Wang Erxiao’s cow ran away. When he noticed, the cow was 300 meters due South of him and was running at a constant speed due West. Wang Erxiao immediately chased the cow. Instead of running in a fixed direction, he always ran towards the cow’s current position at every moment. His speed was 4/3 times the cow’s speed. How far did Wang Erxiao run by the time he caught the cow?
Problem Analysis
At first glance, chasing a cow and tracking a missile seem like two completely unrelated things: one is a trivial matter of daily life, and the other is a physics problem. How can they be linked together?
Think back to how police usually catch a thief. A police officer is not a physicist; they won’t and cannot first study the thief’s escape route function and then design a path for minimum pursuit time. So, without being able to predict the thief’s escape route, how does a police officer catch them? It’s simple: Lock onto him! Yes, as long as you run faster and always head directly toward them, you will eventually catch them. Continuing this line of thought: if you want to use a missile to track and destroy an enemy ship, isn’t this the only method you can adopt? Looking back at the “Cow Pursuit Problem” at the beginning of the article, isn’t it essentially the same? The following is a missile tracking problem proposed by Shanghai Jiao Tong University:
A missile base of a certain army discovered an enemy boat on the sea 120 km due North, traveling due East at a speed of 90 km/h. The base immediately launched a missile to track and pursue the enemy boat. The missile speed is 450 km/h, and the automatic navigation system ensures the missile is aimed at the enemy boat at every moment. At what time and at what location will the missile hit the enemy boat?
Establishing the coordinate system:
Let the missile speed be u and the enemy boat speed be v. We can list: dx^2+dy^2=u^2 dt^2 \tag{1} \frac{dy}{dx}=\frac{H-y}{vt-x} \tag{2}
Luring the Enemy
The general idea for solving a system of differential equations is: elimination and order reduction. If you are unfamiliar with differential equations or are just starting out, you might immediately think of order reduction while neglecting to eliminate variables first, leading to a dead end. This is because our subconscious tells us that higher-order equations are always more troublesome than lower-order ones, and turning a lower-order equation into a higher-order one is looking for trouble. In fact, this is not the case. To eliminate variables, we need to “lure the enemy into a trap”—that is, “increase the order”!
We try to eliminate t by transforming (2) into \frac{dx}{dy}(H-y)=vt-x. Differentiating both sides gives: d\left(\frac{dx}{dy}\right)(H-y)-\frac{dx}{dy}\cdot dy=v dt-dx \tag{3} This expression is equivalent to \frac{d\left(\frac{dx}{dy}\right)}{dy} \frac{dy}{dt}(H-y)=v. From equation (1), we can obtain: \frac{dy}{dt}=\frac{u}{\sqrt{\left(\frac{dx}{dy}\right)^2+1}}
Substituting this into the above equation, we get: \frac{d\left(\frac{dx}{dy}\right)}{dy} \frac{u}{\sqrt{\left(\frac{dx}{dy}\right)^2+1}}(H-y)=v
Direct Attack
After completing the elimination, what we need is order reduction and separation of variables to find the final result. During elimination, one can also compare which unknown variable, when eliminated, makes the form of the differential equation relatively simpler. For example, in the problem above, we obtained an equation where variables can be separated, which saves us a lot of work. If we had chosen to eliminate y, we would not have obtained an equation with separable variables, which would have made our calculations much more complicated.
Let \frac{v}{u}=k, then we have: \frac{d\left(\frac{dx}{dy}\right)}{\sqrt{\left(\frac{dx}{dy}\right)^2+1}}=\frac{k dy}{H-y}
Integrating both sides: \ln\left(\frac{dx}{dy}+\sqrt{\left(\frac{dx}{dy}\right)^2+1}\right)=-k \ln(H-y)+C Substituting the initial conditions \frac{dx}{dy}=0,y=0, we get C=k \ln H. Taking the exponent of each side: \begin{aligned} &\frac{dx}{dy}+\sqrt{\left(\frac{dx}{dy}\right)^2+1}=\left(\frac{H}{H-y}\right)^k \\ &\frac{dx}{dy}=\frac{1}{2}\left[\left(\frac{H}{H-y}\right)^k-\left(\frac{H}{H-y}\right)^{-k}\right] \\ &2dx=[(1-y/H)^{-k}-(1-y/H)^k]dy \end{aligned}
Integrating both sides and substituting the initial values x=0,y=0, we get: x=\frac{1}{2} \left[-\frac{1}{-k+1}(1-y/H)^{-k+1}+\frac{1}{k+1}(1-y/H)^{k+1}\right]+\frac{kH}{1-k^2}
With this, the problem is solved. The time required to destroy the target is T=\frac{Hu}{u^2-v^2}.
However, some friends might ask: what if the speeds of the enemy ship and the missile are not constants but functions of t? How should we solve this type of equation? In this case, we can choose to eliminate x. First, let \dot{x} denote the derivative with respect to the independent variable t, \frac{dx}{dt}. Starting from equation (3), we change \frac{dx}{dy} to \frac{\dot{x}}{\dot{y}}. From equation (1), we solve for \dot{x}=\sqrt{u^2-\dot{y}^2}. After substitution and elimination, here is the result calculated by BoJone: \frac{u\dot{u}\dot{y}-u^2 \ddot{y}}{\dot{y}^2 \sqrt{u^2-\dot{y}^2}}=\frac{\dot{s}}{H-y} where s\equiv s(t) is the distance function of the pursued target. The above is a second-order differential equation regarding y and t.
(Thanks to the Department of Mathematics at Shanghai Jiao Tong University. You can download the SJTU courseware here: Missile Tracking (07.3).rar)
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