roller coaster derivatives: design so transitions are smooth

[sam1238]

Suppose you are asked to design the first ascent and drop for a new roller coaster. By studying photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and the slope of the drop -1.6. You decide to connect these two straight streches y = L[sub:3sgz305d]1[/sub:3sgz305d](x) and y = L[sub:3sgz305d]2[/sub:3sgz305d](x) with part of a parabola y = f(x) = ax[sup:3sgz305d]2[/sup:3sgz305d] + bx + c, where x and f(x) are measured in feet. For the track to be smooth there can't be abrupt changes in direction, so you want the linear segments L[sub:3sgz305d]1[/sub:3sgz305d] and L[sub:3sgz305d]2[/sub:3sgz305d] to be tangent to the parabola at the transition points P (from L[sub:3sgz305d]1[/sub:3sgz305d] to the parabola) and Q (from the parabola to L[sub:3sgz305d]2[/sub:3sgz305d]). To simplify the equations you decide to place the origin at P.

1-a) Suppose the horizontal distance between P and Q is 100 feet. Write equations in a, b, and c that will ensure that the track is smooth at the transition points.

1-b) Solve the equations in part (a) for a, b, and c to find a formula for f(x).

1-c) Plot L[sub:3sgz305d]1[/sub:3sgz305d], f, and L[sub:3sgz305d]2[/sub:3sgz305d] to verify graphically that the transitions are smooth.

1-d) Find the difference in elevation between P and Q.

The solution in Problem 1 might look smooth, but it might not feel smooth because the piecewise defined function (consisting of L[sub:3sgz305d]1[/sub:3sgz305d](x) for x < 0, f(x) for 0 < x < 100, and L[sub:3sgz305d]2[/sub:3sgz305d](x) for x > 100) doesn't have a continuous second derivative. So you decide to improve the design by using a quadratic function q(x) = ax[sup:3sgz305d]2[/sup:3sgz305d] + bx + c only on the interval 10 < x < 90 and connecting it to the linear functions by means of two cubic functions:

g(x) = kx[sup:3sgz305d]3[/sup:3sgz305d] + lx[sup:3sgz305d]2[/sup:3sgz305d] + mx + n for 0 < x < 10

h(x) = px[sup:3sgz305d]3[/sup:3sgz305d] + qx[sup:3sgz305d]2[/sup:3sgz305d] + rx + s for 90 < x < 100

2-a) Write a system of equations in 11 unknowns that ensure that the functions and their first two derivatives agree at the transition points.

2-b) Solve the equations in part (a) with a computer algebra system to find formulas for q(x), g(x), and h(x).

2-c) Plot L[sub:3sgz305d]1[/sub:3sgz305d], g(x), q(x), h(x), and L[sub:3sgz305d]2[/sub:3sgz305d], and compare with the plot in Problem (1-c).

 

[galactus]

Re: roller coaster derivatives

From your diagram and the suggestion in the problem, place the origin at point, (0,0), P. That makes it easier to work with.

Using the derivative of the parabola, we can find the use this and the given slopes:

I am going to use fractions instead of decimals.

\(\displaystyle y'=2ax+b\)

The slope in: \(\displaystyle \frac{4}{5}=2a(0)+b\)

\(\displaystyle b=\frac{4}{5}\)

Slope out: \(\displaystyle 200a+b=-1.6\)

Since we know b=4/5, then \(\displaystyle a=\frac{-3}{250}\)

From the equations of the parabola, we have \(\displaystyle y=ax^{2}+bx+c\)

For point Q, \(\displaystyle -40=\frac{-3}{250}(100)^{2}+\frac{4}{5}(100)+c\)

For point P, \(\displaystyle 0=\frac{-3}{250}(0)^{2}+\frac{4}{5}(0)+c\)

For both of these, we get c=0.

So, the equation of the parabola is \(\displaystyle \boxed{y=\frac{-3}{250}x^{2}+\frac{4}{5}x}\)

You can find the high point(vertex) of the parabola by the standard methods if need be.

Now, I wanted to get you a good start because the part with the cubic is more complicated.

 

[stapel]

For those who can't view the image or who don't want to have to juggle two screens to attempt to assist, the text of the exercise follows:

Suppose you are asked to design the first ascent and drop for a new roller coaster. By studying photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and the slope of the drop -1.6. You decide to connect these two straight streches y = L[sub:27nhvaw1]1[/sub:27nhvaw1](x) and y = L[sub:27nhvaw1]2[/sub:27nhvaw1](x) with part of a parabola y = f(x) = ax[sup:27nhvaw1]2[/sup:27nhvaw1] + bx + c, where x and f(x) are measured in feet. For the track to be smooth there can't be abrupt changes in direction, so you want the linear segments L[sub:27nhvaw1]1[/sub:27nhvaw1] and L[sub:27nhvaw1]2[/sub:27nhvaw1] to be tangent to the parabola at the transition points P (from L[sub:27nhvaw1]1[/sub:27nhvaw1] to the parabola) and Q (from the parabola to L[sub:27nhvaw1]2[/sub:27nhvaw1]). To simplify the equations you decide to place the origin at P.

1-a) Suppose the horizontal distance between P and Q is 100 feet. Write equations in a, b, and c that will ensure that the track is smooth at the transition points.

1-b) Solve the equations in part (a) for a, b, and c to find a formula for f(x).

1-c) Plot L[sub:27nhvaw1]1[/sub:27nhvaw1], f, and L[sub:27nhvaw1]2[/sub:27nhvaw1] to verify graphically that the transitions are smooth.

1-d) Find the difference in elevation between P and Q.

The solution in Problem 1 might look smooth, but it might not feel smooth because the piecewise defined function (consisting of L[sub:27nhvaw1]1[/sub:27nhvaw1](x) for x < 0, f(x) for 0 < x < 100, and L[sub:27nhvaw1]2[/sub:27nhvaw1](x) for x > 100) doesn't have a continuous second derivative. So you decide to improve the design by using a quadratic function q(x) = ax[sup:27nhvaw1]2[/sup:27nhvaw1] + bx + c only on the interval 10 < x < 90 and connecting it to the linear functions by means of two cubic functions:

g(x) = kx[sup:27nhvaw1]3[/sup:27nhvaw1] + lx[sup:27nhvaw1]2[/sup:27nhvaw1] + mx + n for 0 < x < 10

h(x) = px[sup:27nhvaw1]3[/sup:27nhvaw1] + qx[sup:27nhvaw1]2[/sup:27nhvaw1] + rx + s for 90 < x < 100

2-a) Write a system of equations in 11 unknowns that ensure that the functions and their first two derivatives agree at the transition points.

2-b) Solve the equations in part (a) with a computer algebra system to find formulas for q(x), g(x), and h(x).

2-c) Plot L[sub:27nhvaw1]1[/sub:27nhvaw1], g(x), q(x), h(x), and L[sub:27nhvaw1]2[/sub:27nhvaw1], and compare with the plot in Problem (1-c).

 

[ebpunx999]

Can anyone provide me with a little help to the second part of the problem about the cubic please?

 

[Deleted member 4993]

Can anyone provide me with a little help to the second part of the problem about the cubic please?

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

Galactus has provided you with a complete solution of the first part - explaining every step. Follow similar logic and show us YOUR contribution to this project.