roller coaster design

Question

Roller Coaster Design

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. 

Answer 1

Anita:

Use the hint that point P should be placed at the origin. In this case, x_p = 0 and y_p =0

We want the parabola-shaped track to connect smoothly with both linear tracks. In other words, the slope of the parabola must equal the slope of the linear tracks where they both connect.

We know f(x) = a*x²+b*x+c

Since x_p, a point on the parabola, is placed at the origin then f(0) = 0 or c=0 Already we can simplify!

f(x) = a*x² + b* x

Next, we know that the slope of a line equals the derivative of the function at a specific point.

f(x) = a*x² + b* x so f '(x) = 2*a*x + b.

At the origin, f '(0) = 0.7 so 2*a*0+b=0.7 so b=0.7

Point Q has x-coordinate equal to 40 according to the problem text.

f ' (40) = -1.5 so 2*a*(40) + b = 2*a*(40) + 0.7 = 80*a + 0.7 =-1.5

a = -0.0275 solves the equation above.

So our parabola is represented by f(x) = -0.0275*x² + 0.7*x

Remember, point P has y-coordinate equal to 0. The y-coordinate of point Q is f(40).

So how much higher is point P than Q?

Answer 2

quadratic function y =ax^2+bx+c

(0,0)

(40,0)

y'(0) = 0.7

y'(40) = -1.5

four conditions, 3 coefficients