MATH 251 Lecture 17

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Second Derivative Test

The second derivative depends partially on the Taylor series of the function. Given $f(x,y)$,

$T_1(x_0+h,y_0+k)=f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)\cdot h+\frac{\partial f}{\partial y}(x_0,y_0)\cdot k$

$T_2(x_0+h,y_0+k)=T_1(x,y)+\frac{1}{2}(\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)\cdot h^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)\cdot hk+\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)\cdot k^2)$

The partial derivatives of the stuff in "()" are coefficients for variation in $x$ and $y$

From linear algebra, $a{h}^2+2bhk+c{k}^2$ is equal to the following matrix multiplication:

$\left(\begin{array}{cc}h& k\end{array}\right)\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)\left(\begin{array}{c}h\\ k\end{array}\right)$

For small variations in $h$ and $k$, we find the directions in which $f(x,y)$ curves around critical points:

$Q(h,k)=\frac{1}{2}(\lambda h^2+\mu k^2)$, where $\lambda =f_{xx}$ and $\mu =f_{yy}$, assuming $f_{xy}=f_{yx}=0$

Consider the hessian matrix $\left(\begin{array}{cc}\lambda & 0\\ 0& \mu \end{array}\right)$

Conclusive

1. Minimum: (determinant is positive) $\lambda ,\mu >0$
2. Maximum: (determinant is positive) $\lambda ,\mu <0$
3. Saddle Surface: (determinant is negative) $\lambda >0,\mu <0$ or $\lambda <0,\mu >0$

Inconclusive (if determinant of matrix is 0)

1. cylinder: $\lambda \ne 0,\mu =0$ or $\lambda =0,\mu \ne 0$
2. plane: $\lambda =\mu =0$

By the way, in almost all cases, $f_{xy}=f_{yx}$.

Linear Algebra Stuff

If $f_{xy}\ne 0$, then by a rotation of coordinates $x^{\prime }=\cos \theta x+\sin \theta y,y^{\prime }=\dots$, f_{x'y'} = 0

Example

Find and classify any Critical points of the function $f(x,y)=2x^3+x{y}^4+5x^2+y^2+4$

$\begin{array}{rl}{f}_x=6x^2+y^4+10x& =0\\ f_y=2xy+2y& =0\\ y=0\text{ or }x=-1\\ \text{if }y=0,& x=0,-\frac{5}{3}\\ \text{if }x=-1,& y=\pm 2\\ \end{array}$

So the critical points are (-5/3, 0), (0, 0), (-1, 2), (-1, -2)

$\begin{array}{r}{f}_{xx}=12x+10\\ f_{xy}=2y\\ f_{yy}=2x+2\end{array}$

So the Hessian matrix is $\left(\begin{array}{cc}12x+10& 2y\\ 2y& 2x+2\end{array}\right)$, and the determinant at critical points are:

1. $10/3>0$ (min/max); (max since $f_{xx}(-\frac{5}{3},0)<0$)
2. $20>0$ (min/max); (min since $f_{xx}(0,0)>0$)
3. $-16<0$ (Saddle)
4. $-16<0$ (Saddle)

The values at each of the critical points are:

1. 233/27
2. 4
3. 7
4. 7