# Seek-Lock-Strike!

A guided missile is launched to destroy a fighter jet (Fig. 1).

Fig. 1

We introduce a coordinate axes such that at $t=0,$ the missile is at origin $(0, 0)$ and the jet at $(a,b)$. The jet flies parallel to the x-axis with constant speed $v_a$. The missile has locked onto the jet so it is always pointing at the jet as it moves. Its speed is $v_m.$

Find the time and position the missile strikes its target.

“The missile has locked onto the jet so it is always pointing at the jet as it moves” means that the tangent to the missile’s path at any point $(x, y)$ will pass through the position of the jet. The equation of the tangent is $\frac{dy}{dx}\cdot(a + v_a t -x) = b-y.\quad\quad\quad(1)$ $(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2=v_m^2.\quad\quad\quad(2)$

Notice $x, y$ are functions of time $t : x=x(t), y=y(t).$

Differentiate (1) with respecte to $t:$ $\frac{d}{dt}(\frac{dy}{dx})\cdot(a+v_at-x) + \frac{dy}{dx}\cdot\frac{d}{dt}(a+v_at-x) = -\frac{dy}{dt},$ $\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{dy}{dx})\cdot\frac{dx}{dt}=\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}$ $\left(\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\right)\cdot(a+v_at-x) + \frac{dy}{dx}\cdot(v_a -\frac{dx}{dt}) = -\frac{dy}{dt},$ $\left(\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\right)\cdot(a+v_at-x) + \frac{dy}{dx}\cdot v_a -\underbrace{\frac{dy}{dx}\cdot\frac{dx}{dt}}_{\frac{dy}{dt}}= -\frac{dy}{dt},$ $\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\cdot(a+v_at-x) + \frac{dy}{dx}\cdot v_a = 0,$ $\frac{d^2y}{dx^2}= \frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dy}{dx}$ $\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dy}{dx}\cdot\frac{dx}{dt}(a+v_at-x) + \frac{dy}{dx}v_a=0,$ $\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dx}{dt}\cdot\underline{\frac{dy}{dx}(a+v_at-x)} + \frac{dy}{dx}v_a=0.$

By (1), substituting $b-y$ for $\frac{dy}{dx}\cdot(a + v_a t -x)$, $\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dx}{dt}\cdot(b-y) + \frac{dy}{dx}v_a=0.\quad\quad\quad(3)$

Let $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt},$ we express (2) as $(\frac{dx}{dt})^2 + (\frac{dy}{dx}\cdot\frac{dx}{dt})^2=v_m^2,$ $(\frac{dx}{dt})^2\cdot(1+(\frac{dy}{dx})^2)= v_m^2.$

Suppose $a>0$, which implies that $\frac{dx}{dt} > 0$ (see Exercise-4). Solving for $\frac{dx}{dt}$ gives $\frac{dx}{dt} = \frac{v_m}{\sqrt{1+(\frac{dy}{dx})^2}}.\quad\quad\quad(4)$

Submitting (4) into (3), $\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{v_m}{\sqrt{1+(\frac{dy}{dx})^2}}\cdot(b-y) + \frac{dy}{dx}\cdot v_a=0,$ $\frac{d}{dy}(\frac{dy}{dx})\cdot(b-y) + \frac{v_a}{v_m}\cdot\frac{dy}{dx}\cdot\sqrt{1+(\frac{dy}{dx})^2}=0,\quad\quad\quad(5)$

Let $p = \frac{dy}{dx}, r=\frac{v_a}{v_m},$

(5) becomes $\frac{dp}{dy}(b-y) + r\cdot p\cdot\sqrt{1+p^2}=0.$

We solve this non-linear differential equation as follows:

For $y < b$, $\frac{1}{p\sqrt{1+p^2}}\frac{dp}{dy} = \frac{-r}{b-y},$ $\int (\frac{\sqrt{1+p^2}}{p}-\frac{p}{\sqrt{1+p^2}})\frac{dp}{dy} dy = \int \frac{-r}{b-y} dy,$ $\int\frac{\sqrt{1+p^2}}{p}dp-\int\frac{p}{\sqrt{1+p^2}}dp = r\log(b-y)+k_2.\quad\quad\quad(6)$

Since $\int\frac{\sqrt{1+p^2}}{p}dp=\sqrt{1+p^2} - \mathrm{arcsinh}\left(\frac{1}{|p|}\right)$ (See “I vs. CAS“), $\int\frac{p}{\sqrt{1+p^2}}dp=\sqrt{1+p^2},$

(6) gives $-\mathrm{arcsinh}\left(\frac{1}{|p|}\right) = r\log(b-y)+k_1.\quad\quad\quad(7)$

At $t=0, y=0, p=\frac{dy}{dx}=\frac{b}{a},$

(7) yields $k_1 = -\mathrm{arcsinh}\left(\frac{a}{b}\right) - r\log(b).$

Moreover, since $p = \frac{dy}{dx} > 0$, we have $\frac{dy}{dx} = \frac{1}{-\sinh(r\log(b-y)+k_1)},$ $-\sinh(r\log(b-y)+k_1)\cdot \frac{dy}{dx} = 1,$ $\displaystyle\int -\sinh(r\log(b-y) + k_1) \cdot \frac{dy}{dx}\;dx = \displaystyle \int 1\; dx = x +k_2.$

