# Stuck on space curves for vector valued functions

I'm working through the James Stewart Calculus text to prep for school. I'm stuck at this particular point.

How would you sketch the graph for the parametric equations: $x = \cos t$, $y = \sin t$, and $z = \sin 5t$? I understand that if it were the case that $z=t$, I'd merely get a helix around the $z$-axis, as $x$ and $y$ form an ellipse. However, I cannot make the leap to solve more exotic problems such as the problem posed or even the case when $z = \ln(t)$.

Some help and a push in the right direction would be appreciated.

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Here is an animation I made that might help. The left is a plot of $(\cos(t),\sin(t),\sin(5t))$ and the right is a plot of $\sin(5t)$.

For the case of $(\cos(t),\sin(t),\ln(t))$, here is the corresponding animation:

As a sanity check, note that in each animation, you can see that the point on the circle makes its first full revolution as $t=2\pi\approx 6.28$.

Mathematica code for my (and anyone else's) future reference:

size = 1.5

slices = 150

Slice[t_,z_] := {Show[ParametricPlot3D[{Cos[2 Pi*s], Sin[2 Pi*s], z},
{s, 0, 1}, PlotRange -> {{-size, size}, {-size, size}, {-size, size}}],
Graphics3D[{PointSize[Large], Point[{Cos[t], Sin[t], z}]}]],
Show[Plot[Sin[5 s], {s, 0, 2 Pi}, Ticks -> {{0, 2 Pi/5, 4 Pi/5, 6 Pi/5,
8 Pi/5, 2 Pi}}], Graphics[{PointSize[Large], Point[{t, Sin[5 t]}]}]]}

NewSlice[t_,z_] := {Show[ParametricPlot3D[{Cos[2 Pi*s], Sin[2 Pi*s], z},
{s, 0, 1}, PlotRange -> {{-size, size}, {-size, size}, {-2, 2}}],
Graphics3D[{PointSize[Large], Point[{Cos[t], Sin[t], z}]}]],
Show[Plot[Log[s], {s, 0.5, 8}, PlotRange -> {{0, 8}, {-1, 2}},
AspectRatio -> 1/2], Graphics[{PointSize[Large], Point[{t, Log[t]}]}]]}

Export["sin.gif", Table[Slice[2 Pi*t/slices, Sin[5*2 Pi*t/slices]], {t,
0,slices}], "DisplayDurations" -> 0.15]

Export["ln.gif",Table[NewSlice[t, Log[t]], {t, 0.5, 7.5, 7/slices}],
"DisplayDurations" -> 0.15]

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Nice images! (+1) – robjohn Jul 7 '12 at 17:20
Thanks! I recently learned how to do animations in Mathematica so I figured I'd give it a shot. – Zev Chonoles Jul 7 '12 at 17:25
Nice indeed. :) I'll just add the tiny note that you can use GraphicsGrid[]. so that your two frames are bound as an image instead of as a list (which explains) the curly braces present in the animations. – J. M. Jul 8 '12 at 1:59
@J.M.: and the comma in between :-) – robjohn Jul 8 '12 at 7:13
@ZevChonoles: If you're interested, the code for one of my animations is here. – robjohn Jul 8 '12 at 7:16

Hint: Note that the $x$ and $y$ coordinates trace out a circle. As they do, the $z$ coordinate goes through $5$ sinusoidal cycles.

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Thanks. The other question $z=ln(t)$, could you let me know why it makes that asymmetrical "corkscrew" graph in that case? – arkate Jul 7 '12 at 16:42
The $x$ and $y$ coordinates still trace out a circle. As $t$ goes from $0$ to $1$, the $z$ coordinate goes from $-\infty$ to $1$, then as $t$ goes from $1$ to $\infty$, $z$ slows down but goes to $\infty$, just like the curve for $\log$. – robjohn Jul 7 '12 at 17:19