I was trying to replicate the math "How not to land in Lake Tahoe" problem by Barshinger (1992) in the American Mathematical Monthly.
Essentially, he models the landing path of a plane with a cubic polynomial.
The resultant equation is
$y(x)=h({2(x/L)^3+3(x/L)^2})$
The velocity is intuitively the implicit differentiation of the landing path function as stated in the original problem solution:
$\frac{dy}{dt}=\frac{6Uh}{L}((x/L)^2+(x/L))$
$U = \frac{dx}{dt}$ assuming that the after the implicit differentiation, the horizontal velocity is constant.
However, I am stuck when he takes the second derivative to get the vertical acceleration of the plane on the landing path. I understand that in kinematics that acceleration is simply the derivative of the velocity. However, I am unsure how he arrived at the below expression for acceleration.
$\frac{d^2y}{dt^2}=\frac{6U^2h}{L^2}(2(x/L)+1)$
I understand the latter part – $(2(x/L)+1)$ which is just following the normal differentiation rules. However, I don't know how to arrive at the $\frac{6U^2h}{L^2}$ term with the $U^2$ and the $L^2$ components. Why are they raised to the power of 2? Is this some kind of chain rule or product rule that I'm missing?
Best Answer
In this context, $U$, $h$ and $L$ are all constants and $x$ is a function of $t$: $x(t) = Ut$.
Now with $y'(t) = \frac{6Uh}{L}((x(t)/L)^2 + x(t)/L)$, you take the derivative again and get $$\begin{align*} y''(t) & = \frac{6Uh}{L}(2x(t)x'(t)/L^2 + x'(t)/L) \\ & = \frac{6Uh}{L}(2 x(t) U / L^2 + U/L) \\ & = \frac{6U^2h}{L^2}(2 x(t) + 1), \end{align*}$$ so the only thing that is happening is that the constant factor $U/L$ appearing in both terms inside the parentheses is moved outside of the parentheses.