MATH 251 Lecture 19

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Quick Fact

When does $\{\begin{array}{l}ax+by=0\\ cx+dy=0\end{array}$ or $A\cdot X=0$ have a non-zero solution?

in 2 dimensions, there is a unique solution unless the vectors $⟨a,b⟩$ and $⟨c,d⟩$ are scalar multiples (parallel)

in 3 dimensions (3 equations w/r/t $x$, $y$, and $z$), there is a unique solution unless the 3 vectors are coplanar.

Check the area of the 2 vectors (2D) and volume of 3 vectors (3D) by taking the determinant of the matrix formed by the coefficient. If this value is 0, then there are non-zero solutions. If the determiniant is 0, then the only solution is the point (0,0).

Written Homework Sample Problem

Given $Q(x,y)=x^2+2xy+3y^2$,

1. Set up equations satisfied by relative extrema on the unit circle (sphere for 3D)
2. Subject to the condition $g(x,y)=x^2+y^2=1$, set $\mathrm{\nabla }Q=\lambda \mathrm{\nabla }g$ and $g=1$

Part A

Set $\mathrm{\nabla }Q=\lambda \mathrm{\nabla }g$ and our constraint

$\{\begin{array}{l}x+y=\lambda x\\ x+3y=\lambda y\\ x^2+y^2=1\end{array}$

Part B

Find the characteristic polynomial and solve for $\lambda$. Start by putting everything on the left side of the equation:

$\{\begin{array}{l}(1-\lambda )x+y=0\\ x+(3-\lambda )y=0\end{array}$

Find the determinant of the matrix $\left|\begin{array}{cc}1-\lambda & 1\\ 1& 3-\lambda \end{array}\right|=0$

The determinant gives $(1-\lambda )(3-\lambda )-1=0$... our characteristic polynomial.

The solution(s) for $\lambda$ are:

$\lambda =\frac{4\pm \sqrt{16-4(1)(2)}}{2(1)}=2\pm \sqrt{2}$

These are our Lagrange multipliers.

Part C

The values of $\lambda$ (eigenvalues) are the extreme values. Classify $Q$ as positive, negative, mixed, or degenerate.

All values of $\lambda$ are positive, therefore $Q$ is positive.

Part D

Find the solutions $(x,y)$ (eigenvectors) by plugging $\lambda$ into the equations from 'Part A. In order for the solution to be non-zero, the solutions should be redundant, so you can choose one equation and plug it into that one.

$\begin{array}{rl}{\lambda }_1=2+\sqrt{2}& ⟶\frac{1}{\sqrt{4+2\sqrt{2}}}⟨1,1+\sqrt{2}⟩\\ {\lambda }_1=2-\sqrt{2}& ⟶\frac{1}{\sqrt{4+2\sqrt{2}}}⟨1+\sqrt{2},-1⟩\end{array}$

Note: The lines given by plugging $\lambda$ in will always be perpendicular

Part E?

Diagonalize the formula by removing the mixed terms (i.e., the $2xy$)

$Q(x^{\prime },y^{\prime })={\lambda }_1(x^{\prime }{)}^2+{\lambda }_2(y^{\prime }{)}^2$

Multidimensional Integration

Given $f(x,y)$ defined over a rectangle $R=[a,b]\times [c,d]=\{(x,y)|a\le x\le b,c\le y\le d\}$, define $\underset{R}{\iint }f(x,y)\mathrm{d}A$

Back to Calculus 2, the integral $\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$ represents

• Area under a curve $f(x)$
• Total mass of a rod if $f(x)$ is the linear density function
• Probability density of a continuous random variable
• The sum of an infinite family of numbers defined continuously in terms of $x$

In 3 Dimensions, the multidimensional integral $\underset{R}{\iint }f(x,y)\mathrm{d}A$ represents

• The volume under a function
• Total mass in a given volume with mass density of $f(x,y)$
• Probability density of two continuous random variables
• Still an infinite sum, but determined continuously by two variables $x$ and $y$.

Quick Definition

$\underset{R}{\iint }f(x,y)\mathrm{d}A=\underset{|P|\to 0}{\lim }\sum _{i,j=1}^{N}f(x_i,y_j)\mathrm{\Delta }{x}_i\mathrm{\Delta }{y}_j$

$\mathrm{\Delta }{x}_i\mathrm{\Delta }{y}_j$ represents the area of a small rectangle near $(x_i,y_j)$, called a partition ($P$)

Example

Calculate $\underset{R}{\iint }(x-y^2)\mathrm{d}A$, where $R=[0,2]\times [1,3]$.

$\underset{1}{\overset{3}{\int }}\underset{0}{\overset{2}{\int }}(x-y^2)\mathrm{d}x\mathrm{d}y$

Notation is to use Fubini's Theorem: slice the function according to a single variable (by holding the other constant; in this case, $y$)

$\begin{array}{rl}\underset{0}{\overset{2}{\int }}(x-y^2)\mathrm{d}x& ={\frac{1}{2}{x}^2-x{y}^2|}_0^2\\ \underset{1}{\overset{3}{\int }}(2-2y^2)\mathrm{d}y& ={2y-\frac{2}{3}{y}^3|}_1^3=4-\frac{2}{3}26\end{array}$