# 3D parametric equations with polar coordinates

I'm currently studying for my calc 2 midterm and came across this and it completely lost me. I'm not even completely sure where to begin with it. Any ideas?

Put $\langle x[r,t],y[r,t],z[r,t] \rangle = \langle 1,0,1 \rangle + \langle r \cos[t],r \sin[t],0\rangle$. Describe what you get when you plot $\langle x[r,t],y[r,t],z[r,t] \rangle$ for $0\leq r \leq 2$ and $0\leq t \leq 2\pi$ .
Is this a curve or a surface?

Put $\langle x[t],y[t],z[t] \rangle = \langle 1,0,1 \rangle + \langle 2 \cos[t],2 \sin[t],0 \rangle$. Describe what you get when you plot $\langle x[t],y[t],z[t] \rangle$ for $0\leq t \leq 2\pi$.
Is this a curve or a surface? What relation does it have to what you said immediately above?

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Have you tried plotting them to begin with? In general, an $n$-dimensional object represented parametrically requires $n$ parameters to specify. Since a curve is $1$-dimensional and a surface is $2$-dimensional, then... –  Guess who it is. Nov 22 '11 at 3:17

The first of these two objects is a parametric surface. Something described with $2$ parameters will usually give you a surface.

You have ${\bf X}(r,t) = \langle x(r,t),y(r,t),z(r,t) \rangle = \langle 1,0,1 \rangle + \langle r \cos(t),r \sin(t), 0\rangle$ where $0\leq r \leq 2$ and $0\leq t \leq 2\pi$

Notice that the first coordinate is just "$x$" in polar coordinates plus $1$, the second is "$y$" in polar coordinates (both with radius $r$ and angle $t$). The third coordinate says "$z=1$". So the graph is just a piece of the plane $z=1$ where $x,y$ range over the disk of radius $2$ centered at $(1,0)$.

The next object must be a curve since we only have a single parameter.

You have ${\bf X}(t) = \langle x(t),y(t),z(t) \rangle = \langle 1,0,1 \rangle + \langle 2 \cos(t),2 \sin(t),0 \rangle$ where $0\leq t \leq 2\pi$.

Notice that $x(t)=1+2\cos(t)$, $y(t)=2\sin(t)$, and $z=1$. Thus $(x-1)^2+y^2=2^2$ and $z=1$ ($t$ is taking the place of the angle again). So this is the circle of radius $2$ centered at $(1,0,1)$ which is parallel to the $xy$-plane.

So in the end, the first object is a disk and the second object is the circular boundary of that disk.

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