# Working backwards

1. Aug 31, 2006

### programmer

I've taken Calculus 1, but it was a few years ago, so bear with me. I understand how to use derivitaves to find critical numbers, relative max's and min's, points of inflections, incresing, decreasing. all that good stuff.

my question is, if I have two points I want a x^3 line pass through, how can i accomplish this?

2. Aug 31, 2006

### VietDao29

y = x3 is a curve, not a line.
I don't really get what you mean. y = x3 is a specific curve, i.e you cannot change the set of points that it passes through.
If you want to get 2 points on the curve, just choose 2 arbitrary x1, and x2 values, then from there find the corresponding y1, and y2. And you'll have 2 points that the curve passes through.

3. Aug 31, 2006

### programmer

no, i want to MODIFY a x^3 equation to MAKE it pass through the points i already have set.

4. Aug 31, 2006

### programmer

the first point is 0,0, and the 2nd point could be anything, how to i alter the curve to make it pass through my 2nd point?

5. Aug 31, 2006

### chroot

Staff Emeritus
You're trying to curve-fit a cubic to some known data set.

Look up cubic spline interpolation. Numerical Recipes in C has a section on it, with the algorithm coded in C.

http://www.library.cornell.edu/nr/cbookcpdf.html [Broken]

If you only have two points, one of which is (0, 0) and the other is (x0, y0), just solve this equation for a:

$a x_0^3 = y_0$

- Warren

Last edited by a moderator: May 2, 2017
6. Sep 3, 2006

### HallsofIvy

Staff Emeritus
There are many different ways to do this. For example, as chroot said, if you want a formula of the form y= ax3, which necessarily passes through (0,0) for all a, just select a so that y0= ax3: That is $$y= \left(\frac{y_0}{x_0^3}\right)x^3$$ passes through (0,0) and (x0,y0).

Or, you could alter y= x3 to look like y= x3+ ax. In order to have y= y0 when x= x0 we must have $y_0= x_0^3+ ax_0$ or, solving for a, $a= \frac{y_0- x_0^3}{x_0}$. That is, the graph of $y= x^3+ \frac{y_0- x_0^3}{x_0}x$ passes through (0,0) and (x0,y0). There are many other possiblities. The choice is essentially arbitrary unless you have other conditions to fulfill.

Last edited: Sep 8, 2006
7. Sep 7, 2006

### programmer

well....i guess I've forgotten a lot of calculus...what does the I in those equations stand for? Integral?

8. Sep 8, 2006

### HallsofIvy

Staff Emeritus
?? What I are you talking about? Which response does this relate to?

9. Sep 8, 2006

### programmer

nevermind...the X's in those equations look like I's...

that's retarded

10. Sep 8, 2006

### chroot

Staff Emeritus
Mm... that's how all x's are typeset in every math book I've ever seen.

- Warren