Project: Designing a Roller Coaster - AP Calculus

Question

Project: Explain the mathematics of roller coaster construction. 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, L 1 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

Given:

line L1 with slope 0.7 connects with the parabola at the origin.

The parabola connects with the line L2 with slope -1.5 at point Q

THe horizontal distance between P and Q is 40

Then line L1 = 0.7x

the parabola y=ax^2 + bx + c

y' = 2ax + b

THe origin is also on the parabola and at x=0 the tangent line has slope 0.7.

Then c=0, so the parabola becomes y = ax^2 + bx, with the same derivative.

So then 0.7 = 2a(0) + b <-- first derivative of the parabola must be 0.7 at x=0

b = 0.7

The updated equation of the parabola is now y = ax^2 + 0.7x

The horizontal distance between P and Q is 40, with Q BELOW P. So point Q must be at ( _x, -40)

LIkewise, the tangent line of the parabola must be -1.5 at point Q.

So 2a(_x) + 0.7 = -1.5

2a(_x) = -2.2

a(_x) = -1.1 <-- label this equation ALPHA

But point Q must also be on the parabola, so a(_x)^2 + 0.7(_x) = -40

a(_x)(_x) + 0.7(_x) = -40

(-1.1)(_x) + 0.7(_x) = -40 <--- substitutes equation ALPHA

-0.4 (_x) = -40

4(_x) = 400 <-- multiplies by -10

_x = 100

per equation ALPHA, a(100) = -1.1

a = -1.1/100 = -0.011

the parabola is then y = -0.011x^2 + 0.7x

Finally, since the parabola is at y=-40 when x=100,

-1.5*100 = -150, so the equation of line L2 is y = -1.5x + 110

The actual distance between P and Q is sqrt( 100^2 + 40^2) = sqrt( 10000 + 1600) = sqrt(11600) =

sqrt( 116*100) = 20*sqrt(29) which is approximately 107.703

Here are the final functions:

Line #1: y = 0.7x for x<=0

Parabola: -0.011x^2 + 0.7x for 0 <=x <= 100

LIne #2: y = -1.5x + 110

The derivative is -0.022x + 0.7. Note that at x=0, the slope is 0.7 and at x=100, the slope is -1.5.

The max height of this parabola occurs at x = -b/2a = -0.7/ (2 * -0.011) = -0.7 / -0.022

= 700/ 22 <--- multiplies by 1000

= 350/11

= 31 and 9/11

The max height is:

-0.011 (350/11)^2 + 0.7* 350/11 =

(-11/1000)*(122500/121) + (7/10)(350/11) = <--- changes to fractions and squares

-122.5/11 + 245/11 = <-- 121 cancels 11; 1000 cancels 122500; 350 cancels 10

122.5/11

11.136363636...

= 11 and (1/10 + 36/990) <-- 0.36363636... = 36/99 , but it is delayed by 1 digit, so 36/990

= 11 and (99 + 36)/990 <--- common denomiantor

= 11 and 135/990

= 11 and 27/198

= 11 and 9/66

= 11 and 3/22

I have uploaded the graph for you in the resources section.

Answer 2

Using Point P (0, 0) and the fact that Point P is on f(x), we can plug (0, 0) into f(x) = ax² + bx + c to get:

0 = a(0)² + b(0) + c to find that c = 0

So now we have f(x) = ax² + bx. Differentiating f(x), we get f '(x) = 2ax + b and using point P which must have a slope that matched line L1 of 0.7 we can say (since f '(0) must equal 0.7) 0.7 = a(0) + b or b = 0.7

So now we have f(x) = ax² + 0.7x and f '(x) = 2ax + 0.7. Now using point Q, and the fact that the x coordinate of point Q must be 40 (since we are given the horizontal distance between P and Q is 40 meters), we can say (because m for L2 is -1.5 and f ' (40) must be -1.5) -1.5 = 2(40) + 0.7 or a = -0.0275

So f(x) = -0.0275x² + 0.7x and point Q is at (40, -16)

Algebraically, the vertex is found at x = -b/2a = -0.7/(-.055) = 140/11 (or 12.727). So, plugging that into the function gives a max height of 49/11 meters (or 4.455 meters) (height above Point P)