# Thread: Looking for an equation of a curve...

1. ## Looking for an equation of a curve...

Hello there - bit of an unusual question and I'm not quite sure if it belongs in this Calculus section or not. It sorta involves derivatives so here it goes:

I have been looking for monotonic equations that pass through (0,0) and (1,1) with various slopes at x=0 and x=1. For example, I already have:

CLASS 1,2: f(x)=x^n
when n>1: concave up, f'(0) = 0, f'(1) = n
when 0<n<1: concave down, f'(0) = inf, f'(1) = n

CLASS 3,4: f(x) = 1-(1-x)^n
when n>1: concave down, f'(0) = n, f'(1) = 0
when 0<n<1: concave up, f'(0) = n, f'(1) = inf

CLASS 5,6: f(x) = (1-(1-x)^n)^(1/n)
when n>1: concave down, f'(0) = inf, f'(1) = 0
when 0<n<1: concave up, f'(0) = 0, f'(1) = inf

What I am looking for are the equations that will have:
f'(0) = m
f'(1) = n
Thus, 2 different finite slopes at x=0 and x=1 allowing both concave up and down curves

Thank you for any help. I have asked math professors at my college but still no answer. Sorry for the long post...
Terry

2. ## Re: Looking for an equation of a curve...

Partial solution: construct 2 functions g(x) and h(x), to satisfy:

$g(0)=0, g'(0)=m$
$h(1)=1, h'(1)=n+2g(1)-2$ (looks hard to do but isn't, the RHS is just a number)

Now, consider:
$f(x)=x^2h(x) + (1-x^2)g(x)$

$f(0)=0 + g(0)=0$ as required
$f(1)=h(1) + 0=1$ as required

$f'(x)=2xh(x) + x^2h'(x) -2xg(x) +(1-x^2)g'(x)$

$f'(0)=0+0+0+ g'(0)=m$ as required

$f'(1)=2h(1) +h'(1) -2g(1) = 2 + h'(1) -2g(1)$
$= 2 + [n +2g(1)-2] -2g(1) = n$ as required

i cant guarantee f(x) will be monotonic though

3. ## Re: Looking for an equation of a curve...

Thanks for the help! With the above hint as a starting point, I was able to figure out the following general equation for curves with slope 1/n at (0,0) and slope n at (1,1):

y= (x^n^(1/n)) * (x^n) + (1-x^n^(1/n)) * (1-(1-x)^(1/n))

By changing the 'coefficients' from x^2 and (1-x^2) to (x^n^(1/n)) and (1-x^n^(1/n)), I found that the integrals of the curve on both sides of the line y-x+1 become equal although, I am still not convinced that the curve is symmetrical over that line...