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Chapter 10: Problem 1

The function $$k(x, y)=e^{-y^{2}} \cos (5 x)$$ has a critical point at (0,0) . What is the value of \(D\) at this critical point? \(D=\) _________. What type of critical point is it? \(\quad(\square\) maximum \(\square\) minimum \(\square\) saddle point \(\square\) point with unknown behavior)

Short Answer

Expert verified
The value of D at the critical point (0,0) is 50, and the type of critical point is a local maximum.

Step by step solution

01

Find the first partial derivatives of the function with respect to x and y

To find the partial derivatives, we'll differentiate the given function with respect to x and y. k(x, y) = e^(-y^2) * cos(5x) 鈭俴/鈭倄 = -5 * e^(-y^2) * sin(5x) 鈭俴/鈭倅 = -2y * e^(-y^2) * cos(5x)
02

Find the second partial derivatives with respect to x and y

Now, we'll find the second partial derivatives by differentiating the first partial derivatives with respect to x and y. 鈭俕2k/鈭倄^2 = -25 * e^(-y^2) * cos(5x) 鈭俕2k/鈭倅^2 = 4y^2 * e^(-y^2) * cos(5x) - 2 * e^(-y^2) * cos(5x) 鈭俕2k/鈭倄鈭倅 = 10y * e^(-y^2) * sin(5x)
03

Evaluate the second partial derivatives at the critical point (0,0)

Now that we have the second partial derivatives, let's evaluate them at the critical point (0,0). 鈭俕2k/鈭倄^2 = -25 * e^(0) * cos(0) = -25 鈭俕2k/鈭倅^2 = 0 - 2 * e^(0) * cos(0) = -2 鈭俕2k/鈭倄鈭倅 = 0
04

Calculate D

Now, let's calculate D using the formula mentioned in the analysis. D = f_xx * f_yy - (f_xy)^2 D = (-25) * (-2) - (0)^2 = 50
05

Identify the type of critical point

Now that we have the value of D, let's determine the type of critical point: 1. If D > 0 and f_xx > 0, the critical point is a local minimum. 2. If D > 0 and f_xx < 0, the critical point is a local maximum. 3. If D < 0, the critical point is a saddle point. 4. If D = 0, the type of critical point cannot be determined. Since D = 50 > 0, and f_xx = -25 < 0, the critical point is a local maximum. The value of D at the critical point is 50, and it is a local maximum.

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Key Concepts

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

Second Partial Derivative Test
The Second Partial Derivative Test helps determine the nature of critical points for multivariable functions. It involves analyzing the second partial derivatives at a critical point. This test uses a formula to compute a discriminant, denoted by \( D \).
  • To calculate \( D \), use: \( D = f_{xx} \cdot f_{yy} - (f_{xy})^2 \)
  • \( f_{xx} \) and \( f_{yy} \) are the second partial derivatives with respect to \( x \) and \( y \) respectively.
  • \( f_{xy} \) is the mixed partial derivative.
If \( D > 0 \) and \( f_{xx} > 0 \), the point is a local minimum. If \( D > 0 \) and \( f_{xx} < 0 \), it's a local maximum. If \( D < 0 \), it's a saddle point. When \( D = 0 \), the test is inconclusive.
Local Maximum
A local maximum is a point where the multivariable function's value is higher than nearby points. At these points, the function increases to a certain peak and then decreases. Here's how you determine it:
  • If \( D > 0 \) and \( f_{xx} < 0 \), it indicates a local maximum.
  • This shows that the curve bends like a down-facing cup around the point.
In the given function, at the critical point \((0,0)\), \( D = 50 > 0 \) and \( f_{xx} = -25 < 0 \), confirming a local maximum.
Partial Derivatives
Partial derivatives measure the rate of change of a function with respect to one variable while keeping the other variables constant. They're essential for finding critical points in multivariable calculus.
  • To find a partial derivative with respect to \( x \), differentiate as if \( y \) is a constant.
  • To find it with respect to \( y \), treat \( x \) as a constant.
For our function, the first partial derivatives are \( \frac{\partial k}{\partial x} = -5e^{-y^2} \sin(5x) \) and \( \frac{\partial k}{\partial y} = -2y e^{-y^2} \cos(5x) \). These help locate the critical point \((0,0)\).
Multivariable Functions
Multivariable functions involve more than one variable and are common in calculus. They represent surfaces or other higher-dimensional shapes.
  • The function \( k(x, y) = e^{-y^2} \cos(5x) \) is an example, where changes in both \( x \) and \( y \) affect the output.
  • Analyzing such functions requires understanding their behaviors in multiple dimensions.
The critical points of multivariable functions help identify areas where changes occur most significantly, guiding in optimization and analysis of the functions.

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Most popular questions from this chapter

Your monthly car payment in dollars is \(P=f\left(P_{0}, t, r\right),\) where \(\$ P_{0}\) is the amount you borrowed, t is the number of months it takes to pay off the loan, and r percent is the interest rate. (a) Is \(\partial P / \partial t positive or negative? \)\quad(\square\( positive \)\square\( negative) Suppose that your bank tells you that the magnitude of \)\partial P / \partial t is 15 . What are the units of this value? _________ (b) Is \(\partial P / \partial r positive or negative? \)\quad(\square\( positive \)\square$ negative) . What are the units of this value? __________.

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