Problem 28 - Calculus Early Transcendentals

Question

Use a graphing device as in Example 4 (or Newton's method or a rootfinder) to find the critical points of \(f\) correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph. $$f(x, y)=e^{x}+y^{4}-x^{3}+4 \cos y$$

Step-by-Step Solution

Answer

Use a graphing tool or numerical method to approximate critical points; analyze these points using the Hessian for classification.

Step 1: Identify the Partial Derivatives

To find the critical points, we first determine the partial derivatives of the function with respect to each variable, \(x\) and \(y\). The partial derivatives are:\[ f_x = \frac{\partial}{\partial x}(e^x + y^4 - x^3 + 4\cos y) = e^x - 3x^2 \] \[ f_y = \frac{\partial}{\partial y}(e^x + y^4 - x^3 + 4\cos y) = 4y^3 - 4\sin y \]

Step 2: Set Partial Derivatives to Zero

To find the critical points, we set the calculated partial derivatives equal to zero:\[ e^x - 3x^2 = 0 \]\[ 4y^3 - 4\sin y = 0 \]

Step 3: Solve for Critical Points

Solve the equations from Step 2 to find the values of \(x\) and \(y\):- For \(e^x - 3x^2 = 0\), finding an exact solution analytically is complex, so we use numerical methods such as Newton's method or a graphing device to approximate \(x\). We look for points where \(e^x = 3x^2\). - For \(4y^3 = 4\sin y\), simplify to \(y^3 = \sin y\). Use similar numerical methods to find values of \(y\) where this holds true.

Step 4: Classify the Critical Points

Once the critical points \((x, y)\) are found, classify them using the second derivative test. This involves calculating the second partial derivatives and using the Hessian matrix:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = e^x - 6x \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 12y^2 - 4\cos y \]\[ f_{xy} = f_{yx} = 0 \]Construct the Hessian matrix:\[ H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{pmatrix} \] Evaluate the determinant \(\det(H) = f_{xx}f_{yy} - (f_{xy})^2\) to determine if the critical point is a local minimum, local maximum, or saddle point.

Step 5: Find Highest or Lowest Points

Compare the function values \(f(x, y)\) at all critical points determined in previous steps to find the global maximum or minimum on the graph. Consider any boundaries if the domain for \(x\) and \(y\) is restricted in the context.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They involve finding the derivative of a function with respect to one variable while keeping the other variables constant. This is useful in analyzing functions of several variables, like in our given function \( f(x, y) = e^x + y^4 - x^3 + 4 \cos y \).
To find the partial derivatives, we differentiate the function with respect to each variable separately:

• For \( f_x \), the derivative with respect to \( x \), we focus on terms involving \( x \). We get \( f_x = e^x - 3x^2 \).
• For \( f_y \), the derivative with respect to \( y \), we concentrate on terms involving \( y \). We derive \( f_y = 4y^3 - 4\sin y \).

These two derivatives help us locate critical points where both derivatives equal zero.
Critical points occur where the slope of the function is flat in each direction and are candidates for maxima, minima, or saddle points.

Hessian Matrix

The Hessian matrix provides a systematic way to classify critical points in multivariable functions. It is constructed from the second partial derivatives of the function.
For the function \( f(x, y) \), the second partial derivatives are:

• \( f_{xx} = e^x - 6x \)
• \( f_{yy} = 12y^2 - 4\cos y \)
• \( f_{xy} = f_{yx} = 0 \)

The Hessian matrix \( H \) is then formed as:
\[H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{pmatrix} = \begin{pmatrix} e^x - 6x & 0 \ 0 & 12y^2 - 4\cos y \end{pmatrix}\]
The determinant of this matrix, calculated as \( \det(H) = f_{xx}f_{yy} - (f_{xy})^2 \), helps classify the nature of the critical points. A positive determinant can indicate a local minimum or maximum, while a negative determinant suggests a saddle point.

Numerical Methods

Numerical methods are essential for solving equations analytically complex or impossible to solve exactly. In this context, we use them to find approximate solutions for the critical points of our function.
For the equation \( e^x - 3x^2 = 0 \), numerical methods like Newton's method are effective.

• Newton's method involves iterative steps to converge towards a solution. You need an initial guess and use the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).

Similarly, for \( y^3 = \sin y \), graphing devices or rootfinders can plot functions and explore intersecting points to find solutions.
Such methods provide us with critical point estimates, where we can proceed by analyzing them with the Hessian matrix.

Second Derivative Test

The second derivative test is crucial for classifying critical points once they are identified. By examining the second partial derivatives, this test helps us understand what kind of point a critical point is.
When the Hessian determinant \( \det(H) = f_{xx}f_{yy} - (f_{xy})^2 \) is calculated:

• If \( \det(H) > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.
• If \( \det(H) > 0 \) and \( f_{xx} < 0 \), it's a local maximum.
• If \( \det(H) < 0 \), the point is a saddle point, indicating mixed behavior.
• If \( \det(H) = 0 \), the test is inconclusive, and we may need additional analysis.

Knowing how to apply this test aids significantly in understanding the behavior of functions around their critical points, thereby identifying potential peaks, valleys, or points of inflection on the surface graph.