# Problem 30 Solution

Acknowledgements: The following solution was provided by Alan Curry. Most of the words are his, although I have edited his solution in places.

Place the slower boat initially at the origin of a coordinate plane, and equate "north" with the positive y-axis, and east with the positive x-axis. The faster boat starts at the point (50,0). The "curve" shall refer to the path of the faster boat. t shall refer to time since both boats left their starting positions.

There are two ways to indicate the slope at any point along the curve. One is dy/dx. The other is (x-20t)/y, since the faster boat is always pointing towards the slower boat. Thus dy/dx=y/(x-20t).

The above equation can also be derived this way (optional):

Pick a point on the curve and label it (x1,y1). The x1 and the y1 are functions of the current time, t1. The tangent to the curve at point (x1,y1) passes through the point (20t1,0), which is the current location of the 20kph boat. The equation of the line is y=mx+b, where m is the value of dy/dx at (x1,y1) and b is some constant chosen to make sure that (20t1,0) is on the line. b must be -20*t1*y'(x1), where I've used y'(x1) to mean the value of dy/dx evaluated at the point (x1,y1). The equation of the tangent line is therefore:

y = y'(x1)*x - 20*t1*y'(x1)

Rearranging that, and phrasing it as a general statement rather than a particular point (the point x1 was arbitrary, so this applies to every point on the curve).

dy/dx * (x-20t) = y
dy/dx = y/(x-20t)

Something interesting can be found if you treat x as a function of y, instead of the other way around:

dx/dy = (x-20t)/y

This is justifiable, though it may look like abuse of Leibniz, the derivative of the inverse of a function is the reciprocal of the derivative of the function.

Then multiply both sides by y:

y * dx/dy = x - 20t

And here's the killer... differentiate with respect to y, using the product rule on the left hand side:

(1 * dx/dy) + (y * d2x/dy2 ) = dx/dy - (20 * dt/dy)

There's a dx/dy term on each side of the equation, and they now happily disappear.

[1] y * d2x/dy2 = -20 * dt/dy

Now it's time to turn dt/dy into something we can work with. I split it up as follows:

dt/dy = (dt/ds) * (ds/dy)

Where s is the arc length, the distance traveled so far by the chasing boat.

dt/ds is known (it's the reciprocal of the chasing boat's speed ds/dt=30kph), and ds/dy can be expressed in terms of dx and dy by applying the arc length formula:

The formula for arc length is:

s=Integral of [ (1+(dx/dy)2)1/2 ] dy

Take the derivative of both sides:

ds/dy=(1+(dx/dy)2)1/2, remember that the square root can be positive or negative.

Here is another way to derive this:

ds2 = dx2 + dy2
(ds2)/(dy2) = (dx2)/(dy2) + 1
(ds/dy)2 = (dx/dy)2 + 1

sqrt both sides...

|ds/dy| = sqrt((dx/dy)2 + 1)
ds/dy = +/-sqrt((dx/dy)2 + 1)

We need to choose a sign here. Look which way the curve goes. As t increases, y decreases. At the same time, s (the distance traveled by the chasing boat) increases. Therefore, ds/dy must be negative.

ds/dy = -sqrt((dx/dy)2 + 1)

Substituting all this back into the equation marked [1], you get:

y * d2x/dy2 = -20 * (1/30) * -sqrt((dx/dy)2 + 1)
y * d2x/dy2 = 2/3 * sqrt((dx/dy)2 + 1)

That's a second-order differential equation, but not one of the extremely difficult kind. Rewrite it with a new variable u=dx/dy, and it becomes a first-order diffeq:

y * du/dy = 2/3 * sqrt(u2+1)

Breathe a sigh of relief, because that's separable. (I spent most of my solution time needlessly hacking away at this with more advanced diffeq techniques before I noticed it was separable. D'oh!)

```     du         2    dy
----------- = - * ----
sqrt(u^2+1)   3     y
```
Integrate them... (I had to use a table for the left one)

ln(u + sqrt(u2+1)) = 2/3 * ln y + K

I've dispensed with the absolute value stuff, since u+sqrt(u2+1) can't be negative, and y isn't negative either, for the part of the curve that's relevant to the original question.

exp() both sides...

u + sqrt(u2+1) = C*y2/3 K was an integration constant; C is exp(K).

