MATH 251 Lecture 17

« previous | Monday, February 27, 2012 | next »

Second Derivative Test

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_2(x_0+h,y_0+k) = T_1(x,y) + \frac{1}{2} \left( \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 \right)}

The partial derivatives of the stuff in "()" are coefficients for variation in $x$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y}

From linear algebra, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ah^2 + 2bhk + ck^2} is equal to the following matrix multiplication:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}h&k\end{pmatrix} \begin{pmatrix}a&b\\b&c\end{pmatrix} \begin{pmatrix}h\\k\end{pmatrix} }

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

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

Consider the hessian matrix ${\begin{pmatrix}\lambda &0\\0&\mu \end{pmatrix}}$

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: $\lambda \neq 0,\mu =0$ or $\lambda =0,\mu \neq 0$
2. plane: $\lambda =\mu =0$

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

Linear Algebra Stuff

If $f_{xy}\neq 0$, then by a rotation of coordinates $x'=\cos \theta x+\sin \theta y,y'=\ldots$, f_{x'y'} = 0

Example

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

${\begin{aligned}f_{x}=6x^{2}+y^{4}+10x&=0\\f_{y}=2xy+2y&=0\\y=0{\mbox{ or }}x=-1\\{\mbox{if }}y=0,&\ x=0,-{\frac {5}{3}}\\{\mbox{if }}x=-1,&\ y=\pm 2\\\end{aligned}}$

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

${\begin{aligned}f_{xx}=12x+10\\f_{xy}=2y\\f_{yy}=2x+2\end{aligned}}$

So the Hessian matrix is ${\begin{pmatrix}12x+10&2y\\2y&2x+2\end{pmatrix}}$, and the determinant at critical points are:

1. $10/3>0$ (min/max); (max since $f_{xx}(-{\tfrac {5}{3}},0)<0$)
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