# Paramatize a non-linear curve between points

1. Dec 5, 2006

### Tom McCurdy

I was wondering how you could paramatize a non-linear curve between two points (1,0) and (a,b)

I have been trying to do like a parabolic paramatization but I am getting nowhere

I can get the first point alright

x=x
y=1-x^2
1<x<a
but then you get y=1-x^2

Any suggestions

2. Dec 5, 2006

### emi

do you now this?
x*lnx=9
x=?

3. Dec 5, 2006

### arildno

EDIT:
To get a parabola going through (1,0) and (a,b), just set:
$$y=Ax^{2}+Bx+C$$
and determine A,BC by ther equations:
$$0=A+B+C, b=Aa^{2}+Ba+C$$

You'll get many parabolas satisfying your demands.

Last edited: Dec 5, 2006
4. Dec 5, 2006

### emi

but only i need is a answer of that equation, how do you get that ''x'' out of there?

5. Dec 6, 2006

### HallsofIvy

Staff Emeritus
There is only one straight line between any two points but there are an infinite number of non-linear curves between any two points. If you want a parabolic curve between (1, 0) and (a,b), try this: x= t, y= u+ vx+ wt^2. In order that that pass through (1, 0) we must have x= 1 (so t= a), y= =u+ v+ w= 0. In order that it pass through (a, b) we must have x= a, (so t= a), y= u+ va+ va^2= b. That gives two equations for 3 unknown parameters so there are still an infinite number of such parabolic curves through those 2points. Select 1 by take u, say, equal to 0. Then v+ w= 0 and av+ a^2w= b. Multiply the first equation by a to get av+ aw= 0 and subtract: (a^2- a)w= b so w= b/(a^2- a), v= -1 = -b/(a^2-a). x= t, y= (b/(a^2-a)(t^2- t) works(as long as a is not 1 or 0). If a= 1 or 0, then a parabola with vertical axis will not pass through both points but some other parabola will and, of course, there are still an infinite number of other curves, each with an infinite number of possible parametrizations, that pass through (1, 0) and (a,b).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook