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 (x_{1},y_{1}). The x_{1}and the y_{1}are functions of the current time, t_{1}. The tangent to the curve at point (x_{1},y_{1}) passes through the point (20t_{1},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 (x_{1},y_{1}) and b is some constant chosen to make sure that (20_{t1},0) is on the line. b must be -20*t_{1}*y'(x_{1}), where I've used y'(x_{1}) to mean the value of dy/dx evaluated at the point (x_{1},y_{1}). The equation of the tangent line is therefore:y = y'(x

_{1})*x - 20*t_{1}*y'(x_{1})Rearranging that, and phrasing it as a general statement rather than a particular point (the point x

_{1}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 * d^{2}x/dy^{2} ) =
dx/dy - (20 * dt/dy)

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

**[1]** y * d^{2}x/dy^{2} = -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:

ds^{2}= dx^{2}+ dy^{2}

(ds^{2})/(dy^{2}) = (dx^{2})/(dy^{2}) + 1

(ds/dy)^{2}= (dx/dy)^{2}+ 1sqrt 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 * d^{2}x/dy^{2} = -20 * (1/30) * -sqrt((dx/dy)^{2} + 1)

y * d^{2}x/dy^{2} = 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(u^{2}+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 yIntegrate them... (I had to use a table for the left one)

ln(u + sqrt(u^{2}+1)) = 2/3 * ln y + K

I've dispensed with the absolute value stuff, since u+sqrt(u^{2}+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(u^{2}+1) = C*y^{2/3}
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*50^{2/3}

C = 1/50^{2/3}

Put that back in...

u + sqrt(u^{2}+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(u^{2}+1) = (y/50)^{2/3} - u

u^{2}+1 = (y/50)^{4/3} - 2u(y/50)^{2/3} + u^{2}

Lucked out again. Cancel u^{2}'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*1^{5/3} - 1^{1/3} ] + 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.