Problem 34 - 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. $$f(x, y)=\sin (2 \pi x) \cos (\pi y), \text { for }|x| \leq \frac{1}{2} \text { and }|y| \leq \frac{1}{2}$$

Step-by-Step Solution

Answer

In conclusion, the critical points of the given function within the domain |x| ≤ ½ and |y| ≤ ½ are (0, 0), (0, ±½), (±½, 0), and (±½, ±½). Among these points, (0, 0) is a saddle point, (±½, 0) and (0, ±½) are local maxima/minima, and (±½, ±½) are saddle points.

Step 1: Find the first partial derivatives and set them equal to 0.

Calculate the partial derivatives of the function with respect to x and y: $$\frac{\partial f}{\partial x} = 2\pi \cos(2 \pi x)\cos(\pi y)$$ $$\frac{\partial f}{\partial y} = -\pi \sin(2 \pi x)\sin(\pi y)$$ Now set them equal to 0 and solve for x and y: $$2\pi \cos(2 \pi x)\cos(\pi y) = 0$$ $$-\pi \sin(2 \pi x)\sin(\pi y) = 0$$

Step 2: Solve for critical points.

In order to obtain the critical points, we have to solve both equations simultaneously. From the equations, we can determine critical points when either: - sine or cosine terms are equal to zero - or x, y are multiples of one half x-critical points: when $$2\pi x = 0, \pm \pi, \pm 2 \pi, ...$$, then x = 0, ± ½, ±1, ... y-critical points: when $$\pi y = 0, \pm \pi, \pm 2 \pi, ...$$, then y = 0, ±1, ± 2, ... For our domain, |x| ≤ ½ and |y| ≤ ½, the critical points are (x, y) = (0, 0), (0, ±½), (±½, 0), and (±½, ±½).

Step 3: Use the Second Derivative Test.

In order to categorize each critical point, we must calculate the second partial derivatives to perform the Second Derivative Test: $$\frac{\partial^2 f}{\partial x^2} = -(4\pi^2) \sin(2 \pi x)\cos(\pi y)$$ $$\frac{\partial^2 f}{\partial y^2} = -\pi^2 \sin(2 \pi x)\cos(\pi y)$$ $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = -2\pi^2 \sin(2 \pi x)\sin(\pi y)$$ Now, determine the value of the discriminant D at each critical point, by using the formula: $$D = (\frac{\partial^2 f}{\partial x^2} \times \frac{\partial^2 f}{\partial y^2}) - (\frac{\partial^2 f}{\partial x \partial y})^2$$ For all critical points except (0, 0), the value of D is 0, which means the Second Derivative Test is inconclusive and we must rely on graphing to determine the nature of these points. For the critical point (0, 0): $$D = (0) - (-8\pi^4)^2 > 0$$ $$\frac{\partial^2 f}{\partial x^2} = 0 $$ Since D > 0 and $$\frac{\partial^2 f}{\partial x^2} = 0$$, the point (0, 0) is a saddle point.

Step 4: Confirm results using the graphing utility.

By graphing the function within the given domain, we can visually confirm that: - (0, 0) is a saddle point - (±½, 0) and (0, ±½) are local maxima/minima - (±½, ±½) are saddle points This verifies our results from the Second Derivative Test.

Key Concepts

Critical Points

In the realm of multivariable calculus, finding critical points of a function is crucial for analyzing its behavior. Critical points occur where the gradient (or the vector of first partial derivatives) of a function is zero. For a function of two variables, say \( f(x, y) \), this means both partial derivatives with respect to \( x \) and \( y \) must be zero at a critical point.

Let's break this concept down using the function \( f(x, y) = \sin(2 \pi x) \cos(\pi y) \) as an example. To find the critical points, we compute the first partial derivatives:

• \( \frac{\partial f}{\partial x} = 2\pi \cos(2 \pi x)\cos(\pi y) \)
• \( \frac{\partial f}{\partial y} = -\pi \sin(2 \pi x)\sin(\pi y) \)

Both these derivatives are set to zero to find the points where the function's slope is flat in every direction, indicating potential maxima, minima, or saddle points. Like roots in algebra, critical points reveal where changes in a function’s increase or decrease can happen.

In our specific exercise, solving the equations simultaneously gave us critical points such as \((0, 0)\), \((\pm 1/2, 0)\), \((0, \pm 1/2)\), and \((\pm 1/2, \pm 1/2)\), within the prescribed boundaries of \(|x| \leq 1/2\) and \(|y| \leq 1/2\).

Second Derivative Test

Once we identify the critical points of a function, the Second Derivative Test is the next step to classify them. This test helps determine whether these points are local maxima, minima, or saddle points. The test relies on calculating the second derivatives to analyze the curvature near each critical point.

For a function \( f(x, y) \), we compute three second partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} \) - the second derivative with respect to \( x \)
• \( \frac{\partial^2 f}{\partial y^2} \) - the second derivative with respect to \( y \)
• \( \frac{\partial^2 f}{\partial x \partial y} \) or \( \frac{\partial^2 f}{\partial y \partial x} \) - mixed partial derivatives

With these, the \ "discriminant" \( D \) is calculated using the formula:\[D = \left(\frac{\partial^2 f}{\partial x^2} \times \frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2\]This \( D \) value tells us the following for a critical point:

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

For the function in our problem, we found \( D > 0 \) at \((0, 0)\) and because the second derivative with respect to \( x \) was zero, it indicated a saddle point.

Partial Derivatives

Partial derivatives form the backbone of understanding changes in multivariable functions. These derivatives measure how the function changes as one variable changes while keeping all others fixed. In the function \( f(x, y) = \sin(2 \pi x) \cos(\pi y) \), the partial derivatives with respect to \( x \) and \( y \) describe how the function's value changes as \( x \) and \( y \) are individually varied.

To find these, we treat one variable as constant:

• To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant, focusing on changes in \( x \).
• To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant, focusing on changes in \( y \).

The resulting derivatives for this function were:

• \( \frac{\partial f}{\partial x} = 2\pi \cos(2 \pi x)\cos(\pi y) \)
  This shows how the function changes with respect to changes in \( x \).
• \( \frac{\partial f}{\partial y} = -\pi \sin(2 \pi x)\sin(\pi y) \)
  This shows how the function changes with respect to changes in \( y \).

Partial derivatives are essential not only in finding critical points but also in understanding contour maps and the gradient of functions.

Saddle Points

A saddle point is an intriguing concept in multivariable calculus, representing a critical point that is neither a local maximum nor a minimum. Instead, it behaves like a mix where the function increases in some directions and decreases in others. Imagine a saddle on a horse where the center dips downward but the sides curve upward.

In our example, after applying the Second Derivative Test to the function \( f(x, y) \), the critical point \((0, 0)\) was determined to be a saddle point with \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} = 0 \).

Understanding saddle points requires:

• Checking the sign of \( D \). A \( D < 0 \) typically pinpoints a saddle point.
• Analyzing the plots or graphs can visually demonstrate the characteristic 'saddle' shape.

Recognizing saddle points is crucial since they indicate instability in optimization problems or potential points of transition in modeling real-world scenarios.