Sketch Cone z2=x2+y2 in 3D Space + Vector Calc

[rock.freak667]

Homework Statement

Sketch the cone z²=x²+y² in 3D space.

Let (x₀,y₀,z₀)≠(0,0,0) be a point on the given cone. By expressing the fiven equation of the cone in the form f(x,y,z)=a, find a normal vector tot he cone at point (x₀,y₀,z₀)

Find the equation of the tangent plane to the cone at point(x₀,y₀,z₀)

Show that every plane that is a tangent to the cone passes through the origin

Homework Equations

$$\hat{n} =grad(f) \ at \ (x_0,y_0,z_0)$$

The Attempt at a Solution

I was able to do part 2 and 3.

$$\hat{n}= 2x_0 \hat{i} +2y_0 \hat{j} -2z_0 \hat{k}$$

and then tangent plane is

2x₀(x-x₀)+2y₀(y-y₀)-2z₀(z-z₀)=0

so if it passes through the origin x=0,y=0 and z=0, which leaves me with

$$x_0 ^{2}+y_0 ^{2}-z_0 ^{2}=0$$

But since (x₀,y₀,z₀)≠(0,0,0), then doesn't this mean that it does not pass through the origin?


Also, how do I sketch a curve in 3D? Do I just randomly plug in values of (x,y,z) and plot or is there some method in doing it?

 

[gabbagabbahey]

so if it passes through the origin x=0,y=0 and z=0, which leaves me with

$$x_0 ^{2}+y_0 ^{2}-z_0 ^{2}=0$$

But since (x₀,y₀,z₀)≠(0,0,0), then doesn't this mean that it does not pass through the origin?

Remember, the point $(x_0,y_0,z_0)$ lies on the cone, so $z_0^2=x_0^2+y_0^2$.

Also, how do I sketch a curve in 3D? Do I just randomly plug in values of (x,y,z) and plot or is there some method in doing it?

I would plug a few different values of $z$ into the equation of the cone, and sketch the resulting curve at that height (value of $z$), it should become clear pretty quickly what the cone looks like.

 

[rock.freak667]

Remember, the point $(x_0,y_0,z_0)$ lies on the cone, so $z_0^2=x_0^2+y_0^2$.

thanks

I would plug a few different values of $z$ into the equation of the cone, and sketch the resulting curve at that height (value of $z$), it should become clear pretty quickly what the cone looks like.

But I don't get how I would know what it looks like

on the yz plane z²=y² which means y=±z. How does the curve have two straight line portions ?EDIT: yeah this is the first time I've done these things and the person teaching this to me is a bit...hard to talk to, so I can't ask.

 

[gabbagabbahey]

2[/SUP]=y² which means y=±z. How does the curve have two straight line portions ?

Well, let's see...

For $z=0$ we have $x^2+y^2=0$, which is a circle of radius zero, i.e. just a point at the origin.

For $z=1$ we have $x^2+y^2=1$, which is a circle of radius 1 at height $z=1$.

For $z=-1$ we have $x^2+y^2=1$, which is again a circle of radius 1, this time at a height $z=-1$.

For $z=2$ we have $x^2+y^2=4$, which is a circle of radius 2 at height $z=2$.

For $z=-2$ we have $x^2+y^2=4$, which is again a circle of radius 2, this time at a height $z=-2$.

As you get further and further away from z=0, in either direction, the circles get larger and larger...

http://img268.imageshack.us/img268/1341/cone2t.th.jpg

You should see clearly now why there are two lines (in the shape of an X) in the yz plane.

 

[rock.freak667]

oh I see now, so the best thing to do is to try to vary one variable and see what shape is formed by the x and y ( a circle in this case with increasing radius)?

Sketching on paper is going to be a bit odd

thanks!