Using Omega CAS Explorer:

we obtain $\frac{1}{2}\left(\frac{e^{k_1}(b-y)^{r+1}}{r+1} - \frac{e^{-k_1}(b-y)^{1-r}}{1-r}\right) = x + k_2.\quad\quad\quad(8)$

Since $y=0, x=0$, (8) gives $k_2 = \frac{1}{2}\left(\frac{e^{k_1}b^{r+1}}{r+1} - \frac{e^{-k_1}b^{1-r}}{1-r}\right).\quad\quad\quad(9)$

Suppose when $t = t_*,$ the missile hits the target. Then the striking coordinates $(x_*, y_*)$ are the same as that of the fighter jet, i.e., $x_* = a + v_a t_*, \;\;y_*=b.\quad\quad\quad(10)$

Hence, $\lim\limits_{t \rightarrow t_*} x = x_* \overset{(10)}{=} a+v_a t_*, \;\;\lim\limits_{t \rightarrow t_*} y = y_* \overset{(10)}{=}b \implies \lim\limits_{t \rightarrow t_*} b-y = 0.\quad\quad\quad(11)$

For $r \le 1$ (i.e., $v_a \le v_m$), as $t \rightarrow t_*,$ (8) gives (see Exercise-1) $0 = a +v_a t_* + k_2 \implies -k_2-a = v_a t_*.$

That is, $-\underbrace{\frac{1}{2}\left(\frac{e^{k_1}b^{1+r}}{1+r}-\frac{e^{-k_1}b^{1-r}}{1-r}\right)}_{(9):\;\;k_2}-a=v_a t_*.$

By $(\star)$ (see below), $-\frac{1}{2}\left(\frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot\frac{b^{1+r}}{1+r}-b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)\cdot\frac{b^{1-r}}{1-r}\right) - a = v_a t_*,$ $-\frac{1}{2}\left(\frac{1}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot\frac{b}{1+r}-\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)\cdot\frac{b}{1-r}\right) - a = v_a t_*,$ $\frac{\frac{-b}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$ $\frac{\frac{-b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{-1(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$ $\frac{\frac{b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{1+r} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$ $\frac{b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{2(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{2(1-r)} -a = v_a t_*,$ $\frac{a-\sqrt{b^2+a^2}}{2(1+r)} + \frac{a+\sqrt{b^2+a^2}}{2(1-r)} -a = v_a t_*,$ $\frac{(1-r)(a-\sqrt{b^2+a^2}) +(1+r)(a+\sqrt{b^2+a^2})-2a(1-r^2)}{2(1+r)(1-r)} = v_a t_*,$ $\frac{a-\sqrt{b^2+a^2}-ar+r\sqrt{b^2+a^2}+ar+r\sqrt{b^2+a^2}+a+\sqrt{b^2+a^2}-2a + 2ar^2}{2(1-r^2)}=v_a t_*,$ $\frac{2r\sqrt{b^2+a^2} + 2ar^2}{2(1-r^2)} = v_a t_*,$ $\frac{ar^2+\sqrt{b^2+a^2}r}{1-r^2} = v_a t_*,$ $\frac{r(ar + \sqrt{b^2+a^2})}{v_a(1-r^2)} = t_*.$

Since $r = \frac{v_a}{v_m}$, we have $t_* = \frac{v_a}{v_m}\cdot\frac{ar + \sqrt{b^2+a^2}}{v_a(1-r^2)}=\frac{1}{v_m}\cdot\frac{ar + \sqrt{b^2+a^2}}{1-r^2}.$

Namely, $t_* = \frac{\sqrt{b^2+a^2} + a\cdot r}{(1-r^2)\cdot v_m}, \quad\quad r=\frac{v_a}{v_m}.$

It follows that the guided missile strikes the fighter jet at $(a+r\cdot\frac{\sqrt{b^2+a^2}+a\cdot r}{1-r^2}, b).$ $e^{k_1} = e^{-\mathrm{arcsinh}(\frac{a}{b})-r\log(b)}$ $= \frac{1}{e^{\mathrm{arcsinh}(\frac{a}{b})+r\log(b)}}$ $= \frac{1}{e^{\mathrm{arcsinh}(\frac{a}{b})}\cdot e^{r\log(b)}}$ $\mathrm{arcsinh}(\blacksquare) = \log\left(\blacksquare + \sqrt{1+\blacksquare^2}\right)$ (see “Deriving Two Inverse Functions“) $= \frac{1}{e^{\log\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot e^{r\log(b)}}$ $e^{r\log(b)}=e^{\log(b^r)}=b^r$ (see “Introducing Lady L” and “Two Peas in a Pod, Part 3“) $= \frac{1}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b^r}.$

i.e., $e^{k_1} = \frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}.\quad\quad\quad(\star)$

Exercise-1 Show that for $r < 1$, (8) gives $0 = a +v_a t_* + k_2 \implies -k_2-a = v_a t_*.$ (Hint: (10))

Exercise-2 For $r < 1$, what is the total distance traveled by the missile when it strikes the fighter jet? (hint: Don’t make things harder than they are)

Exercise-3 Show that if $r \ge 1$ (i.e., $v_a \ge v_m$), the missile will not strike the fighter jet. Explain.

Excercise-4 Show that $a>0 \implies \frac{dx}{dt} >0$.

Excercise-5 For $a <0$, find the time and position the missile strikes its target .