How to interpret a function in 3 dimentions

[Syrena]

Homework Statement

I have calculus, and a lot of tasks ask for drawing of the functions in 3 dimentions. I have problems thinking of how to do this, often it can be a parabolid or other Objects. Example, one I can't think of how I am suppose to Draw, here I am going to solve for Stokes (that is a pretty easy function) x^2 +y^2 +5z =1 , where z≥0.

Is there any good technique to use for drawing With Three variables?

Homework Equations

I don't know what to use here

The Attempt at a Solution

I think that if I put z=0, you get a circle in the xy-plane With a radius 1, but when z=1, I get X^2 +y^2 =-4, so the New radius is -4 ( or is it √-4?) i don't think I am thinking correctly here.

 

[Mark44]

Homework Statement

I have calculus, and a lot of tasks ask for drawing of the functions in 3 dimentions. I have problems thinking of how to do this, often it can be a parabolid or other Objects. Example, one I can't think of how I am suppose to Draw, here I am going to solve for Stokes (that is a pretty easy function) x^2 +y^2 +5z =1 , where z≥0.

Is there any good technique to use for drawing With Three variables?

Homework Equations

I don't know what to use here

The Attempt at a Solution

I think that if I put z=0, you get a circle in the xy-plane With a radius 1, but when z=1, I get X^2 +y^2 =-4, so the New radius is -4 ( or is it √-4?) i don't think I am thinking correctly here.

That's right - you aren't.
For this surface, you can get a lot of information from cross-sections that are parallel to the x-y plane. To do this, rewrite the equation as x² + y² = 1 - 5z.

Since x² + y² ≥ 0 for all real x and y, then it must be that 1 - 5z ≥ 0 as well. If you substitute z = 1, you don't get a circle with an imaginary radius. It means that there are no points on the surface for which z = 1.

If 1 - 5z = 0, what you get is a degenerate circle - a single point. Try substituting a few values of z for which 1 - 5z > 0. All of these will give you cross-sections at various heights that will help you recognize what the surface looks like.

Another technique that can be used is to look at the three traces, cross-sections of the curve in the x-y plane (set z = 0), x-z plane (set y = 0), and y-z plane (set x = 0).

 

[Syrena]

Thanks! But how do you see that 1-5z=0 is a single point?
And to get 1 - 5z > 0, dosent z have to be negative? to have z=, -1, -2 -3 , but that can't be becous z≥0. and does this gives us the radius of the parabolid?

 

[haruspex]

Thanks! But how do you see that 1-5z=0 is a single point?

It clearly fixes the value of z. If you substitute that value in the original equation, what do you get? I think you'll find there is only solution for x and y as well.

And to get 1 - 5z > 0, dosent z have to be negative?

No, think that through again.