I'm having a hard time solving this problem:
Let there be a town $A$ in a shore of a river. Let $x=0$ be the shore. Let $(0,0)$ be the location of the town. Let $B$ be another town, in the oppossite shore, $x=b$, and let the town be in $(b,0)$. Suppose a person from town $A$ goes in a boat with velocity $v$ to $B$, always poiting at $B$ (see the picture), and let the river flow in the positive direction of $y$ with velocity $u$. Find the curve that gives the person's trayectory over time.
In an arbitrary point of the curve there will always be a system of three vectors, $v$, $u$ and $w=v+u$ in the following manner, where $w$, the tangent vector, is $u+v$. Their modulus is constant, thus since $v+u$ varies, the variables here are the angles and the modulus. Remember that $v$ always points at $B$. IN the image, the lengths of $v$ and $v_1$ and of $u$ and $u_1$ should be the same (since they're the same vector).
You can see that $v$ is "pushing" to get to $B$ but the vector $u$ will always modify $v$'s direction (and thus the man's), making the tangent actually be $w = u+v$. The dotted parallel lines are a reference to the tangent angle which is that between the dotted line and $w$. My approach is:
$ \tan \theta = \frac{dy}{dx}$
and the modulus of $v+u$ will satisfy, being a function of time.
$|| v+u||(t) = \frac{ds}{dt}$
I just need then to find a way of relating the modulus to the angle to find the solution.
Another thing that comes to mind is thinking as the solution in a parametric way, thus first finding $\frac{dy}{dt} = f(t) $ and $\frac{dx}{dt} = g(t) $ then taking the quotient and integrating.