Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.