Solving a 1st-Order Diff Eq for Boating Path

[PhysicsMark]

Homework Statement

The following problem involve the setting up and solving of a first-order differential equation for a physical situation. Once, derived the equation itself is not difficult to solve.

A boater rows across a straight river of constant width "w", always heading (i.e., pointing the front of the boat) toward the position on the bank directly opposite the starting point. If the river flows with uniform speed "v" and if the speed with which the boater can row is also "v", find the equation of the path of the boat. How far downstream does the boater finally land? [Hint: if x is the cross-stream position of the boat and y is its downstream position, find an expression for dy/dx. Then solve for y(x).]

Homework Equations

All knowledge gained up to this point.

The Attempt at a Solution

Thanks for taking the time to read this. I would first like to point out (I assume most will understand this, but I didn't at first) that the path of the boat is a parabolic curve, not a straight line. I have been working at this for a couple of days. I believe I am extremely close. I think it would take a lot to explain all my steps, so I am going to try and write down the "important" parts and take it from there. If something is unclear, or you see something you do not agree with, I will explain what I did in detail and we can take it from there.Ok, drawing a picture and breaking down the velocities into components ( I drew a triangle here) I have the following EQ for dy/dx (This part wasn't obvious to me, though it may be for you):

$${\frac{dy}{dx}={\frac{sin(\theta)}{1-cos(\theta)}$$

Where:

$$cos(\theta)={\frac{y}{\sqrt{(w-x)^2+y^2}}}$$

and

$$sin(\theta)={\frac{w-x}{\sqrt{(w-x)^2+y^2}}}$$

After getting here, I got stuck for awhile. I tried different algebraic manipulations but in the end was not able to get the y^2 term out. Based on that I started to think the equation was not separable (by methods that I know). So I made a substitution.

$$z=w-x$$

$$dz=-dx$$

First I inserted the "z" substitution here(I made more than a few manipulations with the equation but I don't think they're hard to follow):

$${\frac{dy}{dx}={\frac{z}{\sqrt{z^2+y^2}-y}}$$

Then I separated the Equation and made the dz=-dx substitution.

$$(\sqrt{z^2+y^2}-y})dy=-zdz$$

Ok so here is where I am now. First; I am not sure if everything from the original dy/dx equation until now is correct. I am sure about the dy/dx (at least according to my professor. I am of course open to criticism). Second; I do not know where to go from here. I was thinking of using a method we have learned about solving inexact differentials. I am not sure if I can use this because my book seems to imply that you must have functions of 2 variables where I do not have functions of 2 variables. If someone has an idea, I would love to hear it. Thanks again.

 

[willem2]

What you did is correct. Actually a function that depends only on z is also a function of 2 variables: f(y,z) = -z
I don't think you can solve it as an exact differential equation.

I wrote the differential equation in polar coordinates. (with the target point as the origin)
and got a differential equation for r and $\phi$ that was separable.
You should be able to solve your equation by substituting

$$r = \sqrt{y^2 + z^2}$$ and

[tex] \cos {\phi} = \frac {y}{r} [/itex]

 

[PhysicsMark]

What you did is correct. Actually a function that depends only on z is also a function of 2 variables: f(y,z) = -z
I don't think you can solve it as an exact differential equation.

I wrote the differential equation in polar coordinates. (with the target point as the origin)
and got a differential equation for r and $\phi$ that was separable.
You should be able to solve your equation by substituting

$$r = \sqrt{y^2 + z^2}$$ and

[tex] \cos {\phi} = \frac {y}{r} [/itex]

Do you mean changing to polar from the beginning?