# Prove that curve lies on a cone?

I have a curve given by this equation: $$c(t) = (t\cos t,t \sin t,t)$$

I need to prove that this curve lies on a cone, and draw that cone and curve in Sage.

I've read somewhere that I could prove it by taking some random point and then put it into equation. But is there a more general proof, formal way of proving it lies on a cone?

How should I prove it (the easiest way?)

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It's obvious that $z^2=x^2+y^2$ and that's the equation of a (double) cone. –  kiwi May 20 '12 at 23:42
@kiwi Can you add that as answer so that this question gets an answer? Thanks! –  user17762 May 21 '12 at 0:22

In this case since you have your parametrized curve given by $\gamma(t) = (t\cos{t}, t\sin{t}, t)$, one way of trying to find an algebraic equation satisfied by any point on your curve is writing $x = t\cos{t}$, $y = t\sin{t}$ and $z = t$. Now since you have sine and cosine, you have to think about the relations that these functions satisfy.

The easiest one, which you obviously know, is $\cos^2{t} + \sin^2{t} = 1$. Now in terms of $x, y, z$ you see that

$$\cos{t} = \frac{x}{z} \quad \text{and} \quad \sin{t} = \frac{y}{z}$$

Then this tells you that

$$\left( \frac{x}{z} \right )^2 + \left( \frac{y}{z} \right )^2 = 1$$

and this implies that $\ x^2 + y^2 = z^2 \$ for any point $(x, y, z) = (t\cos{t}, t\sin{t}, t) \$ on the curve $\gamma$.

So this tells you that the curve $\gamma$ is contained in the surface of equation $x^2 + y^2 = z^2$, which is the equation of a cone as already pointed out in the comments.

Finally since you needed to draw the curve in SAGE, the following is a plot of the curve and the cone. You can see how the curve wraps around the cone.

In case you want it, the SAGE code I used is the following.

var('x, y, z, t')

implicit_plot3d(x^2 + y^2 - z^2, (x, -15, 15), (y, -15, 15), (z, -15, 15), opacity = 0.5, color = 'blue', aspect_ratio = 1) + parametric_plot3d([t*cos(t), t*sin(t), t], (t, -5*pi, 5*pi), aspect_ratio = 1, thickness = 5, color = 'red')

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