Hello hooke Originally Posted by

**hooke** A canal of width 2a has parallel straight banks and the water flows due north . The points A and B are on opposite banks and B is due east of A , with the point O as the midpoint of AB . THe x-axis and y-axis are taken in the east and north directions respectively with O as the origin . The speed of the current in the canal , $\displaystyle v_c=v_o(1-\frac{x^2}{a^2}) $, where $\displaystyle v_o$ is the speed of the current in the middle of the canal and x the distance eastwards from the middle of the canal . A swimmer swims from A towards the east at speed $\displaystyle v_r $, relative to the current in the canal . Taking y to denote the distance northwards travelled by the swimmer , show that

$\displaystyle

\frac{dy}{dx}=\frac{v_o}{v_r}(1-\frac{x^2}{a^2})

$

Draw a triangle of velocities such that the velocity of the swimmer relative to the ground = velocity of swimmer relative to the water plus the velocity of the water relative the ground.

This is a right-angled triangle, in which the velocity of the swimmer relative to the ground has a component $\displaystyle v_r$ eastward and $\displaystyle v_c$ northward. So, if $\displaystyle (x,y)$ is his position relative to the ground at time $\displaystyle t$, we get: $\displaystyle \frac{dx}{dt}= v_r$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{v_r}$

and$\displaystyle \frac{dy}{dt}=v_c=v_o\left(1-\frac{x^2}{a^2}\right)$

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$ $\displaystyle =\frac{v_o}{v_r}\left(1-\frac{x^2}{a^2}\right)$

Grandad