# Parametric equation/cone

• Mar 16th 2009, 07:12 PM
arbolis
Parametric equation/cone
Hi,
Here is my problem: Show that the curve with the parametric equation $x=t \cos (t)$, $y=t \sin (t)$, $z=t$ is contained in the cone $z^2=x^2+y^2$.
My attempt : I don't really know how to show this. However I've showed that $x$, $y$ and $z$ satisfy the equation of the cone. Does that mean it forms it? Does that mean that the parametric curve is in the cone? What does this exactly mean?
• Mar 16th 2009, 08:50 PM
Reckoner
Quote:

Originally Posted by arbolis
My attempt : I don't really know how to show this. However I've showed that $x$, $y$ and $z$ satisfy the equation of the cone. Does that mean it forms it? Does that mean that the parametric curve is in the cone? What does this exactly mean?

Your cone is defined as all of the points (and only those points) whose coordinates satisfy the given equation. That is, a point is on the cone if and only if the coordinates satisfy $z^2=x^2+y^2.$

So, if the given parametric equations satisfy the equation of the cone for all $t,$ then what does that tell you about the points on the curve formed by these parametric equations?
• Mar 17th 2009, 06:57 AM
arbolis
This tells me that they're on the cone.
But do they form the entire cone? Or they're just a part of the cone?
• Mar 17th 2009, 07:34 AM
HallsofIvy
The curve is one-dimensional, the cone is two-dimensional. (It is possible that a one-dimensional graph passes through every point of a two-dimensional surface but not this simple.)
• Mar 17th 2009, 07:52 AM
arbolis
Quote:

Originally Posted by HallsofIvy
The curve is one-dimensional, the cone is two-dimensional. (It is possible that a one-dimensional graph passes through every point of a two-dimensional surface but not this simple.)

Ok. So the curve doesn't form the entire cone, but rather some part of it.
I thought the cone was 3 dimensional. If I take z=constant then I get $k=x^2+y^2$ which is a circle. That is, if I "cut" the cone in a certain way, I can get a circle. Hence isn't it a 3 dimensional figure?
• Mar 17th 2009, 12:22 PM
Reckoner
Quote:

Originally Posted by arbolis
Ok. So the curve doesn't form the entire cone, but rather some part of it.
I thought the cone was 3 dimensional. If I take z=constant then I get $k=x^2+y^2$ which is a circle. That is, if I "cut" the cone in a certain way, I can get a circle. Hence isn't it a 3 dimensional figure?

The solid cone contained by the surface is three-dimensional. The surface itself is two-dimensional, but it is embedded in three-dimensional Euclidean space.

As for your original question, it might help to graph it. You should see that the curve represented by the parametric equations (and it is a curve, not a surface) spirals around the cone. This makes sense if you take a look at the equations for $x$ and $y.$ You basically have a circle, except that there is a factor of $t$ in the front that causes the "radius" to increase as the circle is traced out, forming a spiral. Then, $z=t$ means that this spiral gradually rises as $t$ increases.
• Mar 17th 2009, 02:06 PM
arbolis
Ok I totally get it. So in fact the curve is on the surface of the cone rather than inside.
However if you give me the equation of a curve which doesn't satisfy $z^2=x^2+y^2$, I think I wouldn't be able to say whether it's inside or outside the cone. I guess I should know this, but I don't have an example.
• Mar 17th 2009, 02:34 PM
Reckoner
Quote:

Originally Posted by arbolis
Ok I totally get it. So in fact the curve is on the surface of the cone rather than inside.
However if you give me the equation of a curve which doesn't satisfy $z^2=x^2+y^2$, I think I wouldn't be able to say whether it's inside or outside the cone. I guess I should know this, but I don't have an example.

For a fixed $z,$ the points inside the circular cross section are given by $x^2+y^2 and the points outside the circle by $x^2+y^2>z^2.$ (from the distance formula--the points on the circle are those that are exactly $z$ units from the center of the circle, those with greater distance from the center are outside the circle, and so on). Curves which always satisfy one of these two inequalities will lie entirely inside or entirely outside of the cone.
• Mar 17th 2009, 02:41 PM
arbolis
That makes sens, thank you.