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Chapter 15: Problem 42

Inconclusive tests Show that the Second Derivative Test is inconclusive when applied to the following functions at (0,0) Describe the behavior of the function at (0,0) $$f(x, y)=\sin \left(x^{2} y^{2}\right)$$

Short Answer

Expert verified
Based on the provided solution, we found that the Second Derivative Test is inconclusive for the given function at the point (0,0). This means we cannot definitively determine whether the function has a local maximum, local minimum, or saddle point at this point using the Second Derivative Test. However, through analysis of the function's behavior, we noticed that the function will have an oscillatory behavior around (0,0) due to the sine function being bounded between -1 and 1 and the argument of the sine function approaching zero as x and y approach zero.

Step by step solution

01

Find the first partial derivatives

Calculate the partial derivatives of the function with respect to x and y. $$ \frac{\partial f}{\partial x} = -2xy^2\cos\left(x^2y^2\right) $$ $$ \frac{\partial f}{\partial y} = -2x^2y\cos\left(x^2y^2\right) $$
02

Determine the critical point

Set both partial derivatives equal to zero and solve for x and y. $$ -2xy^2\cos\left(x^2y^2\right) = 0 $$ $$ -2x^2y\cos\left(x^2y^2\right) = 0 $$ The critical point occurs at (0,0).
03

Find the second partial derivatives

Calculate the second partial derivatives of the function with respect to x and y. $$ \frac{\partial^2 f}{\partial x^2} = -4y^{4}\cos\left(x^{2}y^{2}\right)+8x^{2}y^{4}\sin\left(x^{2}y^{2}\right) $$ $$ \frac{\partial^2 f}{\partial y^2} = -4x^{4}\cos\left(x^{2}y^{2}\right)+8x^{4}y^{2}\sin\left(x^{2}y^{2}\right) $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 4x^{3}y^{3}\sin\left(x^{2}y^{2}\right)+8x^{2}y^{2}\cos\left(x^{2}y^{2}\right) $$
04

Apply the Second Derivative Test

Evaluate the second partial derivatives at the critical point (0,0) and calculate the discriminant. $$ D = \left(\frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 $$ At (0,0): $$ \frac{\partial^2 f}{\partial x^2} = 0 $$ $$ \frac{\partial^2 f}{\partial y^2} = 0 $$ $$ \frac{\partial^2 f}{\partial x \partial y} = 0 $$ Therefore, $$ D = (0)(0) - (0)^2 = 0 $$
05

Analyze the results

Since D = 0, the Second Derivative Test is inconclusive. This means we cannot decisively determine whether the function has a local maximum, local minimum, or saddle point at (0,0) through this method.
06

Describe the behavior of the function at (0,0)

Although the Second Derivative Test is inconclusive, we can still analyze the function's behavior at (0,0). By observing the function, we see that the sine function is bounded between -1 and 1. Additionally, as x and y approach zero, the argument of the sine function approaches zero as well. Thus, the function will have an oscillatory behavior around (0,0).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Critical Points
Critical points of a function occur where its first partial derivatives are equal to zero or are undefined. They are important because they can help identify regions where a function reaches local maxima, minima, or saddle points. To find these points, substitute the solutions of the system of equations formed by setting each first partial derivative equal to zero. In the exercise, the critical point at (0,0) is found when setting the partial derivatives of the given function to zero.
Exploring Partial Derivatives
Partial derivatives measure how a function changes as one of the variables is varied while keeping other variables constant. They are crucial in calculus for functions of several variables, allowing us to study the function's behavior along different axes. In our exercise, we found the first and second partial derivatives with respect to both x and y. This step was vital for applying the Second Derivative Test, where these derivatives help us construct the function's Hessian matrix 鈥� a key part in understanding local behavior.
Understanding the Discriminant
The discriminant plays a pivotal role in the Second Derivative Test. It is defined as \( D = f_{xx} f_{yy} - (f_{xy})^2 \). The discriminant helps classify the nature of a critical point by comparing the magnitude of the product of the second partial derivatives \( f_{xx} \) and \( f_{yy} \) to the square of the mixed partial derivative \( f_{xy} \). If \( D > 0 \), the point might be a local extremum; if \( D < 0 \), it's indicative of a saddle point. However, in this exercise, \( D = 0 \), rendering the test inconclusive.
Unveiling Oscillatory Behavior
Oscillatory behavior refers to a function's tendency to repetitively move up and down or back and forth around a certain value. In certain functions, such as those involving sine or cosine, this behavior is common. As seen with the function \( f(x, y) = \sin(x^2 y^2) \), the sine term causes the function to oscillate, especially as its arguments approach zero. Around (0,0), this oscillation makes it challenging to determine a clear local extremum or saddle point solely using the Second Derivative Test.

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Suppose you are standing at the center of a sphere looking at a point \(P\) on the surface of the sphere. Your line of sight to \(P\) is orthogonal to the plane tangent to the sphere at \(P\)b. At a point that maximizes \(f\) on the curve \(g(x, y)=0,\) the dot product \(\nabla f \cdot \nabla g\) is zero.

Give the value of the utility function at the optimal point. $$U=f(\ell, g)=10 e^{1 / 2} g^{1 / 2} \text { subject to } 3 \ell+6 g=18$$

Using gradient rules Use the gradient rules of Exercise 85 to find the gradient of the following functions. $$f(x, y)=x y \cos (x y)$$

Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$H(x, y)=\sqrt{x^{2}+y^{2}}$$

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) $$ \text { b. } f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}} $$ This property has the following interpretation. Suppose a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May \(1985,\) and Calculus and Analytical Geometry, 2 nd ed., Philip Gillett, 1984 )
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