On a nice summer's day a stream carries water downstream at a uniform speed of a. The banks of the stream are parallel and are separated by a distance c. Captain Jack Sparrow is on one side of the bank with his ship, the Black Pearl, perpendicular to it. Directly across from his ship, on the other bank is a pot of gold.
Jack Sparrow pulls out a compass which will always point towards the pot of gold. He keeps the the Black Pearl always headed in that direction (despite the stream pushing his ship downstream) and sets off with constant speed b.
What is the path of the boat, considering that the boat may move faster, slower, or at the same speed as the stream?
Note: The boat's path can be modeled using the Cartesian coordinate system: let the y axis be the bank with the gold and let the line x=c be the opposite bank, with the Black Pearl starting at (c,0) and the gold at (0,0).
let y(x) be the function defining sparrows path, then we have y(c)=0. now doing some quick vector analysis we can come up with these 2 equations
dy/dt=(by/sqrt(x^2+y^2))a and
dx/dt=bx/sqrt(x^2+y^2)
then dy/dx=(dy/dt)/(dx/dt)
dy/dx=(y/x)(a/b)*sqrt(x^2+y^2)/x
now let f(x,y)=(y/x)(a/b)*sqrt(x^2+y^2)/x
f(k*x,k*y)=f(x,y) so this is a homogenous differential equation and can be solved using z=y/x and f(x,y)=f(1,z) and
dy/dx=x*(dz/dx)+z=f(1,z)
x*(dz/dx)+z=z(a/b)*sqrt(1+z^2)
x*(dz/dx)=(a/b)*sqrt(1+z^2)
1/sqrt(1+z^2) dz= (a/b)*(1/x) dx
arcsinh(z)+k=(a/b)*ln(x)
arcsinh(z)=k(a/b)*ln(x)
z=sinh(k(a/b)*ln(x))
y=z*x
y=x*sinh(k(a/b)*ln(x))
y(c)=0 so
0=c*sinh(k(a/b)*ln(c))
k(a/b)*ln(c)=0
k=(a/b)*ln(c) so thus we finally have
y(x)=x*sinh[ (a/b)*ln(c/x) ]

Posted by Daniel
on 20081222 16:39:08 