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### Catching a Pig

```
Date: 05/09/97 at 07:57:08
From: Udo Guenthner
Subject: When does the farmer catch the swine

Hi,

This question is from Martin Gardner's book:

A farmer wants to catch a pig which is x meters away from him. They
start running at the same time.

The pig runs straight ahead, perpendicular to the line farmer-pig, at
constant speed. The farmer always runs in the direction where he sees
the pig (also at constant speed).  If we know the speed of the farmer
and the pig and the distance x between them, how long does it take the
farmer to catch the pig?

_   ---------------------> pig
|                       .> farmer
|             .
|         .
x      .
|    .
|   .
_  .

The solution in Martin Gardner's book was: calculate the time when the
pig runs straight up, then calculate the time when the pig runs
straight down and take the average. That's it!

So my question: Why is it exactly the average? What if the pig runs at
an angle of 73 degrees? In other words, what is the underlying equation?

I tried to solve this using differential calculus and complex numbers
with no success.
```

```
Date: 05/09/97 at 11:33:18
From: Doctor Anthony
Subject: Re: When does the farmer catch the swine

I would prefer to use x as a variable and also to give some constant
ratio for the speed of the man and the pig (say 3:1).

Draw the following figure to see how I do the calculation. Let the pig
start at O the origin and move along the x axis. Let the man start at
point B on the y axis such that OB = b   At time t the pig will have
moved to the point T on the x-axis, and the man will have moved along
the arc BP such that the tangent to the arc at P is directed at the
point T. Let s = arc length of BP. Because of the relative speeds, the
arc BP will be three times the length of OT (arc BP = 3(OT)). Draw the
perpendicular PN from P to the x axis.

If (x,y) are the coordinates of the man, we have NP/NT = -dy/dx since
the slope of curve is given by NP/NT and the slope is negative.

Then OT = ON + NT = ON + NP(NT/NP)
=  x - y(dx/dy)    (note dx/dy not dy/dx)

OT = (1/3)s  = x - y(dx/dy)
s = 3(x-y(dx/dy))

If we differentiate with respect to y:

ds/dy = 3(dx/dy - dx/dy - y.d^2x/dy^2)

= - 3y.d^2x/dy^2

But ds/dy = -sqrt(1+(dx/dy)^2), the minus being taken because y
decreases as s increases.

If we write q for dx/dy this gives:

sqrt(1+q^2) = 3y.dq/dy

dy/3y = dq/sqrt(1+q^2)      Integrating

(1/3)ln(ky) = sinh^(-1)(q)      k is constant of integration

When y = b, q = 0, so kb = 1 or k = 1/b.

Therefore, q = sinh[ln(y/b)^(1/3)]       sinh(x) = (e^x - e^-x)/2

2q = (y/b)^(1/3) - (y/b)^(-1/3)

Since q = dx/dy a second integration gives:

2(x+c) = (3/4)b^(-1/3)y^(4/3) - (3/2)b^(1/3)y^(2/3)

When x = 0, y = b, so that 2c = -(3/4)b and the equation of the curve
becomes:

b^(1/3)(8x-3b) = 3(y^(4/3)-2b^(2/3)*y^(2/3))

The man catches the pig when y = 0, that is when 8x-3b = 0  or
x = (3/8)b

If you know the actual speed of the pig, say u m/sec, then the time to
catch the pig is given by (3/8)b/u (b is the initial distance they are
apart).

This result of course depends on the ratio of their speeds.  This
ratio must be greater than 1 if the man is to catch the pig at all.
If the ratio is near to 1 the time will increase without limit.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 05/10/97 at 06:16:29
From: Doctor Mitteldorf
Subject: Re: When does the farmer catch the swine

Dear Udo,

I wrote this before reading Dr. Anthony's reply.  He has gotten much
further with the problem than I have, but I'm sending my reply along
in case you find my approach interesting as well.

Does Martin Gardner tell you any more about why the answer comes out
to the average of the two times (pig runs toward farmer) and (pig runs
away from farmer)?  It's not obvious to me why this should be so, but
it's a reasonable hypothesis. I say it's reasonable because in simple
cases it gives the right answer: If the farmer moves much faster than
the pig, it clearly gives the right answer. If the farmer moves the
same speed as the pig, then the answer comes out infinite, as it
should.

In problems like this, there's a brute force method and there's the
"aha!" method. The brute force method always works. The aha! method
can save you a lot of time if you can see through to something about
the question that's being asked that makes it simpler than it at first
appears. I must admit that I have yet to have my aha! experience for
this puzzle, so I have approached it with brute force. The answer I
get seems to confirm Gardner's averaging rule, but I can't tell you
why.

Incidentally, Gardner has a book of puzzles called _Aha!_ in which is
included a similar problem: Four turtles begin at the corners of a
square. Each turtle proceeds to walk at constant speed toward the
turtle on his right. Gradually, the square twists and shrinks until
the turtles meet at the center of the square. How long does it take
for the turtles to meet? This puzzle has a simple answer that depends
on a somewhat sophisticated insight: since the turtles are always in
the shape of a square, and since each one is always walking
perpendicular to the direction of his pursuer's approach, he neither
hinders nor helps the pursuer in reaching him. Therefore the time to
the center is the same as the time it takes one turtle to walk across
a side of the square.

This insight doesn't apply to the farmer and the pig, and I haven't
thought of any way to make that problem simple, so I've taken the
brute force approach:

Let the pig start at t = 0, x = 0, y = 0 and proceed to run at speed v
along the x-axis, so x = vt.

Let the farmer start at t = 0, x = 0, y = -b with a constant speed w.
I'll represent the x speed as x' = dx/dt and y' = dy/dt. Then we have
two equations for x' and y':

x'^2 + y'^2 = w^2
y'/x' = y/(vt-x)

The first equation says that the vector sum of the x and y velocities
has constant magnitude w, and the second equation says that the
velocity of the farmer is always directed along a line from the
farmer's position (x,y) to the pig's position (vt,0).

Now, having taken the brute force approach and written these
differential equations, I must admit further that I don't know how to
solve them! So again I've taken the brute force approach and written
a computer program to solve the equations numerically. The program
just starts with the initial positions, calculates the x and y
velocities at each step, and moves the farmer along with increments
dx = x'dt and dy = y'dt in each time step dt. I think you'll be able
to follow it even if you're not familiar with Pascal. I've used the
variable R to mean y/(vt-x), and solved the two equations
simultaneously to give:

x' = w / sqrt(1+R^2) and
y' = wR / sqrt(1+R^2)

The program gives an answer in a fraction of a second, and to an
accuracy of three decimal places it agrees with Gardner's averaging
rule for the two cases I've tried (farmer runs two times pig speed and
three times pig speed).

Please write again if you have any more insight into this puzzle, or
if there's anything else in Gardner's account of it that might help us
toward an aha!

-Doctor Mitteldorf,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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