An airplane is flying at altitude H when it begins its descent to an airport run

Question

An airplane is flying at altitude H when it begins its descent to an airport runway that is horizontal ground distance L from the airplane, as shown. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax^3 + bx^2 + cx + d, where y(-L) = H and y = 0. What is dy/dx at x = 0? What is dy/dx at x = -L? Use the values for dy/dx at x = 0 and x = -L? Use the values for dy/dx at x = 0 and x = -L together with y (0) = 0 and y (-L) = H to show that y(x) = H[2(x/L)^3 + 3(x/L)^2].

Explanation / Answer

y=ax^3+bx^2+cx+d
y(0) = 0 = d => d = 0

For the airplane to arrive at its landing point in a controlled landing, the slope of the path, dy/dx, must be zero - the airplane must arrive at the runway surface moving tangentially to the runway;

dy/dx = 3ax + 2bx + c
dy(0)/dx = 0 = c

y = ax³ + bx²

At the point, x = -L, when the airplane begins its descent, it's moving in level flight, so dy(-L)/dx must also be 0.

dy/dx = 3ax² + 2bx
dy(-L)/dx = 0 = 3aL² - 2bL
3aL = 2b
b = 3aL/2

y = ax²(x + 3L/2)

we are given that at x = -L, y = H. That is the final condition needed to evaluate the remaining unknown coefficient, a:
y(-L) = aL²(-L + 3L/2) = H
aL³(3/2 - 1) = H
aL³/2 = H
a = 2H/L³

y = (2Hx²/L³)(x + 3L/2)

Rewriting this equation, we have:

y/H = 2(x/L)²[(x/L) + 3/2]
y = H[2(x/L)³ + 3(x/L)²]

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