Question

Suppose that the temperature \(T\) on the circular plate \(\left\\{(x, y): x^{2}+y^{2} \leq 1\right\\}\) is given by \(T=2 x^{2}+y^{2}-y .\) Find the hottest and coldest spots on the plate.

Short answer

The hottest spot is \((1,0)\) with \(T=2\), and the coldest is \((0,\frac{1}{2})\) with \(T=-\frac{1}{4}\).

Step-by-step solution

Set up the objective function

The given temperature function is \(T(x, y) = 2x^2 + y^2 - y\). We need to find the extrema of this function on the boundary described by \(x^2 + y^2 \leq 1\), which represents a circle of radius 1 centered at the origin.

Find critical points in the interior

To find critical points inside the circle, take the partial derivatives of \(T\) with respect to \(x\) and \(y\) and set them equal to zero:\(\frac{\partial T}{\partial x} = 4x = 0 \) implies \(x = 0\),\(\frac{\partial T}{\partial y} = 2y - 1 = 0\) implies \(y = \frac{1}{2}\).Thus, the critical point inside the circle is \((0, \frac{1}{2})\). Verify that this point lies within the circle by checking \(0^2 + (\frac{1}{2})^2 = \frac{1}{4} \leq 1\).

Evaluate critical point

Evaluate the temperature function at the critical point: \[T(0, \frac{1}{2}) = 2(0)^2 + (\frac{1}{2})^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}.\]

Consider the boundary of the circle

Parameterize the boundary of the circle with \(x = \cos(\theta)\), \(y = \sin(\theta)\) where \(0 \leq \theta < 2\pi\). Substitute these into \(T(x, y)\), giving:\[T(\cos(\theta), \sin(\theta)) = 2\cos^2(\theta) + \sin^2(\theta) - \sin(\theta).\]

Simplify and find extrema on boundary

Simplify the expression from Step 4 using the identity \(\cos^2(\theta) = 1 - \sin^2(\theta)\):\[T(\cos(\theta), \sin(\theta)) = 2(1 - \sin^2(\theta)) + \sin^2(\theta) - \sin(\theta) = 2 - \sin^2(\theta) - \sin(\theta).\] To find critical points, take the derivative with respect to \(\theta\), set equal to 0, and solve. After solving, check values at critical points and endpoints.

Evaluate potential extrema on boundary

Solve the derivative of \(-\sin^2(\theta) - \sin(\theta)\) to find critical values of \(\theta\) using trigonometric identities and solve. Evaluate \(T\) at these points, checking values at endpoints \((x, y) = (1, 0), (-1, 0), (0, 1), (0, -1)\) and any others found by solving.

Compare temperature values

Compare the temperature values obtained from the interior point and the boundary. The highest and lowest temperatures found determine the hottest and coldest points on the plate.

Key concepts

Critical Points

Finding critical points is a fundamental technique in optimization problems in calculus. It involves determining where the derivative of a function is zero or undefined, as these are potential locations for maxima or minima of the function.

In our exercise, the function is defined by the temperature on the plate, expressed as \(T(x, y) = 2x^2 + y^2 - y\). We want to find the critical points of this function within the interior of a circle, where \(x^2 + y^2 \leq 1\).

To find these critical points, we take the partial derivatives \(\frac{\partial T}{\partial x}\) and \(\frac{\partial T}{\partial y}\):

• \(\frac{\partial T}{\partial x} = 4x\)
• \(\frac{\partial T}{\partial y} = 2y - 1\)

Setting these partial derivatives to zero gives us the critical points' coordinates where potential extrema could occur. In this problem, solving these equations led to the critical point \((0, \frac{1}{2})\). This point needs to be verified as a within the circle by checking \(0^2 + (\frac{1}{2})^2 = \frac{1}{4} \leq 1\).

Thus, the inside of the circle contains at least one critical point, essential for finding extrema.

Partial Derivatives

Partial derivatives are an extension of derivatives into multi-variable calculus. They assess how a function changes as one of its variables changes while keeping other variables constant.

For the two-variable function \(T(x, y)\), we find the partial derivatives with respect to \(x\) and \(y\). These are crucial in locating critical points:

• \(\frac{\partial T}{\partial x} = 4x\) gives us the changes in \(T\) along the \(x\) direction.
• \(\frac{\partial T}{\partial y} = 2y - 1\) gives changes along the \(y\) direction.

It's through setting these partial derivatives equal to zero, that we solve for specific values of \(x\) and \(y\). Doing so helps us find points like \((0, \frac{1}{2})\) where \(T(x, y)\) doesn't change direction. Essentially, these are candidates for local maxima, minima, or saddle points that could help identify the hottest or coldest spots on the plate.

Beyond just finding critical points, partial derivatives also tell us about the behavior of \(T\) near these points.

Boundary Evaluation

Boundary evaluation is a key part of optimization in constrained domains, as it considers potential extrema on the edges of the region.

In the given problem, the boundary of the circle, defined by \(x^2 + y^2 = 1\), must also be checked because extrema can occur there too. We do this by parameterizing the boundary using a trigonometric substitution:

• \(x = \cos(\theta)\)
• \(y = \sin(\theta)\)

Substituting these into our temperature function \(T(x, y)\), we reparameterize it as \(T(\cos(\theta), \sin(\theta))\). Simplifying this expression offers insights into the temperature variation along the circle's boundary.

To locate boundary extrema, we differentiate this new expression with respect to \(\theta\), solving for values that make the derivative zero. Evaluating the temperature at these critical angle values and endpoints like \((1, 0), (0, 1), (0, -1), (-1, 0)\), we identify maximum and minimum temperature spots dictated by boundary constraints.

Extrema

Extrema, referring to maximum and minimum values of a function, are key in optimization problems. To comprehensively determine where the function attains its highest and lowest values, consider both interior critical points and boundary evaluations.

In the context of the given exercise, we have already identified a critical point \((0, \frac{1}{2})\) on the interior.

We calculated \(T(0, \frac{1}{2}) = -\frac{1}{4}\).

Additionally, after evaluating the expression along the boundary, we need to examine boundary critical values. Analyzing temperature values at angles \(\theta\) when \( \text{d}T(\theta)/\text{d}\theta = 0\), plus predefined checkpoints, helps us find the absolute extrema.

Ultimately, comparing these temperature values from the set of critical points provides the precise position for the planet's hottest and coldest points. This thorough approach ensures no extreme is overlooked due to boundary peculiarities.