Second partial derivative test

Second partial derivative test
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Second_partial_derivative_test".

In mathematics, the second partial derivatives test is a method in multivariable calculus used to determine if a critical point (x, y) is a minimum, maximum or saddle point.

Suppose that

$M = f_{xx}(a,b)f_{yy}(a,b) - \left( f_{xy}(a,b) \right)^2$

or in other words the determinant of a 2×2 Hessian matrix,

$M = \begin{vmatrix}f_{xx}(a,b) &f_{xy}(a,b)\\f_{yx}(a,b) &f_{yy}(a,b)\end{vmatrix}$ .

If M > 0 and $f x x (a, b) > 0$ then f(a, b) is a local minimum.

If M > 0 and $f x x (a, b) < 0$ then f(a, b) is a local maximum.

If M < 0 then f(a, b) has a saddle point.

If M = 0 then the second derivatives test is indecisive.

M and $f x x (a, b)$ are the leading principal minors of the Hessian. The conditions listed above (the sign of these values) are the conditions for the definiteness of the Hessian.

We only test values for which $f x$ = 0 and $f y$ = 0. This is rationalized because the function must on its trace along the xz-plane have its derivative equal to zero and the same is true for a trace on the yz-plane.

Geometric Interpretation

Assuming that all derivatives are evaluated at (a,b), and that the value of the first derivatives vanish there.

If M < 0 then $f_{xx}f_{yy} < f_{xy}^2$ . If either $f x x$ or $f y y$ are negative, the other must be positive and thus the concavities of the x cross section (the yz trace) and the y cross section (the xz trace) are in opposite direction. This is clearly a saddle point.

If M > 0 then $f_{xx}f_{yy} > f_{xy}^2$ , which implies that $f x x$ and $f y y$ are the same sign and sufficiently large. For this case the concavities of the x and y cross sections are either both up if positive, or both down if negative. This is clearly a local minimum or a local maximum, respectively.

This leaves the last case of M < 0 so $f_{xx}f_{yy} < f_{xy}^2$ and $f x x$ and $f y y$ having the same sign. The geometric interpretation of what is happening here is that since $f x y$ is large it means the slope of the graph in one direction is changing rapidly as we move in the orthogonal direction and overcoming the concavity of the orthogonal direction. So for example, let's take the case of all second derivatives are positive and (a,b) = (0,0). In the case of M > 0 it would mean that any direction in the x-y plane we move from the origin, the value of the function increases--a local minimum. In the M < 0 case ( $f x y$ sufficiently large), however, if we move at some direction between the x and y axis into the second quadrant, for example, of the xy plane, then despite the fact that the positive concavity would cause us to expect the value of the function to increase, the slope in the x direction is increasing even faster, which means that as we go left (negative x-direction) into the second quadrant, the value of the function ends up decreasing. Additionally, since the origin is a stationary point by hypothesis, we have a saddle point.

Examples

Find and label the critical points of the following function:

z = (x + y)(x y + x y 2)

To solve this problem we must first find the first partial derivatives with respect to x and y of the function.

$\frac{\partial z}{\partial x} = y(2x +y)(y+1)$

$\frac{\partial z}{\partial y} = x \left( 3y^2 +2y(x+1) + x \right)$

Looking at

$\frac{\partial z}{\partial x} \left( x \right) = 0$

we see that y must equal 0, −1 or $- 2 x$ .

We plug this into the next equation, we get

$\frac{\partial z}{\partial y} = x \left( 3y^2 +2y(x+1) + x \right) = x^2$

There were other possibilities for y, so we have

$\frac{\partial z}{\partial y} = x \left( 3 -2(x+1) + x \right) = 0 = x(1-x)$

So x must be equal to 1 or 0.

$\frac{\partial z}{\partial y} = x \left( 3(-2x)^2 +2(-2x)(x+1) + x \right) = 4x^2(2x-1)$

So x must equal 0 or $\frac{1}{2}$

Let's list all the critical values now.

$(x,y) \in {(0,0), (0, -1), (1,-1), (\frac{1}{2}, -1)}$

Now we have to label the critical values using the second derivative test.

$D = f_{xx}(a,b)f_{yy}(a,b) - \left( f_{xy}(a,b) \right)^2 = 2y(y+1)(2(3y+x+1))x - (3y^2+y(4x+2)+2x)^2$

Now we plug in all the different critical values we found to label them.

At (0, 0) we have D = 0, at (0, −1); D = −1, at (1, −1); D = −1, at

$(\frac{1}{2}, -1)$ D = 0.

So we can now label some of the points, at (0, −1) and (1, −1) f(x, y) has a saddle point. At the other two points we need higher order tests to find out what exactly the function is doing.

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