Problem 25 - Calculus with Concepts in Calculus

Question

Find the extreme values of \(f\) on \(R\). $$ f(x, y)=x^{2}-y^{2} ; R \text { is the disk } x^{2}+y^{2} \leq 1 $$

Step-by-Step Solution

Answer

The maximum value is 1 and the minimum value is -1.

Step 1: Check for Critical Points Inside the Disk

Find the first partial derivatives of the function: \[ f_x = \frac{\partial}{\partial x}(x^2 - y^2) = 2x \]\[ f_y = \frac{\partial}{\partial y}(x^2 - y^2) = -2y \]Set these equal to zero to find critical points:\[ 2x = 0 \quad \Rightarrow \quad x = 0 \]\[ -2y = 0 \quad \Rightarrow \quad y = 0 \]So, the critical point is \((0, 0)\).

Step 2: Evaluate the Function at the Critical Point

Substitute the critical point into the function to find its value:\[ f(0, 0) = 0^2 - 0^2 = 0 \]Therefore, the value of the function at this critical point is 0.

Step 3: Explore the Boundary of the Disk

The boundary of the disk is given by \(x^2 + y^2 = 1\). Use a parametrization to explore it:Let \(x = \cos\theta\) and \(y = \sin\theta\), with \(0 \leq \theta < 2\pi\).Substitute these into the function:\[ f(x, y) = \cos^2\theta - \sin^2\theta \]We know \(\cos^2\theta - \sin^2\theta = \cos(2\theta)\). So, \[ f(x, y) = \cos(2\theta) \]

Step 4: Find Extreme Values on the Boundary

The function \(f(x, y) = \cos(2\theta)\) attains its maximum value of 1 when \(2\theta = 0, \pi, 2\pi, \ldots\) (i.e., \(\theta = 0, \frac{\pi}{2}\)), and its minimum value of -1 when \(2\theta = \frac{\pi}{2}, \frac{3\pi}{2},\ldots\) (i.e., \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)).Thus, the maximum value is 1 at points like \((1, 0)\) and \((0, -1)\), and the minimum value is -1 at points like \((0, 1)\) and \((-1, 0)\).

Step 5: Conclusion

The critical point inside the disk gives a value of 0. Comparing this with the boundary values, the maximum value on the region \(R\) is 1 and the minimum value is -1.

Key Concepts

Partial Derivatives

Partial derivatives are like specialized tools for functions with multiple variables. They measure how the function changes as we vary one variable, keeping others constant. For a function like \( f(x, y) = x^2 - y^2 \), finding the partial derivatives means calculating just how much \( f \) changes when \( x \) or \( y \) changes a little.

- **Partial derivative with respect to \( x \):** We assume \( y \) stays the same and differentiate with respect to \( x \). This gives \( f_x = 2x \).
- **Partial derivative with respect to \( y \):** Here, we keep \( x \) constant and differentiate with respect to \( y \), giving \( f_y = -2y \).

These derivatives help us understand the function's behavior in any direction. When partial derivatives are zero, as seen here at point \((0, 0)\), the function isn't changing at that point in the \( x \) or \( y \) direction.

Critical Points

Critical points are special places on the graph of a function where a lot of interesting things might happen. Specifically, these are points where the derivative of the function is zero or undefined, indicating potential peaks, dips, or flat areas.

For the function \( f(x, y) = x^2 - y^2 \), we've calculated its partial derivatives and set them to zero:

• \( f_x = 2x = 0 \) which leads to \( x = 0 \)
• \( f_y = -2y = 0 \) which leads to \( y = 0 \)

Putting these together, \((0, 0)\) is a critical point. This means at the center of our disk, the function has a special property, though by itself, the critical point does not tell us if it's a minimum, maximum, or a saddle point.

Parametrization

Parametrization is a handy technique used to describe curves and paths using parameters like \( \theta \). It's like using one variable to walk around the edge of a shape.

For the boundary of our disk \( x^2 + y^2 = 1 \), we use trigonometric functions, setting \( x = \cos \theta \) and \( y = \sin \theta \). This parametrizes the circle, allowing us to explore the entire boundary as \( \theta \) ranges from 0 to \( 2\pi \).

By plugging these into the function, we find:

• \( f(x, y) = \cos^2\theta - \sin^2\theta \), which simplifies to \( \cos(2\theta) \).

This transformation highlights how trigonometric identities can simplify the evaluation of functions on circles, making our task more manageable.

Boundary Values

Boundary values help identify where a function reaches its highest and lowest on a constrained region, like the edge of a disk.

For the function \( f(x, y) = \cos(2\theta) \) defined on the boundary \( x^2 + y^2 = 1 \), we found that:

• The maximum value is 1, occurring at angles like \( \theta = 0 \) or \( \pi \).
• The minimum value is -1, appearing at \( \theta = \frac{\pi}{2} \) or \( \frac{3\pi}{2} \).

So as you trace along this boundary, the function swings between these extremes. Comparing with the critical point within the disk which was 0, it shows the importance of checking both the interior and boundary of a region when searching for extreme values.