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Let's say I have the following vector function:

$\mathbf{r}(t) = t \cos t\,\mathbf{i} + t\,\mathbf{j} + t \sin t\,\mathbf{k}$

What properties of this function will allow me to sketch a curve drawn by this function?

I know that:

$x = t\cos t$

$y = t$

$z = t \sin t$

What is a general approach that I can take to solve problems that want me to sketch curves drawn by vector functions?

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Drawing a graph in 3-space is usually much more trouble than it's worth, but if you're determined, I'd start be looking at the projections of your curve onto the $xy$-, $yz$- and $xz$- planes. –  Avi Steiner Feb 2 '13 at 19:20
    
Did the answer below resolve your question? Do you know about upvoting and/or accepting answers? –  JohnD Mar 9 '13 at 6:11

1 Answer 1

Trace out the space curve given by the parametrization $x=t\cos t$, $y=t$, $z=t\sin t$, for example, as $t$ varies over $[0,2\pi]$:

enter image description here

The vector-valued function $\mathbf{r}(t)=\langle t\cos t,t,t\sin t\rangle$ would be the function that at time $t$ outputs the vector $\langle t\cos t,t,t\sin t\rangle$, so at time $t$, its tip would be at the point $(t\cos t,t,t\sin t)$ on the space curve shown.

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Rotating the graphic in Mathematica, the tail of that vector is indeed at the origin. Must be a bit of an optical illusion here based on the 3D viewpoint. –  JohnD Feb 2 '13 at 19:59

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