# Calculus 3 curvature question

## Homework Statement

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

## The Attempt at a Solution

Not sure even where to start

## Answers and Replies

Dick
Science Advisor
Homework Helper
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.

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