Question

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 figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function \(y=a x^{3}+b x^{2}+c x+d\) where \(y(-L)=H\) and \(y(0)=0 .\) (a) What is \(d y / d x\) at \(x=0 ?\) (b) What is \(d y / d x\) at \(x=-L ?\) (c) Use the values for \(d y / d x\) at \(x=0\) and \(x=-L\) together with \(y(0)=0\) and \(y(-L)=H\) to show that $$y(x)=H\left[2\left(\frac{x}{L}\right)^{3}+3\left(\frac{x}{L}\right)^{2}\right]$$

Short answer

The derivative of the function at \(x = 0\) is \(c\), and the derivative at \(x = -L\) is \(3aL^{2} - 2bL + c\). After substituting these derivatives along with the original function evaluations at \(y(0) = 0\) and \(y(-L) = H\), it can be shown that \(y(x) = H[2(\frac{x}{L})^{3} + 3(\frac{x}{L})^{2}]\)

Step-by-step solution

Find the derivative of the function

The derivative of the function \(y = ax^{3} + bx^{2} + cx + d\) with respect to \(x\) is \(y' = 3ax^{2} + 2bx + c\)

Values of derivative at \(x = 0\) and \(x = -L\)

Substituting \(x = 0\) in the derivative function provides \(y'(0) = c\). Substituting \(x = -L\) in the derivative function gives \(y'(-L) = 3aL^{2} - 2bL + c\)

Substitute values of \(y(x)\) from the problem

From the problem's constraints, we know that \(y(0) = 0\) therefore \(d = 0\). We also know \(y(-L) = H\) therefore \(H = aL^{3} + bL^{2} - cL\)

Solve the Equation

Substitute \(y(0) = 0\) and \(y(-L) = H\) into the expression to prove that \(y(x) = H[2(\frac{x}{L})^{3} + 3(\frac{x}{L})^{2}]\)