# Thread: Sketching vector-valued functions

1. ## Sketching vector-valued functions

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).

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

3. 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.

4. 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.)