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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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2 Answers 2

up vote 3 down vote accepted

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 $\sin(5t)$.

enter image description here

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

enter image description here

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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1  
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

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