# Calculus word problem

1. Feb 22, 2014

### physics=world

1.A line segment extends horizontally to the left from the point (-1, -1). Another line segment extends horizontally to the right from the point (1, 1). Find a curve of the form
y = ax^5 + bx^3 + x
that connects the points (-1, -1) and (1, 1) so that the slope and curvature of the curve are zero at the endpoints.

2. This is from chapter 12 of calculus in my book. This chapter is about vector valued function.

3. I tried plugging in values for the constant a, b, and c.

Which I tried a = -1, b = 1, c = 1.

and this works. It give an equation that connects the endpoints,

but it does not give me zero for the curvature of the curve.

Here is my work:

the formula for curvature that I used:

K = |y"| / [1 + (y')^(2)]^(3/2)

y' = -5x^(4) + 3x^(2) + 1
y" = -20x^(3) + 6x

If I plug in x = 1 (because of the point (1,1))

I get:

y' = -1
y" = 14

Plugging in those values into the formula gives me 14.

2. Feb 22, 2014

### SammyS

Staff Emeritus
Many combinations of a & b will make the graph of y vs. x pass through those two points.

All that's required is that a + b + 1 = 1 and -a - b - 1 = -1 .

Don't choose values for a & b until you look at requirements for y' and y'' .

Oh, I see that you left c out of
y = ax5 + bx3 + x​

I assume that should be y = ax5 + bx3 + cx

Therefore, to get y = x at x= 1, -1 , you need

a + b + c = 1

and

-a - b - c = -1

Then conditions on y' and y'' should nail down a, b, c .

Last edited: Feb 22, 2014
3. Feb 22, 2014

### physics=world

I'm stuck at what to do next. I been trying a lot of things like trying to get values for a, b, c and setting one thing to another. I need some advice on what to do next.

4. Feb 22, 2014

### SammyS

Staff Emeritus

I Edited it shortly after posting it.

5. Feb 22, 2014

### physics=world

If the equation is of the form

y = ax^5 + bx^3 + cx

How would I put conditions on the equation so that the derivative of the second order produces a zero?

6. Feb 22, 2014

### SammyS

Staff Emeritus
a, b, and c, are constants.

What is y' for that function?

What is y'' for that function?

7. Feb 22, 2014

### physics=world

y' = 5ax^(4) + 3bx^(2) + c

y" = 20ax^(3) + 6bx

8. Feb 22, 2014

### SammyS

Staff Emeritus
OK.

Now set y'' to 0 for x = 1 and/or x = -1 . Won't that answer your question regarding the second order being zero?

9. Feb 22, 2014

### physics=world

If I set y" to zero and have x = 1 it produces

20a + 6b = 0

10. Feb 22, 2014

### SammyS

Staff Emeritus
That's a start.

Now continue using what is required of y' and y . - - both at x = 1 . (x = -1 gives the same results.)

11. Feb 22, 2014

### physics=world

okay now at x = 1

y = a + b + c
y' = 5a + +3b + c

12. Feb 22, 2014

### Ray Vickson

In order to have y = -1, y' = 0 and y" = 0 at x = -1, the polynomial y = y(x) must have the form $y = -1 + a(x+1)^3 + b(x+1)^4 + c(x+1)^5$ (because terms in $x+1$ and $(x+1)^2$ do not have both derivatives = 0 at x = -1). In order to have y = 1, y' = 0 and y'' = 0 at x = +1, y must have the form $y = 1 + a'(x-1)^3 + b'(x-1)^4 + c'(x-1)^5$. If we set x = z+1 (so z = x-1) in the first form, we can expand it and get it in terms of z alone; the second form is already in terms of z alone. So, the two forms must match, which really means that when we express the first form in terms of z the constant must = +1 and the coefficients of $z$ and $z^2$ must vanish. Can you see the consequences of that?

13. Feb 22, 2014

### SammyS

Staff Emeritus
Sure, and what must be the value of y' at x = 1 ?