# Homework Help: Calculus 3 curvature question

1. Feb 16, 2012

### bdh2991

1. The problem statement, all variables and given/known data
A line segment extends horizontally to the left from (-1,-1) and another line extends horizontally to the right from the point (1,1) find a curve of the form y=ax^5+bx^3+cx that connects the two endpoints so that the curvature and slope are zero at the endpoints

2. Relevant equations

3. The attempt at a solution
Not sure even where to start

2. Feb 16, 2012

### Dick

If y=ax^5+bx^3+cx is going to connect the two endpoints, then if you substitute 1 for x what should you get for y? You should get an equation that has to be satisfied by a, b and c. You want to get enough equations in those three unknowns to determine them. The slope is y'. Use that. What's an expression for the curvature? Use that too.

3. Mar 8, 2012

### ozzyglez112

y = ax^5 + bx^3 + cx

You take the first derivative which will give you the slope of the tangent line
Y' = 5ax^4 +3bx^2 +c

You know that Y' = 0

Take the second derivative which relates to curvature
Y'' = 20ax^3 + 6bx

You should know Y'' = 0

If you consider the point (1,1) your x = 1 and your y = 1
You can now set up three equations
a+b+c = 1
5a + 3b + c = 0
20a +6b = 0

if you solve these equations for a, b, and c
you get
a = 3/8
b = -5/4
c = 15/8

so the equation is
y = 3/8 x^5 - 5/4 x^3 + 15/8 x