A man swims across a river 100ft wide, always heading for a tree directly across

Question

A man swims across a river 100ft wide, always heading for a tree directly across from his starting point. he can swim at a rate of 3ft/sec.

A) find equation of his path if the current is carrying him downstream at the rate of 1ft/sec, 3ft/sec, and 4ft/sec

B) draw graph of each equation.

answer as given by my book: y = 50[(x/100)2/3 - (x/100)4/3 ] , x2 =-200(y-50) , y = 50[(x/100)-1/3 - (x/100)7/3 ]

please show all work and assumptions in getting these answers, i have no idea how to do it lol

thank you!

Explanation / Answer

This is a nontrivial problem that results in a fairly "ugly" differential equation. The solution, however, ends up being a simply equation that has an interesting symmetry.

Set up this problem so that in rectangular coordinates, the river is parallel to the x-axis, with the far side of the river coincident with x axis. Initially, the swimmer is on the side of the river defined by the line y = W, where W is the width of the river. The initial coordinates of the swimmer are therefore y(t=0) = +W, x(t=0) = 0, and he wants to swim to the other side. His velocity vector is always aimed at a tree located at the origin. The river's current has a speed of C ft/sec in the positive x direction.

Let T(t) be the time-varying angle between the positive x axis and the line connecting the swimmer with the origin. Let Vm be the maximum speed at which the swimmer can swim.

The y-component of the swimmer's velocity is then given by:

dy/dt = -Vm * sin(T) = -(Vm * y)/sqrt(x^2 + y^2)

Similarly, the x-component of the swimmer's velocity is given by:

dx/dt = -Vm * cos(T) + C = -(Vm * x)/sqrt(x^2 + y^2) + C

These two differential equations (along with the initial conditions) specify the trajectory of the swimmer in parametric form (i.e., x and y are given as functions of t). What we are really interested in is the y position of the swimmer as a function of x, or vice versa. As it turns out, because of the way I've set up the coordinate system, it's easier to express x as a function of y in this case.

We can divide dx/dt by dy/dt to obtain:

dx/dy = [(Vm*x)/sqrt(x^2+y^2) - C]/[(Vm*y)/sqrt(x^2+y^2)]

dx/dy = x/y - (C/Vm)*sqrt((x^2 + y^2)/y))

dx/dy = x/y - (C/Vm)*sqrt((x/y)^2 + 1)

Now make the substitution u = x/y

x = u*y

dx/dy = u + y*du/dy

With this substitution, the differential equation becomes:

y * du/dy = -(C/Vm)*sqrt(u^2 + 1)

This is a separable equation:

du/sqrt(1+u^2) = -(C/Vm) * dy/y

The left hand side is one of those integrals that one looks up in a table (at least I do) . See source.

asinh(u) = -(C/Vm)*ln(y) + B = ln(y^(-C/Vm)) + B

where B is a constant of integration.

The inverse hyperbolic sin can also be written as (see source):

asinh(z) = ln(z + sqrt(1+z^2))

So we can also write the solution as:

ln(u + sqrt(1+u^2)) = ln(y^(-C/Vm)) + B

u + sqrt(1+u^2)) = B*y^(-C/Vm)

sqrt(1+u^2)) = B*y^(-C/Vm) - u

Square both sides:

1 + u^2 = (B^2)*y^(-2C/Vm) - 2*u*B*y^(-C/Vm) + u^2

(B^2)*y^(-2C/Vm) - 1 = 2*B*y^(-C/Vm)

[B*y^(-C/Vm) - (1/B)*y^(C/Vm)]/2 = u

Now back substitute for u:

x/y = [B*y^(-C/Vm) - (1/B)*y^(C/Vm)]/2

x = y*[B*y^(-C/Vm) - (1/B)*y^(C/Vm)]/2

x = [B*y^(1-C/Vm) - (1/B)*y^(1+C/Vm)]/2

This is the general solution. Now use the initial condition (x = 0 when y = W) to solve for B:

0 = [B*W^(1-C/Vm) - (1/B)*W^(1+C/Vm)]

0 = (B^2)*W^(1 - R) - W^(1 + R)

where R = C/Vm

B^2 = W^(1 + R - 1 + R) = W^(2R)

B = W^R = W^(C/Vm)

The solution that satisfies the initial condition is therefore:

x = [(W^(C/Vm) * y^(1-C/Vm) - W^(-C/Vm) * y^(1+C/Vm)]/2

x = [(W^(R) * y^(1-R) - W^(-R) * y^(1+R)]/2

The fact that C and Vm only appear in the solution as a ratio (R), indicates that only their relative, not their absolute magnitudes are important to the behavior of the solution.

You can plug in the appropriate values of W (100 ft), and the different values of Vm (1, 3, and 4 ft/sec) to get the different trajectories.

You can easily calculate how far downstream of the tree the swimmer will be when he finally reaches the opposite bank. At that point, y = 0, so:

x(y = 0) = [(W^(R) - W^(-R)]/2

Note that only in the case of C = 0 (so R = 0) does the swimmer actually end up at the tree when he reaches the opposite shore.

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