Sketching vector-valued functions

**SyNtHeSiS** · February 21st 2011, 12:40 AM

Sketch $r(t) = t^2i + t^4j + t^6k$

Attempt:

I tried to find a surface that this curve lies on so I said:

$x = t^2$ therefore:

$y = x^2$ and $z = x^4$

I then drew these surfaces, then tried to see what happens when t = large positive number and then a large negative number. I got that the graph lies in the first octant, but I am not sure where on the surfaces $y = x^2$ and $z = x^4$ the curve lies on (I learnt that you use these surfaces as a guide to draw the curve).

**emakarov** · February 21st 2011, 01:49 AM

**SyNtHeSiS** · February 21st 2011, 02:36 AM

How do you know that the curve, curves a bit (the part which looks like a little turning point) near the origin?

**emakarov** · February 21st 2011, 02:58 AM

Well, it has to produce the projections drawn in dashed lines. Actually, this graph curves everywhere, the curvature is just higher near the origin because z does not grow as fast yet.

**HallsofIvy** · February 21st 2011, 03:53 AM

If $\vec{r}(t)= t^2\vec{i}+ t^4\vec{j}+ t^6\vec{k}$ , then
$x= t^2$ , $y= t^4$ , and $z= t^6$ .

Those certainly satisfy $xy= t^2t^4= t^6= z$ for all t.

That is, this curve lies on the surface xy= z.

(Note that any curve lies on an infinite number of surfaces.)

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