Answer
a. $y'(0)=c=0$.
b. $y'(-L)=3aL^2-2bL=0$.
c. See explanations.
Work Step by Step
a. Given the equation $y=ax^3+bx^2+cx+d$ and $y(-L)=H, y(0)=0$, we have $y'=3ax^2+2bx+c$ and at $x=0$, we have $y'(0)=c$. We can further evaluate the parameters. $y(0)=0$ gives $d=0$. Because the airplane will glide horizontal when landing, we have $y'(0)=0$, which gives $c=0$.
b. At $x=-L$, we have $y'(-L)=3a(-L)^2+2b(-L)=3aL^2-2bL$. Because we can assume the airplane travels horizontally before descent for landing, $y'(-L)=0$, which gives $3aL=2b$. As $y(-L)=H$, we have $a(-L)^3+b(-L)^2=H$ and $-aL^3+bL^2=H$ or $-2aL^3+3aL^3=2H$ which gives $a=\frac{2H}{L^3}$ and $b=\frac{3H}{L^2}$
c. Using the above results, we have $y=ax^3+bx^2+cx+d=(\frac{2H}{L^3})x^3+(\frac{3H}{L^2})x^2+(0)x+0=H[2(\frac{x}{L})^3+3(\frac{x}{L})^2]$
