Chapter 4: Problem 7
Given \(z=\left(y-x^{2}\right)\left(y-2 x^{2}\right),\) show that \(z\) has neither a maximum nor a minimum at \((0,0),\) although \(z\) has a minimum on every straight line through (0,0).
Short Answer
Expert verified
z has neither a maximum nor a minimum at (0,0), but shows a minimum along each line through (0,0).
Step by step solution
01
- Express the function
We have the function given by: \[ z = (y - x^2)(y - 2x^2) \]
02
- Find the first partial derivatives
Compute the partial derivatives with respect to x and y using the product rule: \[ z_x = \frac{\text{d}}{\text{d}x}[(y - x^2)(y - 2x^2)] = (y - x^2)(-4x) + (y - 2x^2)(-2x) \] \[ z_y = \frac{\text{d}}{\text{d}y}[(y - x^2)(y - 2x^2)] = (y - x^2) + (y - 2x^2) \]
03
- Simplify the first partial derivatives
After simplification, get: \[ z_x = -4x (y - x^2) - 2x (y - 2x^2) = -6xy + 8x^3 \] \[ z_y = y - x^2 + y - 2x^2 = 2y - 3x^2 \]
04
- Evaluate the first partial derivatives at (0,0)
Substitute (0,0) into the first partial derivatives to check if they are zero: \[ z_x(0,0) = -6(0)(0) + 8(0)^3 = 0 \] \[ z_y(0,0) = 2(0) - 3(0)^2 = 0 \]
05
- Find the second partial derivatives
Compute the second partial derivatives: \[ z_{xx} = \frac{\text{d}}{\text{d}x}(-6xy + 8x^3) = -6y + 24x^2 \] \[ z_{yy} = \frac{\text{d}}{\text{d}y}(2y - 3x^2) = 2 \] \[ z_{xy} = \frac{\text{d}}{\text{d}y}(-6xy + 8x^3) = -6x \]
06
- Evaluate the second partial derivatives at (0,0)
Substitute (0,0) into the second partial derivatives: \[ z_{xx}(0,0) = -6(0) + 24(0)^2 = 0 \] \[ z_{yy}(0,0) = 2 \] \[ z_{xy}(0,0) = -6(0) = 0 \]
07
- Use the second derivative test
Apply the second derivative test for a function of two variables. Calculate the determinant of the Hessian matrix at (0,0): \[ H = \begin{vmatrix} z_{xx}(0,0) & z_{xy}(0,0) \ z_{xy}(0,0) & z_{yy}(0,0) \ \text{ = } \begin{vmatrix} 0 & 0 \ 0 & 2 \ = 0 \] Since the determinant is zero, the test is inconclusive.
08
- Analyze along straight lines through (0,0)
Check the behavior of z along a straight line through the origin. Let y = kx, substitute into z: \[ z(x, kx) = (kx - x^2)(kx - 2x^2) = kx^2(k - 3x) \ For k > 0, as x approaches 0, z(x, kx) is positive, confirming a minimum. \ For k < 0, same logic applies \]
09
- Conclude the analysis
Since the test is inconclusive and every line through (0,0) shows a minimum but different values at (0,0), z has neither a maximum nor a minimum at (0,0).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives extend the concept of derivatives to functions of several variables. If you have a function \(z = f(x, y)\), partial derivatives measure how \(z\) changes as one variable changes while keeping the other constant. For example, the partial derivative with respect to \(x\) is written as \(\frac{\partial z}{\partial x}\) or \(f_x\). This derivative shows how \(z\) changes if \(x\) changes, but \(y\) remains the same.
To find these derivatives, you usually treat the other variables as constants and apply the usual rules of differentiation. In the exercise, we started with the function \(z = (y - x^2)(y - 2x^2)\) and computed its first partial derivatives:
To find these derivatives, you usually treat the other variables as constants and apply the usual rules of differentiation. In the exercise, we started with the function \(z = (y - x^2)(y - 2x^2)\) and computed its first partial derivatives:
- \(z_x = -6xy + 8x^3\)
- \(z_y = 2y - 3x^2\)
Second Derivative Test
The second derivative test helps determine if a function has a local maximum, local minimum, or saddle point at a critical point. For a function \(z = f(x, y)\), you find the critical points where the first partial derivatives are zero. In our example, at \((0,0)\), we found that \(z_x = z_y = 0\).
The next step involves finding the second partial derivatives and forming the Hessian matrix:\[ H = \begin{pmatrix} z_{xx} & z_{xy} \ z_{xy} & z_{yy} \end{pmatrix} \] In the exercise, we calculated:
The next step involves finding the second partial derivatives and forming the Hessian matrix:\[ H = \begin{pmatrix} z_{xx} & z_{xy} \ z_{xy} & z_{yy} \end{pmatrix} \] In the exercise, we calculated:
- \(z_{xx} = 0\)
- \(z_{yy} = 2\)
- \(z_{xy} = 0\)
- \(H = 0\)
Multivariable Calculus
Multivariable Calculus deals with functions of more than one variable. It includes concepts like partial derivatives and integrals over multiple dimensions. One core idea is analyzing the behavior of these multivariable functions in terms of their extrema (maxima and minima).
In this exercise, we see how analyzing functions in multiple variables involves several steps: finding partial derivatives, evaluating them at specific points, and applying second derivative tests. We checked \(z(x, y)\) at \((0,0)\), explored along lines through the origin, and concluded it had neither maximum nor minimum at \((0,0)\).
This reveals a critical aspect of multivariable calculus: understanding the detailed behavior along different curves and lines. By exploring how \(z\) changes along straight lines like \(y = kx\), we showed that \(z\) seems to always have a minimum along these lines even when the overall test was inconclusive. This thorough examination illustrates the depth of analysis that multivariable calculus allows for complex functions.
In this exercise, we see how analyzing functions in multiple variables involves several steps: finding partial derivatives, evaluating them at specific points, and applying second derivative tests. We checked \(z(x, y)\) at \((0,0)\), explored along lines through the origin, and concluded it had neither maximum nor minimum at \((0,0)\).
This reveals a critical aspect of multivariable calculus: understanding the detailed behavior along different curves and lines. By exploring how \(z\) changes along straight lines like \(y = kx\), we showed that \(z\) seems to always have a minimum along these lines even when the overall test was inconclusive. This thorough examination illustrates the depth of analysis that multivariable calculus allows for complex functions.