To find C, use the point (0,50) on the curve. This is at t=0, when the target boat is due south of the chasing boat. The curve's tangent at that point is vertical, so u=dx/dy=0. Plug in u=0 and y=50:

0+1 = C*502/3
C = 1/502/3

Put that back in...

u + sqrt(u2+1) = (y/50)2/3

We've still got some differential equations work to do.. u is a derivative! argh, a derivative in a radical. But it works itself out, just rationalize it:

sqrt(u2+1) = (y/50)2/3 - u
u2+1 = (y/50)4/3 - 2u(y/50)2/3 + u2

Lucked out again. Cancel u2's...

1 = (y/50)4/3 - 2u(y/50)2/3
2u(y/50)2/3 = (y/50)4/3 - 1
2u = (y/50)2/3 - (y/50)-2/3
u = 1/2*[ (y/50)2/3 - (y/50)-2/3 ]
dx/dy = 1/2*[ (y/50)2/3 - (y/50)-2/3 ]

A simple integration:

x = 1/2 * [ 3/5*(y/50)5/3*50 - 3*(y/50)1/3*50 ] + K

Factor out those 3's and 50's

x = 75 * [ 1/5*(y/50)5/3 - (y/50)1/3 ] + K

Again look at the starting point. When x is 0, y is 50.

0 = 75 * [ 1/5*15/3 - 11/3 ] + K
0 = 75 * [ 1/5 - 1 ] + K
0 = -60 + K
60 = K

x = 75 * [ 1/5*(y/50)5/3 - (y/50)1/3 ] + 60

That's the path the chasing boat follows. If that curve has a name, I don't know what it is.

The chasing boat intercepts the other boat when y=0.

x = 75 * [ 1/5*0 - 0 ] + 60
x = 60

The boats meet at the point (60,0). Since the slower boat is traveling due east at 20kph, it must have taken 3 hours to get there.

It takes 3 hours for the chasing boat to catch the other boat.

Below is another solution provided by Tristan Simbulan:

let t = s/30, distance traveled by first boat is 20 * s/30 = 2/3 * s dy/dx=y/( x - 2/3 * s), at t = 0 dy/dx is 50/0 is undefinable.

At t=0 dx/dy=0/50=0 is definable.

dx/dy = (x-2/3*s)/y, y * dx/dy = x - 2/3 * s, take second derivative.

Let first derivative dx/dy = p and second derivative dp/dy, ds/dy = sqrt(1 + (dx/dy)^2) = sqrt( 1+ p^2 ) dy/y = -3/2 * dp/sqrt(1+ p^2), lny + C = -3/2 * ln( p + sqrt( 1 + p^2 )), substitute initial condition y = 50, p = 0.

C = -ln 50. p = dx/dy = 0 y = { p + sqrt(1 + p^2 ) }^(-3/2) * 50 = { dx/dy + sqrt( 1 + (dx/dy)^2) }^(-3/2) * 50

Integration result: 2 * 50^(2/3)x =50^(4/3) * y^(1/3) * 3 - { y^(5/3) * 3/5 } + C C = 12/5 * 50^(5/3) at initial x = 0 y = 50 substitute value of C and you get equation of the curve.

At y = 0, x = 60 . divide distance by rate = time = 60/20 = 3 hours

Note: For the general case in which the slow boat travels at a k.p.h, the fast boat travels at b k.p.h., and the initial distance is d kilometers the formula of the curve is:

x = 1/2*[ db/(a+b)*(y/d)(a+b)/b - db/(b-a)*(y/d)(b-a)/b + db/(b-a) - db/(a+b) ]

The time to interception is [db/(b-a) - db/(a+b)]/2a.

Michael Shackleford, A.S.A.