The problem I am trying to figure out is as follows:
A man initially standing at the point O walks along a pier pulling a rowboat by a rope of length L. The man keeps the rope straight and taut. The path followed by the boat is a curve called a tractrix and it has the property that the rope is always tangent to the curve. Show that the if the path followed by the boat is the graph of the function $$y=f(x),$$ then $$f'(x) = \dfrac{\sqrt{L^2-x^2}}{x}.$$ 
I am not really sure what I am supposed to do here. The picture above is the picture of the problem. So do I basically just have to find the original equation f(x) and do this by integration? Any help would be appreciated.
Best Answer
let $t$ be the angle that the rope makes with the $x$-axis. wee know that $$\tan t = f'(x), \cos t = \dfrac xL. $$ we can eliminate $t$ between these by using $$\frac 1{\cos^2 t }= 1 + \tan^2 t $$ which gives us $$ \frac{L^2}{x^2} = 1 + (f'(x))^2$$ if you clean this up, you should get an nice equation for $f'$ use the initial condition $f(L) = 0.$
$\bf edit:$ we can integrate the equation $$f'(x) = - \frac{\sqrt{L^2 - x^2}}{x}, \, f(L) = 0 $$ to get $$\begin{align}f(x) &= \int_x^L\frac{\sqrt{L^2 - x^2}}x \, dx \\ &= \int_t^0\frac{\sqrt{L^2 - L^2\cos^2 t}}{L\cos t}\left( \, -L\sin t \right) \,dt\\ &= L\int_0^t \frac{\sin^2 t}{\cos t} \, dt = L\int_0^t \frac{1}{\cos t} \, dt - L \sin t\\ &=L\ln (\sec t + \tan t) - L\sin t \end{align}$$