Connecting Two Endpoints with Zero Curvature and Slope Using a Cubic Curve

Click For Summary
SUMMARY

The discussion focuses on finding a cubic curve of the form y=ax^5+bx^3+cx that connects the endpoints (-1,-1) and (1,1) with zero curvature and slope at both endpoints. The solution involves setting up three equations based on the conditions of the endpoints and the derivatives of the curve. The values obtained for the coefficients are a = 3/8, b = -5/4, and c = 15/8, resulting in the equation y = 3/8 x^5 - 5/4 x^3 + 15/8 x.

PREREQUISITES
  • Understanding of polynomial functions, specifically cubic curves
  • Knowledge of calculus, including derivatives and curvature
  • Ability to solve systems of equations
  • Familiarity with the concept of endpoints in curve analysis
NEXT STEPS
  • Study polynomial interpolation techniques
  • Learn about curvature and its applications in curve fitting
  • Explore the implications of boundary conditions in calculus
  • Investigate higher-order derivatives and their significance in curve analysis
USEFUL FOR

Mathematicians, engineering students, and anyone involved in curve modeling or optimization in calculus.

bdh2991
Messages
102
Reaction score
0

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


Homework Equations





The Attempt at a Solution


Not sure even where to start
 
Physics news on Phys.org
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
 

Similar threads

Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K