Name that curve! What curve will a kayak describe if the paddler aims her bow at an object on a distant shore ahead and keeps the bow pointing to that object as she paddles toward it with constant velocity, in the presence of sideways drift due to constant current or wind? It's definitely not a straight line.  Is it a common curve? If not, has it got a name?
 A: $\def\r{{\bf r}}
\def\v{{\bf v}}$
As mentioned by Peter Tamaroff, this problem is a slight generalization of this one. 
Without loss of generality, assume the shore is at the origin, the initial $x$ coordinate is $x_0=1$, and the current is in the $y$ direction. 
The relevant differential equation is coupled and nonlinear, 
$$\begin{eqnarray*}
\frac{d\r}{dt} &=& \v_c + \v_r \\
&=& \v_c - v_r \frac{\r}{r},
\end{eqnarray*}$$
where $\r$ is the position of the paddler, $\v_c$ is the velocity of the current, and $\v_r$ the velocity of the rower. 
The condition $\v_r = - v_r \frac{\r}{r}$ is imposed since the rower is always rowing towards the origin. 
Note that $\v_c$ and $v_r$ are constant, but that $\r$ and $r$ (and thus $\v_r$) vary with time.
In components, 
$$\begin{eqnarray*}
\frac{d x}{d t} &=& - v_r \frac{x}{r} \\
\frac{d y}{d t} &=& v_c - v_r \frac{y}{r}.
\end{eqnarray*}$$
Thus, 
$$\begin{equation*}
\frac{d y}{d x} = \frac{y - \rho r}{x},\tag{1}
\end{equation*}$$
where $\rho = v_c/v_r$. 
Recall that $r = \sqrt{x^2+y^2}$.
With a clever change of variables, this differential equation is not too hard to solve. 
(Let $y(x) = x u(x)$ and integrate.) 
See here, for example. 
The result is 
$$y(x) = x \sinh(\sinh^{-1}(y_0) - \rho\log x).$$
This can be massaged somewhat.
We find 
$$\begin{equation*}
y(x) = \frac{x}{2} \left(\sqrt{1+y_0^2}\left(x^{-\rho}-x^\rho\right)
+ y_0\left(x^{-\rho}+x^\rho\right)\right).\tag{2}
\end{equation*}$$
As mentioned by Rahul Narain in the comments, this is a simple example of a pursuit curve.
Some key facts: 
$$\begin{eqnarray*}
y(1) &=& y_0 \\
y(x) &=& y_0 x, \textrm{ if }\rho=0 \textrm{ (we row in a straight line if } v_c = 0) \\
y(0) &=& 0, \textrm{ if } |\rho| < 1 \textrm{ (we make it to the shore if } v_c<v_r) \\
y(0) &=& \frac{1}{2}\left(y_0 + \sqrt{1+y_0^2}\right), \textrm{ if } \rho =1 
\textrm{ (we don't make it to the shore if } v_c\ge v_r) \\
y(x) &=& \frac{x}{2} \left(x^{-\rho}-x^\rho\right), \textrm{ if } y_0 = 0.
\end{eqnarray*}$$
The last result is equivalent to the one derived by robjohn in his excellent answer to the question referred to above. 
Below we give some sample trajectories. 
The current is the the positive $y$ direction. 

Figure 1. $y_0 = 0$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$


Figure 2. $y_0 = 0.15$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$


Figure 3. $y_0 = 2$, $\rho = 0,0.125,0.25,\ldots,1, 1.125,1.25$
