Problem 32 - Calculus Early Transcendentals

Question

Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. $$f(x, y)=x e^{-x-y} \sin y, \text { for }|x| \leq 2,0 \leq y \leq \pi$$

Step-by-Step Solution

Answer

Question: Determine the critical points of the function \(f(x, y)=x e^{-x-y} \sin y\) and their natures using the Second Derivative Test. Solution Steps: 1. Compute the first partial derivatives with respect to x and y. 2. Set the partial derivatives equal to zero and solve the resulting system of equations to find the critical points. 3. Compute the second partial derivatives with respect to x and y. 4. Determine the nature of the critical points using the Second Derivative Test. 5. Confirm the results using a graphing utility.

Step 1: Compute the partial derivatives

Compute the first partial derivatives of the given function with respect to x and y. $$\frac{\partial f}{\partial x}=e^{-x-y}(-x\sin y + \sin y)$$ $$\frac{\partial f}{\partial y}=x e^{-x-y} (\cos y)$$

Step 2: Find critical points

Set the partial derivatives equal to zero and solve the resulting system of equations. $$e^{-x-y}(-x\sin y + \sin y) = 0$$ $$x e^{-x-y} (\cos y) = 0$$ From the second equation, we get either \(x=0\) or \(\cos y = 0\). If \(x=0\), then the first equation simplifies to: $$\sin y = 0$$ which yields critical points \((0, 0)\) and \((0, \pi)\). If \(\cos y = 0\), then \(y = \frac{\pi}{2}\) or \(y = \frac{3\pi}{2}\). Since the domain restricts \(0 \leq y \leq \pi\), we only consider \(y = \frac{\pi}{2}\). And plugging this value into the first equation, we get \(x = 1\). Therefore, we have a critical point at \((1, \frac{\pi}{2})\).

Step 3: Compute second partial derivatives

Compute the second partial derivatives of the given function with respect to x and y. $$\frac{\partial^2 f}{\partial x^2}=e^{-x-y}(-x\sin y + 2\sin y)$$ $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}= e^{-x-y}(x\cos y - \cos y)$$ $$\frac{\partial^2 f}{\partial y^2}=-x e^{-x-y} (\sin y)$$

Step 4: Determine the nature of critical points using the Second Derivative Test

Evaluate the second partial derivatives at each critical point and compute the determinant of the Hessian matrix. $$D(x,y) = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right)-\left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$$ At \((0, 0)\): $$D(0, 0)=(-2(0)) - (0)^2= 0$$ Since \(D(0, 0)=0\), we cannot determine the nature of the critical point using the Second Derivative Test. At \((0, \pi)\): $$D(0, \pi)=(-2(0)) - (0)^2= 0$$ Since \(D(0, \pi)=0\), we cannot determine the nature of the critical point using the Second Derivative Test. At \((1, \frac{\pi}{2})\): $$D(1, \frac{\pi}{2})=(2(1)) - (1-1)^2=2$$ Since \(D(1, \frac{\pi}{2})>0\) and \(\frac{\partial^2 f}{\partial x^2}(1, \frac{\pi}{2})>0\), the critical point \((1, \frac{\pi}{2})\) corresponds to a local minimum.

Step 5: Confirm the results using a graphing utility

Plot the function $$f(x, y)=x e^{-x-y} \sin y$$ with the domain \(|x| \leq 2\) and \(0 \leq y \leq \pi\) to confirm the results from Steps 2-3.

Key Concepts

Partial Derivatives

Partial derivatives provide a method to measure how a function changes as one of its variables is altered while keeping others constant. When dealing with a function of two variables, such as \( f(x, y) \), there are two primary partial derivatives: one with respect to \( x \) (\( \frac{\partial f}{\partial x} \)) and one with respect to \( y \) (\( \frac{\partial f}{\partial y} \)).
These derivatives are crucial in the process of finding critical points, as they help us understand the slope of the function in each direction, essentially measuring how steeply the function rises or falls along the \( x \) and \( y \) axes, respectively.

• To calculate \( \frac{\partial f}{\partial x} \), take the derivative of \( f \) while treating \( y \) as a constant.
• Similarly, to compute \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant.

In the given problem, we computed the partial derivatives as:

• \( \frac{\partial f}{\partial x} = e^{-x-y}(-x \sin y + \sin y) \)
• \( \frac{\partial f}{\partial y} = x e^{-x-y} \cos y \)

This step sets the stage for identifying potential critical points, where the slope along either variable direction is zero.

Critical Points

Critical points occur where the gradient (a vector of the function's partial derivatives) is zero. This means both first partial derivatives are zero.
Finding these points is critical (pun intended!) in determining locations where the function might reach local maxima, minima, or saddle points.

Steps to Identify Critical Points

• Set \( \frac{\partial f}{\partial x} = 0 \) to find conditions for \( x \).
• Set \( \frac{\partial f}{\partial y} = 0 \) to find conditions for \( y \).
• Solve this system of equations to find the critical points.

For the function \( f(x, y) = x e^{-x-y} \sin y \), solving the equations \( e^{-x-y}(-x \sin y + \sin y) = 0 \) and \( x e^{-x-y} \cos y = 0 \) leads to the critical points: \( (0, 0) \), \( (0, \pi) \), and \( (1, \frac{\pi}{2}) \).
These points are potential candidates for local maxima, minima, or saddle points, pending further verification using the Second Derivative Test.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. For a function \( f(x, y) \), it helps determine the nature of critical points by indicating whether they are local minima, maxima, or saddle points.

Components of the Hessian Matrix

• \( \frac{\partial^2 f}{\partial x^2} \): Second derivative with respect to \( x \).
• \( \frac{\partial^2 f}{\partial y^2} \): Second derivative with respect to \( y \).
• \( \frac{\partial^2 f}{\partial x \partial y} \) or \( \frac{\partial^2 f}{\partial y \partial x} \): Mixed derivative.

The determinant of the Hessian matrix \( D(x, y) = \left( \frac{\partial^2 f}{\partial x^2} \right)\left( \frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 \) is used in the Second Derivative Test.

Using the Second Derivative Test

• If \( D(x, y) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the point is a local minimum.
• If \( D(x, y) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the point is a local maximum.
• If \( D(x, y) < 0 \), the point is a saddle point.
• If \( D(x, y) = 0 \), the test is inconclusive.

For this specific function, the Hessian determinant at \((0, 0)\) and \((0, \pi)\) is inconclusive. However, at \((1, \frac{\pi}{2})\), the determinant is positive, indicating a local minimum.