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Equation for a Curved Line

Date: 01/19/97 at 23:46:08
From: Aaron Hardinger
Subject: Formula for a curved line


I would like to ask about a problem concerning curved lines. I would 
like some sort of a formula or process that would allow me to enter 
a starting point and a starting slope and then an ending point and an 
ending slope. I would like to make it so that the starting slope 
changes in a gradual manner to the ending slope and in the process 
make sure that the ending point of the curve is in a specific 
position, the position supplied as the ending point. This curve would 
in essence start at the begining point with the beginning slope. Then 
the slope would curve and also the points would translate to produce a 
curved 'line'. Then when it was done the curve end would be at an 
appointed ending point with the slope of an appointed ending slope. 

Thank you,
Aaron Hardinger

Date: 01/21/97 at 11:08:00
From: Doctor Wilkinson
Subject: Re: Formula for a curved line

Good question!  In general there will be a lot of curves that will 
satisfy these conditions, but probably the simplest will be a cubic 

 ax^3 + bx^2 + cx + d

This has four coefficients, and you have four conditions that you can 
use to make equations for a, b, c, and d. For example, suppose you 
want the curve to go through the points (0,0) and (1,1), and you want 
the slope at the first point to be 1 and the slope at the second point 
to be 0. Then you would get the equations:

 d = 0  
 a + b + c + d = 1
 c  = 1
 3a + 2b + c = 0

You can solve this to get: 

 a = -1
 b = 1
 c = 1
 d = 0

Your curve would be:

 y = -x^3 + x^2 + x

-Doctor Wilkinson,  The Math Forum
 Check out our web site!   

Date: 01/23/97 at 21:16:43
From: Aaron Hardinger
Subject: Re: Formula for a curved line


Thank you for your help, but one more question.  I do not know 
calculus yet and need help with how you derived those formulas such 
as: d = 0, a + b + c + d = 1, c = 1, and 3a + 2b + c = 0.

How did you derive the formulas for a, b, c, and d based on the points 
that were given and those slopes?

Thank you very much,
Aaron Hardinger

Date: 01/24/97 at 11:54:18
From: Doctor Wilkinson
Subject: Re: Formula for a curved line

The first two equations I got by substituting into the assumed 

y = ax^3 + bx^2 + cx + d

The first point was (0,0) which gives:

0 = a * 0^3 + b * 0^2 + c * 0 + d


d = 0

The second point was (1,1) which gives:

1 = a * 1^3 + b * 1^2 + c * 1 + d


a + b + c + d = 1

The other two equations come from substituting for the slopes. For a 
curved line, slope is a concept from calculus, so it's a little hard 
to explain without any calculus. It turns out the the formula for the 
slope of:

y = ax^3 + bx^2 + cx + d 


3ax^2 + 2bx + c

I substitued 0 and 1 into this and set the results equal to the given

-Doctor Wilkinson,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Calculus

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