Equation of ellipsoid and graph

[toforfiltum]

Homework Statement

Equation of ellipsoid is:

$\frac{x^2}{4} + \frac{y^2}{9} + z^2 = 1$

First part of the question, they asked to graph the equation. I have a question about this, I know that $-1\leq z \leq 1$. So what happens when the constant 1 gets smaller after minusing some value of $z^2$? Does it's "radius" get smaller?

Second part of the question is:

Is it posiible to find a function $f(x,y)$ so that this ellipsoid may be considered to be the graph of $z=f(x,y)$?

Homework Equations

The Attempt at a Solution

For the second part, I answered no, because the graph $z=f(x,y)$ is a function of the level curve of the graph of the ellipsoid.

Am I right?

Thanks.

 

[mfb]

So what happens when the constant 1 gets smaller after minusing some value of $z^2$? Does it's "radius" get smaller?

1 is 1, it cannot get smaller. What do you mean by "what happens", and what do you call radius?

Is it posiible to find a function $f(x,y)$ so that this ellipsoid may be considered to be the graph of $z=f(x,y)$?
[...]
For the second part, I answered no, because the graph $z=f(x,y)$ is a function of the level curve of the graph of the ellipsoid.

I don't understand your answer, but it does not contain the crucial point. If there would be such a function: What is e. g. f(0,0)? It needs a unique z-value.

 

[toforfiltum]

1 is 1, it cannot get smaller. What do you mean by "what happens", and what do you call radius?

How do I graph this ellipsoid? I was trying to use the level curves method. By looking at the equation, I know that the value of $z$ cannot be more than 1. I was trying to set the value of $z$ to plot the graph. So I was trying to plot the level curves for $x$ and $y$. That's why the value of 1 changes when I do this. Is what I'm doing wrong?

I don't understand your answer, but it does not contain the crucial point. If there would be such a function: What is e. g. f(0,0)? It needs a unique z-value.

I think I now get it. There's no such unique function for $z=f(x,y)$ because it is $z^2$ in the original equation. Hence there needs to be 2 functions, $z = \pm \sqrt{[1 - \frac{x^2}{4} - \frac{y^2}{9}]}$.

 

[LCKurtz]

@toforfiltum: In response to your question of how to sketch an ellipsoid. Your example was$$
\frac{x^2}{4}+ \frac{y^2} 9 + z^2 = 1$$ I'm going to assume that you know that if you have the equation$$
\frac{x^2}{a^2}+ \frac{y^2} {b^2} = 1$$in the $xy$ plane, that gives an ellipse with $x$ intercepts $(\pm a,0)$ and $y$ intercepts $(0,\pm b)$. You could lightly draw a rectangular box square with the axes through those four points and sketch a nice looking ellipse in that box.
To sketch a 3D ellipsoid like the one you gave, you can just draw the traces in the coordinate planes. For example in the plane $x=0$ your equation would be $\frac{y^2} 9 + \frac {z^2} 1 = 1$. Sketch that in the $yz$ plane with $y$ intercepts of $\pm 3$ and $z$ intercepts of $\pm 1$. Do the other two traces similarly.

 

[mfb]

So I was trying to plot the level curves for x and y.

That is a possible approach. z cannot exceed 1 (and cannot be smaller than -1), sure, so you get curves for |z|<1. Add curves in the other planes for a nicer 3D illustration? Or use some predefined plotting algorithm that can draw the ellipsoid.