## An Interesting Problem July 23, 2009

Posted by genericme in Uncategorized.

Though this post isn’t necessarily characteristic of what I’d like to post on in the future, I have an interesting problem that I must share with you all.  To encourage everyone to try it, I will not include a solution.

Two points, $A$ and $B$, sit on the $x$-axis.  Point $A$ is at the origin, and point $B$ is at $(L, 0)$.  At time $t = 0$, $B$ starts moving at a constant speed $u$ perpendicular to the $x$-axis, while $A$ starts moving with constant speed $v$ such that its velocity vector is always pointing in the direction of $B$.  Assume $v > u$.  At what time will the points meet?

This problem is much harder than it seems.  It comes from a book of physics problems, but I’d really consider it a math problem.  I’d like to note that I know of two solutions: an extremely clever solution that requires only a trivial amount of calculus, and a long solution that requires quite a bit of calculus (but that still requires a clever substitution).  The latter solution is due to Eric Larson, who solved this problem a few days after I gave it to him.  However, I know of no one who has independently thought of the clever solution, and I’ll be extremely impressed if you do!  If you’ve seen this problem before, please don’t reveal the solution.

1. hilbertthm90 - July 23, 2009

Alright. I was curious, and I have a long calculus way. I thought that once I saw the answer, it would give insight into how to do it a slicker way, but it doesn’t immediately suggest anything. I’ll think about it some more though.

2. some_one - July 24, 2009

Reminds me a nice variation :
three points particles(A,B,C) are located at the three vertices of a equilateral triangle at t=0 .
They start moving towards each other( say A moving towards B, B towards C and C towards A) with equal and constant speed v.
Describe their trajectory.

genericme - July 24, 2009

hilbertthm90: That was quick! Even a long calculus method, in my experience, isn’t easy. The really clever solution involves integral rather than differential equations.

some_one: Yes, describing the trajectories is an interesting problem, though in this case, finding the time at which the three particles meet is much easier and requires no calculus.

3. Omar - July 24, 2009

Well I can’t figure out any clever way to do it, but it is perfectly straightforward by just writing out the equations and using a tiny bit of calculus:

Letting prime denote time derivatives, the way the points move means that A + f A’ = B for some unknown scalar function f. (I will consistently suppress time dependence to ease notation.) Differentiating we get that (1+f’)A’ + f A” = B’.

Now, we also know that B = (L, ut) and that |A’|=v, so B’=(0,u) and =0. So taking the inner product of that last equation with A’ we get (1+f’)v^2 = = u y’ –where we’ve set A=(x,y).

Integrating produces (t+f)v^2 = u y + K for some constant K. Setting K=0 it is easy to see K = L v (because when t=0, A=(0,0), A’=(v,0) so f = L/v).

Let A and B meet at time t=T. At that moment, f=0 (since A=B), and y = second coordinate of A = second coordinate of B = u T, so solving for T we get K / (v^2-u^2) = L v / (v^2 – u^2).

4. Omar - July 24, 2009

My previous comment used less than and greater than signs for inner products and they got parsed as HTML delimiters or something. It’s only the third paragraph that got mangled (although the remaining ones did turn into links or something). Here’s the third paragraph corrected (feel free to fix the previous comment and delete this one if you prefer):

Now, we also know that B = (L, ut) and that |A’|=v, so B’=(0,u) and < A’, A” > =0. So taking the inner product of that last equation with A’ we get (1+f’)v^2 = < B’, A’ > = u y’ –where we’ve set A=(x,y).

genericme - July 24, 2009

Wow, I’ve never seen that solution, and it’s hardly more effort than the “clever” one. Apparently, this problem isn’t hard enough for you guys.

I’ll probably post the solution using integral equations soon.

5. hilbertthm90 - July 25, 2009

I actually accidentally simplified the problem on my initial solution so I didn’t have the correct solution when I posted that. Although there is a neat way to do it by just writing down the constant vector field of magnitude v on R^2 that is pointing towards the point B at any given time. This is a nice and simple time dependent vector field.

Now we can just solve for the integral curve that goes through the point (x,y) at time 0. This gives the equation of the curve that A travels if it starts at (x,y). So this is more general than making A start at the origin.

Of course, to explicitly solve for that integral curve probably amounts to precisely the integral equations you refer to (it just is a nice setup to get to them).

6. genericme - July 26, 2009

Well, here’s the slick solution using integral equations, anyway (which I didn’t think of):

Denote by $\theta$ the angle between the velocity vectors of A and B, and let $T$ be the time after which they meet. For their coordinates to be equal at time $T$, a condition in the vertical direction is $\int_0^T v\cos\theta dt = uT$. Considering the components of their velocities along the line joining them, another condition is $\int_0^T (v - u\cos\theta) dt = L$. Taking a linear combination of these two integrals to get rid of the annoying $\theta$ term, and evaluating the resulting trivial integral, gives the answer.

7. 一个有趣的物理问题 « Liu Xiaochuan’s Weblog - July 30, 2009

[…] 来源：https://deltaepsilons.wordpress.com/2009/07/23/an-interesting-problem/ […]

8. mysteve - August 7, 2009

I think if $T$ is the time they meet, point B must run $Tu$. $\theta$ is the angle of $\measuredangle BOL$. So $\theta =\tan ^{-1}(\frac{Tu}{L})$ and $\cos \theta =\cos (\tan ^{-1}(\frac{Tu}{L}))$. Point A runs $Tv$ and meet point B. It must satisfy the equation $\cos \theta Tv=L$, and we have $T=\frac{L}{v\cos \theta }=\frac{L}{v\cos (\tan ^{-1}(\frac{Tu}{L}))}$. Then we use fixed point to solve $T$. Check the graph: http://img44.imageshack.us/img44/3228/ss01.gif

9. mysteve - August 7, 2009
genericme - August 7, 2009

Unfortunately, your condition $Tv\cos\theta = L$ is incorrect because the direction of the velocity vector of $A$ changes in time.

10. mysteve - August 8, 2009

Yet, I think Point $latex A$ is running on the ray line $latex \overrightarrow{OA}$, and this ray line chages its angle as time go by. We can think a bridge connect $latex O$ and $latex B$, and this bridge will extend its lentgh and change its angle in time. Point $latex A$ just run on this way. So I think the condition must be satisfied. This is my point of view. I used program to check my view point, and obtain figures inss01.gif. The last figure of ss02.gif shows that Point $latex B$ run $latex Tu$, i.e. $latex \overline{LB}=Tu$. If Point $latex A$ run on $latex x$-axis, we denoted Point $latex A’$ to represent it. We can find Point $latex A$ run $latex Tv$, as $latex \overline{OA’}=Tv$.

genericme - August 8, 2009

I see what you mean by point A’ now. You have some nice ideas, but there are still a few things that aren’t consistent in what you’re saying. You’ve defined $\theta$ as both a constant angle and one that varies in time. Also, points O, A, and B aren’t necessarily collinear for all time. In fact, if they were, it would imply that the velocity vector of point A is always parallel to its position vector, which clearly isn’t true.

11. mysteve - August 8, 2009

Ya, there are some problem in my solution. The auto-rotate bridge does not exist, this rotation will increase the distance from point A to point B. We can see that the green path is the trace from A to B, but its length is greater then $latex Tv$. The additional length come from the auto-rotate bridge.