February 2023

We see from “Seek-Lock-Strike!” Again that given the missile’s position $(x, y),$

$\frac{dx}{dy}=\frac{a+v_a t-x}{b-y}$

where $x$ and $y$ are themselves functions of time $t.$

It means

$\frac{dx}{dy} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{a+v_a t-x}{b-y}\implies \frac{dx}{dt} = \frac{a+v_a t-x}{b-y}\cdot\frac{dy}{dt}.$

That is, let $\kappa(x,y,t) = \frac{a+v_a t-x}{b-y},$

$\frac{dx}{dt}=\kappa\cdot \frac{dy}{dt}.\quad\quad\quad(1)$

We also have (see “Seek-Lock-Strike!”)

$v_m = \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\implies v_m = \sqrt{1+\kappa^2}|\frac{dy}{dt}| .$

Since $\frac{dy}{dt} >0,$

$\frac{dy}{dt} = \frac{v_m}{\sqrt{1+\kappa^2}}.\quad\quad\quad(2)$

Substitute (2) into (1) yields

$\frac{dx}{dt} = \frac{v_m\cdot \kappa}{\sqrt{1+\kappa^2}}.\quad\quad\quad(3)$

It follows that $(x(t), y(t))$ , the position of the missile satisfies the initial-value problem

$\begin{cases} \frac{dx}{dt} = \frac{\kappa\cdot v_m}{\sqrt{1+\kappa^2}} \\ \frac{dy}{dt} = \frac{v_m}{\sqrt{1+\kappa^2}} \\x(0)=0, y(0)=0\end{cases}\quad\quad\quad(4)$

To obtain the missile’s trajectory, we solve (4) numerically using the Runge-Kutta algorithm. It integrates (4) from $t=0$ to $t= t_*$ (see “Seek-Lock-Strike!”).