Question

56. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the accompanying figure. 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)=0.
a. What is dy/dx at x = 0?
b. What is dy/dx at x = -L?
c. 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]

Accepted Answer

Step 1: Begin by understanding the problem. The airplane's landing path is modeled by a cubic polynomial function y = $$ax^3$$ + $$bx^2 + cx + d. $$The given conditions are y(-L) = H (altitude at horizontal distance -L) and y(0) = 0 (altitude at the airport). Additionally, dy/dx at x = 0 and x = -L are required to derive the specific form of the polynomial. Step 2: To find dy/dx at x = 0, differentiate the polynomial y = $$ax^3$$ + $$bx^2 + cx + d $$with respect to x. The derivative is dy/dx = $$3ax^2 + 2bx + c. $$Substitute x = 0 into this derivative to find dy/dx at x = 0, which simplifies to c. Step 3: To find dy/dx at x = -L, use the same derivative dy/dx = $$3ax^2 + 2bx + c. $$Substitute x = -L into this derivative to find dy/dx at x = -L, which simplifies to $$3a(-L)^2 + 2b(-L) + c. $$Step 4: Use the boundary conditions y(-L) = H and y(0) = 0 to solve for d and relate the coefficients a, b, c, and d. Substituting x = -L into y = $$ax^3$$ + $$bx^2 + cx + d $$gives H = $$a(-L)^3$$ + $$b(-L)^2 + c(-L) + d. $$Substituting x = 0 into y = $$ax^3$$ + $$bx^2 + cx + d $$gives 0 = d. Step 5: Combine the results for dy/dx at x = 0 and x = -L, along with the boundary conditions y(-L) = H and y(0) = 0, to derive the specific form of the polynomial. After simplification, the landing path is shown to be y(x) = H[$$2(x/L)^3$$ + $$3(x/L)^2$$].

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Key Concepts

Cubic Polynomial Functions

Derivative and Slope

Boundary Conditions