MATH 251 Lecture 19

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Quick Fact

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

in 2 dimensions, there is a unique solution unless the vectors $\left\langle a,b\right\rangle$ and $\left\langle c,d\right\rangle$ 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 $\nabla Q=\lambda \nabla g$ and $g=1$

Part A

Set $\nabla Q=\lambda \nabla g$ and our constraint

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

Part B

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

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

Find the determinant of the matrix ${\begin{vmatrix}1-\lambda &1\\1&3-\lambda \end{vmatrix}}=0$

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.

${\begin{aligned}\lambda _{1}=2+{\sqrt {2}}&\longrightarrow {\frac {1}{\sqrt {4+2{\sqrt {2}}}}}\left\langle 1,1+{\sqrt {2}}\right\rangle \\\lambda _{1}=2-{\sqrt {2}}&\longrightarrow {\frac {1}{\sqrt {4+2{\sqrt {2}}}}}\left\langle 1+{\sqrt {2}},-1\right\rangle \end{aligned}}$

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 $2xy$)

$Q(x',y')=\lambda _{1}(x')^{2}+\lambda _{2}(y')^{2}$

Multidimensional Integration

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

Back to Calculus 2, the integral $\int _{a}^{b}f(x)\,\mathrm {d} x$ 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 $\iint _{R}f(x,y)\,\mathrm {d} A$ 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

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

$\Delta x_{i}\Delta y_{j}$ represents the area of a small rectangle near $(x_{i},y_{j})$, called a partition ($P$)

Example

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

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

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

${\begin{aligned}\int _{0}^{2}(x-y^{2})\,\mathrm {d} x&=\left.{\frac {1}{2}}x^{2}-xy^{2}\right|_{0}^{2}\\\int _{1}^{3}(2-2y^{2})\,\mathrm {d} y&=\left.2y-{\frac {2}{3}}y^{3}\right|_{1}^{3}=4-{\frac {2}{3}}26\end{aligned}}$