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I have a question from one my problems that ask me to graph this space curve using the "appropriate parameter range" to find the true nature of the graph. $$r(t) = (t, \exp(t), \cos(t)).$$

On Maple I did

with(plots):
spacecurve([t, exp(t), cos(t)], t = 0 .. 2*Pi, axes = boxed)

I chose $2\pi$ because that's the domain of $\cos(t)$, but when I extend this from $-2\pi$ to $2\pi$, I get a better view. SO what does it mean to have a "true nature of the curve"?

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    $\cos(t)$ has domain all real numbers, not just $[0,2\pi]$. I think you meant the *period* of $\cos(t)$. But since the first two coordinates of $r(t)$ are not periodic, the fact that the third one is doesn't really matter all that much, except to give an upper and lower bound for the $z$-coordinate on which the curve lies.2012-01-20
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    Right, period not domain sorry. I originally intended to use [-1, 1] instead for that last bit because was the range of cos(t). But my x-coordinate was bothering me since i got something that looks like a parabola2012-01-20
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    the point is that the period of the third coordinate does not really tell you enough, because neither the first nor the second coordinates are periodic. You're going to need more than just a couple of up-and-downs to see what is happening to $r(t)$. Selecting $[0,2\pi]$, $[-2\pi,2\pi]$, or $[-1,1]$ does not give you something very representative, because none of those intervals are representative for your *second* coordinate.2012-01-20
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    Without any more context, the thing that makes the most sense is that they want you to give a unit-speed paramaterization of the curve. This allows you to glean the most information from the curve in the sense that it's curvature, torsion, and so it's Frenet frame can be easily computed.2012-01-20
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    With t=-100..100, I got a cosinudal and a crooked (due to the nature of x = t I guess) exponential curve.2012-01-20

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About the true nature of the graph: There are two totally different regimes for $t\ll-1$ and $t\gg1$, and there is a zone of transition in between, where nothing spectacular happens. For $t\ll-1$ one has $y(t)=e^t\doteq0$; therefore $$r(t)\doteq(t,0,\cos t)\qquad (t\ll-1)\ ,$$ which is an ordinary cosine curve in the left half of the $(x,z)$-plane, extending to $-\infty$ in the $x$-direction.

For $t\gg1$ it is obvious that $y\gg1$ is the prominent variable; therefore we choose $y$ as new parameter. In this way we obtain the new parametrization $$\tilde r(y)=\bigl(\log y, y, \cos(\log y)\bigr)\qquad(y\gg1)\ .$$ Looking from the point $(\infty,0,0)$ we see in the $(y,z)$-plane the curve $z=\cos(\log y)$ which is an oscillation in $z$-direction becoming ever slower as $y\to\infty$. At the same time the moving point on the space curve $y\mapsto\tilde r(y)$ is also increasing its $x$-coordinate towards $\infty$, but at ever decreasing speed.

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If you think about each coordinate separately, the first is linear, the second is exponential with scale 1, and the third is periodic with period $2\pi$. You would like the range to show these. Without the specification of space curve, I would just graph each one on a 2D plot because they aren't related. I think you could defend $[0,2\pi]$ to $[-2\pi,4\pi]$ or anywhere in between. You could try a few and see what you like.