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 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 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 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 g(x,y) = x^2+y^2=1} , set 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 \nabla Q = \lambda \nabla g} 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 g=1}

Part A

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

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{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 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 \lambda} . Start by putting everything on the left side of the equation:

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

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

The determinant gives 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 (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, $y$)

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