Roller Coaster Design

Question

Imagine you are a roller coaster designer, and are asked to mathematically represent the curve of the first drop of a new ride. By looking at some of your favorite coasters, you determine that the optimal slope for the ascent, when the car is being pulled up the slope by a chain, is 0.7. The slope of the first drop that will be the most thrilling (without being dangerous) is –1.5. You decide to connect these two straight stretches, y = L1 and y = L2, with a parabola of the form y = f(x) = ax^2 + bx + c, where x and f(x) are measured in meters. In order for the curve of the track to remain smooth, you must make sure that the linear segments, L1 and L2, have the same slope at each endpoint of the parabola, points P and Q. This ensures that the parabola is differentiable, as well as continuous at the endpoints. Choose point P, the left endpoint of your parabola, as your origin, to simplify calculations.

a) Draw a rough sketch of L1 with slope 0.7, ending at point P. Then sketch L2 with slope –1.5, starting at point Q a short distance to the right and below point P. Finally, sketch the parabola that would connect P and Q while matching the slopes of L1 and L2. 

b) If the horizontal distance between P and Q is 40 meters, write equations in a, b, and c that make sure the track is smooth at the connections P and Q.

c) Solve the equations for a, b, and c to find the equation of the parabola that connects P and Q.

d) Using your calculator or a computer, plot the parabola f(x). Compare this to your original sketch in part a) by noting any similarities or differences.

e) How much higher is point P than point Q?

f) Find the maximum height of the coaster algebraically. Verify your answer by graphing f(x) on your calculator. 

Can anybody help with work??

Answer 1

Hi Aiden,

I will give you some hints to help you get started.

The key to the problem is realizing that the derivative for the parabola is: y' = 2ax + b. When x = 0 (point P), y' = .7 (the same as the slope of L1. That gives equation .7 = 2a(0) + b or b = .7.

At point Q where x = 40, y' = -1.5. Which results in the equation -1.5 = 2a(40) + .7. Solve for a and we are getting close to having the equation for the parabola. How to find c? Assuming point P at the origin means when x=0, y=0. Substitute those values for x and y in the equation for the parabola. That results in c = 0.

Note that the equation for L1 is y = .7x. The equation for L2 is? How can we determine the coordinates of point Q? We already know the x coordinate is 40. Find the y value by substituting 40 for x in the equation for the parabola. Now you have enough information to write the equation for L2 using point slope.

Hopefully that is enough information to help you determine the answers to all parts of this question. My suggestion is to use the Desmos online graphing calculator to graph L1, L2, and the parabola restricting the domain to values that make sense. For example L1: y=.7x {-50 < x< 0}. For the parabola restrict the domain to {0<x<40}. Similarly for L2.