Question

1) You must design an ascent and drop for a new roller coaster. 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 stretches y Li(x) and y La(x) with part of a parabola y fx)-ax2+ 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 Li and L2 to be tangent to the parabola at the transition points P and Q (See figure below). To simplify the equations, you decide to place the origin at P 6 Suppose that the horizontal distance between P and Q is 100 ft. Write equations in a. a, b, and c that will ensure that the track is smooth at the transition points. b. Solve the equations in part (a) for a, b, and c to find a formula for f(x). c. Plot Li, f, and L2 to verify graphically that the transitions are smooth. d. Find the difference in elevation between P and Q. e. Bonus: The solution in Problem 1 might look smooth, but it might not feel smooth because the piecewise defined function [consisting of Li(x) for x < 0, f(x) for 0 x 100, and L2(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)-ad-bt e only on the interval 10 xs90 and conneeting it to the linear functions by means of two cubic functions: g(x) = kx3 + 1x2 + mx + n 0S$<10. i. Write a system of equations in 11 unknowns that ensure that the functions and their first two derivatives agree at the transition points L1 L2

I’m sort of lost with my question. it will extremelly be appreciated if the work is very detailed. Thank you.

Answer 1

a)

Consider the function f(x) = ax2 + bx + c . The derivative of the function is, f(x)=ax2+bx + c f' (x) = 2ax + b The three conditions are, Location at Pis, f(x) =ax2 +bx+c s (0)-a(0)+b(0)+c Slope at Pis, f'(x)=2ax + b (0)-2a(0)+b 0.8 b

Slope at Ois f' (x) = 2ax + b f (100) 2a (100)+b -1.6 = 200a + 0.8 α -0.012

So a = -0.012 , b = 0.8 and c = 0

b)

f(x) = ax^2 + bx + c

= -0.012x^2 + 0.8x + 0

= -0.012x^2 +0.8x

So the equation is f(x) = -0.012x² +0.8x

d)

The elevation of point Pis 0 given The elevation of point Ois, f(x)0.012x +0.8x (100) -0.012(100) +0.8(100) =-40 feet

So the difference in elevation between P and Q is 40 feet

c)

The equation for Llis = 0.8x+ 0 0.8x The equation for L2is, (y-yl)-m(x-%) (+40)--16(x-100) -1.6x+160 y =-1.6x+120

The equation for L2 is y = -1.6x + 120

50 y -1.6x+120 y=0.8x -50 100 50 yー-0.01 2x^3+0.8x -50