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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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