Thus, the critical point is (\left(\frac{1}{2}, 0\right)).

3. Evaluate the Temperature at Critical Point

   Substitute the critical point into the temperature function to find the temperature: $$ T\left(\frac{1}{2}, 0\right) = \left(\frac{1}{2}\right)^2 + 2(0)^2 - \frac{1}{2} $$ $$ = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} $$

4. Evaluate the Temperature on the Boundary

   The boundary is defined by (x^2 + y^2 = 1). We can parameterize the boundary using trigonometric identities:

   • Let (x = \cos(\theta)) and (y = \sin(\theta)), where (\theta \in [0, 2\pi)).

   Substitute into the temperature function: $$ T(\cos(\theta), \sin(\theta)) = \cos^2(\theta) + 2\sin^2(\theta) - \cos(\theta) $$ Simplifying this, we get: $$ T(\theta) = \cos^2(\theta) + 2\sin^2(\theta) - \cos(\theta) = 1 + \sin^2(\theta) - \cos(\theta) $$ We need to find the maximum and minimum of (T(\theta)) on ([0, 2\pi)).

5. Find Extrema on the Boundary

   Taking derivative: $$ \frac{dT}{d\theta} = 2\sin(\theta)\cos(\theta) + \sin(\theta) - \sin(\theta) $$ Setting to zero: $$ \frac{dT}{d\theta} = 0 $$

   This may require evaluating at key angles: (\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}).

6. Calculate Temperatures at Key Angles

   Evaluating (T) at these angles:

   • At (\theta = 0): $$ T(0) = 1 + 0 - 1 = 0 $$

   • At (\theta = \frac{\pi}{2}): $$ T\left( \frac{\pi}{2} \right) = 0 + 2 - 0 = 2 $$

   • At (\theta = \pi): $$ T(\pi) = 1 + 0 + 1 = 0 $$

   • At (\theta = \frac{3\pi}{2}): $$ T\left( \frac{3\pi}{2} \right) = 0 + 2 - (-1) = 3 $$

7. Compare All Values

   The computed temperatures are:

   • Critical point: $-\frac{1}{4}$
   • On boundary: $0$, $2$, and $3$

   Thus, the hottest point is at (\theta = \frac{3\pi}{2}) with temperature (3) and the coldest is at (\left(\frac{1}{2}, 0\right)) with temperature (-\frac{1}{4}).