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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

Hello,

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
polynomial:

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

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

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

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

Hello,

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
equation:

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

or

d = 0

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

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

or

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

is

3ax^2 + 2bx + c

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

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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