Find the temperatures at the hottest and coldest points on the plate.
Understand the Problem
The question is asking to find the temperatures at the hottest and coldest points on a flat circular plate with a specified temperature function and boundary conditions defined by the equation of a circle.
Answer
The hottest temperature is \(3\) and the coldest temperature is \(-\frac{1}{4}\).
Answer for screen readers
The hottest temperature is (3) and the coldest temperature is (-\frac{1}{4}).
Steps to Solve
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Identify the Temperature Function
The temperature at any point ((x, y)) on the plate is given by the function: $$ T(x, y) = x^2 + 2y^2 - x $$
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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)).
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Evaluate the Temperature at Critical Point
Substitute the critical point into the temperature function to find the temperature: $$ T\left(\frac{1}{2}, 0\right) = \left(\frac{1}{2}\right)^2 + 2(0)^2 - \frac{1}{2} $$ $$ = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} $$
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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)).
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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}).
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Calculate Temperatures at Key Angles
Evaluating (T) at these angles:
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At (\theta = 0): $$ T(0) = 1 + 0 - 1 = 0 $$
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At (\theta = \frac{\pi}{2}): $$ T\left( \frac{\pi}{2} \right) = 0 + 2 - 0 = 2 $$
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At (\theta = \pi): $$ T(\pi) = 1 + 0 + 1 = 0 $$
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At (\theta = \frac{3\pi}{2}): $$ T\left( \frac{3\pi}{2} \right) = 0 + 2 - (-1) = 3 $$
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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}).
The hottest temperature is (3) and the coldest temperature is (-\frac{1}{4}).
More Information
The analysis involved finding critical points from the temperature function and examining the boundary conditions based on the circular domain defined by (x^2 + y^2 \leq 1).
Tips
- Forgetting to evaluate the temperature at both critical points and boundaries.
- Failing to properly set derivatives equal to zero and solve for critical points.
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