A difficult boat and river problem

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SUMMARY

The discussion focuses on solving a physics problem involving a boat crossing a river with a constant speed, u, while the river flows at speed nu. The key equations derived include dr/dt = u(1 - n sin(theta)) and two additional equations relating the velocities in polar coordinates: nu sin(theta) - u = dr/dt and nu cos(theta) = r d(theta)/dt. The participants clarify the interpretation of velocities and provide insights into the integration process necessary to find the equation of the path, r = f(theta).

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Homework Statement


A boat is rowed with constant speed u starting from a point A on
the bank of a river of width d , which flows with a constant speed
nu . The boatman always points the boat at a point O on the other
side of the bank opposite to A. Find the equation of the path
r = f (theta ) for the boat.

Homework Equations


dx/dt = v

The Attempt at a Solution


I was able to find dr/dt in terms of theta. here is my expression dr/dt= u(1-nsin(theta)). The problem is when I integrate it i have to express dt in terms of d(theta) which I cannot find.
 

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I don't understand the problem much. Is it the boat's velocity relative to the ground, or its velocity relative to the water, which always points towards the water? I guess it's the latter, because if n>1 then there is no chance that the boatman can guide the boat towards O.

Let's assume so. I think your equation is a little bit incorrect (just a little bit, most is fine). First, you should write down the formula of velocity in polar coordinates, i.e. write down \vec{v} in term of r , \theta and the unit vectors \vec{e}_r and \vec{e}_{\theta}. Then use the condition that \vec{v} is the sum of 2 velocity: u in the direction towards O and nu in the direction to the left. You should arrive at these 2 equations:

nusin\theta-u=\frac{dr}{dt}

nucos\theta=r\frac{d\theta}{dt}

From here, you can solve for r(\theta).
 

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can u explain me how u arrive at the second equation. I don't understand it
 
Can you do step 1, i.e. writing down the formula of velocity in polar coordinates?
 
oh ok I got it. I didnt think of v=r(omega). Thx...
 

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