Ellipsoid

Graph the following equation. Note that it is not actually a function, since it isn't one-to-one. $\dfrac{x^2}{9}+\dfrac{y^2}{16}+\dfrac{z^2}{25} = 1$

Solution

One way to graph an equation like this is to use traces, which are curves in a plane parallel to one of the coordinate planes. For instance, if we let $z=0$, we get a graph in the $xy$ plane: $\dfrac{x^2}{9}+\dfrac{y^2}{16}+\dfrac{0^2}{25} = 1 \longrightarrow \dfrac{x^2}{9}+\dfrac{y^2}{16} = 1$

This is an ellipse with major and minor axes of $4$ and $3$. For other values of $z$, we graph ellipses in planes above or below $z=0$, until $z=±5$, where the ellipse shrinks to a point.

The graph looks like the following (kind of like an M&M shape):

Paraboloid

Graph the following function. $z=f(x,y)=x^2+y^2$

Solution

Consider different values of $z$ again: $\begin{array}{l l l} z & \textrm{Curve} & \textrm{Description}\ \hline -1 & x^2+y^2=-1 & \textrm{Impossible}\ 0 & x^2+y^2=0 & \textrm{Point at the origin}\ 1 & x^2+y^2=1 & \textrm{Circle of radius 1 centered at the origin}\ 2 & x^2+y^2=2 & \textrm{Circle of radius 2 centered at the origin}\ 4 & x^2+y^2=4 & \textrm{Circle of radius 4 centered at the origin}\ 9 & x^2+y^2=9 & \textrm{Circle of radius 9 centered at the origin}\ \end{array}$

We'll use Matlab to see the graph of this function:

>> fsurf(@(x,y) x.^2+y.^2)

This is called a paraboloid, and as the name suggests, it looks like a parabola in both directions.

The domain of this function is $D = {(x,y)\ :\ -\infty < x < \infty,\ -\infty < y < \infty}$ and the range is $R = {z\ :\ 0 \leq z < \infty}.$

Describe the level curves of the following function. $f(x,y) = y-x^2-1$

Solution

If we hold the function constant ($f(x,y)=z=z_0$), we get curves like $y-x^2-1=z_0 \longrightarrow y=x^2+1+z_0$

Since $1+z_0$ is a constant, these are parabolas.

The graph of the function and a few of the level curves are shown below.

Describe the level curves of the following function. $f(x,y) = 2+\sin (x-y)$

Solution

Hold $z$ constant: $2+\sin (x-y)=z_0 \longrightarrow y=x-\sin^{-1}(z_0-2)$

Note that $\sin^{-1}(z_0-2)$ is a constant, so these are straight lines $y=x−k$.

The graph of the function is shown below, and you should be able to see that the function stays constant along straight lines.

Electric Potential

The electric potential function for two positive charges, one at $(0,1)$ with twice the charge of one at $(0,1)$, is given by $\phi (x,y) = \dfrac{2}{\sqrt{x^2+(y-1)^2}} + \dfrac{1}{\sqrt{x^2+(y+1)^2}}$

First, we'll graph this function using Matlab:

>> fsurf(@(x,y) 2./sqrt(x.^2+(y-1).^2) + 1./sqrt(x.^2+(y+1).^2))

1. For what values of $x$ and $y$ is $\phi$ defined?
2. Is the electric potential greater at $(3,2)$ or $(2,3)$?
3. Describe how the electric potential varies along the line $y=x$

Solution

1. Domain: where is the denominator in each term $0$? $\begin{aligned} x^2+(y-1)^2 = 0 &\longrightarrow (0,1)\ x^2+(y+1)^2 = 0 &\longrightarrow (0,-1) \end{aligned}$

Therefore, the domain is every point except for these two (the potential approaches infinity near these points). $\ans{D = {(x,y)\ :\ \mathbb{R}^2 \setminus {(0,1) \cup (0,-1)}}}$

2. Plug these two points into the function: $\begin{aligned} \phi (3,2) &= \dfrac{2}{\sqrt{9+1}} + \dfrac{1}{\sqrt{9+9}} \approx 0.868\ \phi (2,3) &= \dfrac{2}{\sqrt{4+4}} + \dfrac{1}{\sqrt{4+16}} \approx 0.931 \end{aligned}$

Therefore, the potential is greater at $\ans{(2,3).}$

3. Replace $y$ with $x$: $\begin{aligned} \phi &= \dfrac{2}{\sqrt{x^2+(x-1)^2}} + \dfrac{1}{\sqrt{x^2+(x+1)^2}}\ &= \ans{\dfrac{2}{\sqrt{2x^2-2x+1}} + \dfrac{1}{\sqrt{2x^2+2x+1}}} \end{aligned}$

Economics

The output $Q$ of an economic system subject to two inputs, such as labor $L$ and capital $K$, can be modeled by the Cobb-Douglas production function $Q (L,K) = cL^aK^b$

where $a$, $b$, and $c$ are positive real numbers.

Suppose $a=\dfrac{1}{3}$, $b=\dfrac{2}{3}$, and $c=40$. We can graph this with Matlab:

>> fsurf(@(x,y) 40.*x.^(1/3).*y.^(2/3))

If $L$ is held constant at $10$, find the dependence of $Q$ on $K$.

Solution

$Q=40(10^{1/3})K^{2/3}$

This is a trace along a vertical plane.