# Thread: surfaces and contour maps

1. ## surfaces and contour maps

hi im new to the forum . i needed help on the following question. sorry if i posted this in the wrong area.

1. draw a level curve diagram (with atleast 3 contours) for the surface: $\displaystyle z^2$ = 9$\displaystyle x^2$ + 4$\displaystyle y^2$

2. sketch a graph of the surface: $\displaystyle z^2$ = 9$\displaystyle x^2$ + 4$\displaystyle y^2$

any help much appreciated

2. ## Contour maps

Imagine the xy-plane as being the ground, and the z-axis represents height. So, z=0 is "ground" level, z=1 is one unit above the ground, z=2 is two units up, etcetera. What we are doing is taking horizontal cross-sections of the 3D graph.

First, rewrite the function as $\displaystyle 1=\left(\frac{x}{z/3}\right)^2+\left(\frac{y}{z/2}\right)^2$

So each slice of the graph is a vertical ellipse with axes z/2 tall and z/3 wide.

$\displaystyle z=0 \to 0=9x^2+4y^2$ The only point (x,y) satisfying this is the origin, (0,0)
$\displaystyle z=\pm 6 \to 1=\left(\frac{x}{2}\right)^2+\left(\frac{y}{3}\rig ht)^2$
$\displaystyle z=\pm 12 \to 1=\left(\frac{x}{4}\right)^2+\left(\frac{y}{6}\rig ht)^2$

See if you can visualize it from here.

3. Hi im doing a similar question but confused on how you rewrote the equation into the ellipse standard form. Can you please help?

4. ## Basic Algebra

$\displaystyle z^2=9x^2+4y^2$
$\displaystyle z^2=(3x)^2+(2y)^2$
$\displaystyle z^2=\left(\frac{x}{1/3}\right)^2+\left(\frac{y}{1/2}\right)^2$
$\displaystyle 1=\left(\frac{x}{z/3}\right)^2+\left(\frac{y}{z/2}\right)^2$

Since an ellipse has the form: $\displaystyle 1=\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\rig ht)^2$, then the above equation represents an ellipse with major axis $\displaystyle z/2$ and minor axis $\displaystyle z/3$.

5. ## hyperbola contours

Thanks that helps, but im doing it and iv realised its a hyperbola. Can i ask you if what iv done so far is correct.

z^2=4+9x^2-y^2
z^2=c
c-4=9x^2-y^2

for which c>0 and -infinity< z < infinity

(3x)^2 - (y)^2 = c-4

(x/(1/3))^2 - (y)^2 = c-4

(x/(sqrt(c-4)/3)^2 - (y/sqrt(c-4))^2 = 1

So then i need values of c to draw the initial rectangles, which i can then draw the the curves. The problem is on the question it says for
c>=4 and c<4. But since its a square root which i got c cant be less than 4. Its infinity but thats not the solution. Theres a hyperbola for c less than 4. If you can, any help would be really appreciated.

Thanks so much.