# Parametric equation of a basic cubic function

1. Sep 1, 2014

### pondzo

1. The problem statement, all variables and given/known data

Find a parametric form for the part-cubic curve with equation y = x3, 0 ≤ y ≤ 8; starting point (2, 8),

3. The attempt at a solution

The question beforehand was the exact same but the starting and ending points reversed. My answer for that was; r(t) = (t, t^3) for t an element of [0,2], ill call this equation A.

So for this question is set up a few linear equations and solved them and came up with the parametric vector function; r(t) = (2-t, 8-t^3), ill call this equation B. B satisfies the starting and ending point conditions but, to me, it looks like A and B are describing different curves. When i graph them, they look different. And im not sure how to interpret it. Is it that my answer is wrong? or is it that when you reverse the order of the start and end points as you do in B, its almost as if you you are looking at the graph upside down and considering (2,8) as the origin (if that makes any sense at all). Thank you for your time.

2. Sep 1, 2014

### HallsofIvy

By "starting point" you mean the point given by y= 8? And by "the starting and ending points reversed" you mean starting at y= 0 and ending at y= 8?

So for that first problem, starting at y= 0 and ending at y= 8, you got $\vec{r}(t)= (t, t^3)$ for t in [0, 2]. Yes, that is correct. Note that if t= 0, you have (0, 0), if t= 1 you have (1, 1), and for t= 2, (2, 8).

If you change that to $\vec{r}(t)= (2- t, 8- t^3)$ then it certainly is true that for t= 0 you get (2, 8), for t= 2 you get (0, 0) but for t= 1 you get (1, 7) not (1, 1). No, that is not the same:

In the first case you are using $\vec{r}= (t, t^3)$ and in the second case you are using $(2- t, 8- t^3)= (2, 8)- \vec{r}(t)$. No, that is NOT the same curve.

Instead just reverse t: replace t, rather than x, with 2- t: instead of $(t, t^3)$, use $(2- t, (2- t)^3)$.

3. Sep 1, 2014

### pondzo

Thats exactly how the question was stated in the tutorial, but you got the right interpretation. Thanks for the reply that makes sense now.

If you dont mind, I have another question on parametric equations for polynomials. Could I post it on this thread?