How do you graph the surface $\vec{r}(u,v) = \langle e^{u}\cos v, e^{u}\sin v, e^{u}\rangle, 1 \leq u \leq 2, 0 \leq v \leq 2\pi$?

Question

Graph the surface $\vec{r}(u,v) = \langle e^{u}\cos v, e^{u}\sin v, e^{u}\rangle, 1 \leq u \leq 2, 0 \leq v \leq 2\pi$

How do I graph this? Is there an easy way to do it? Like zeroing out one of the variables, and see x and y only and how that looks like and etc.

In my notes, used $x^2 + y^2$ but why?

Could someone show how to graph this without using graphing devices? How's this a cone?

Answer 1

HINT: $$ (e^u \cos v)^2 + (e^u \sin v)^2 = (e^u)^2 $$ in other words, any point $(x,y,z)$ on this surface satisfies: $$ x^2 + y^2 = z^2. $$

Answer 2

Think of cylindrical coordinates, $(r, \theta, z)$, where $$ \left\{ \begin{align} x &= r \cos \theta \\ y &= r \sin \theta \\ \end{align} \right. $$

You're parametrized surface has $r = z = e^u$ and $\theta = v$. In the $(r, z)$-plane (say when $\theta = 0$), you have the parametrized line segment $r = z$ for $r \in [e, e^2]$. This segment is now "spun" around the $z$-axis, since $\theta$ can vary. What you get is a truncated cone.

Here's a picture:

Answer 3

Break it into

$$\langle e^u \cos(v), e^u \sin(v), 0 \rangle + \langle 0, 0, e^u \rangle.$$

The first part parametrizes an annulus with inner $e$ and outer radius $e^2$. The second part increases the $z$ coordinate proportionally to radius of the annulus. Thus, you get part of a cone.

Here's the plot with Mathematica:

ParametricPlot3D[E^u {Cos[v], Sin[v], 1}, {u, 1, 2}, {v, 0, 2 Pi},
  BoundaryStyle -> Black]