Connecting Points with a Curved Equation: Solving a Calculus Word Problem

[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.

I need some help/advice.

 

[SammyS]

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.

I need some help/advice.

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'' .

Added in Edit:

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 .

 

[physics=world]

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 .

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.

 

[SammyS]

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.

See my above reply.

I Edited it shortly after posting it.

 

[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?

 

[SammyS]

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?

a, b, and c, are constants.

What is y' for that function?

What is y'' for that function?

 

[physics=world]

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

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

 

[SammyS]

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?

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

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

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?

 

[physics=world]

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

20a + 6b = 0

 

[SammyS]

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

20a + 6b = 0

That's a start.

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

 

[physics=world]

okay now at x = 1

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

 

[Ray Vickson]

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.

I need some help/advice.

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?

 

[SammyS]

okay now at x = 1

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

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