Question

Four flies sit at rhe corners of a squre table facing inwards. They start walking simultaneously at the same rate each directing it's motion steading towards the fly on its right .find the plane of each fly .

Answer 1

Each fly walks with a velocity v always towards the fliy at its right: always. Side of the square is 2a. half diagonal is a√2.

The answer is

$r = a\sqrt{2} e^{-\theta\sqrt{2}}, \ \ \ \ \ \ \frac{dr}{dt} = - v, \ \ \ \ \frac{d\theta}{dt} = v/r\sqrt{2} \\ \\ r\ is\ the\ distance\ of\ fly\ from\ the\ center\ of\ Square, \\ \theta\ is\ the\ angle\ made\ by\ position\ vector\ r\ of\ the\ fly\ with\ x\ axis.$

Initially, the fly at left bottom of square as shown in the diagram walks v Δt along the right side. The fly to the right walks upwards by v Δt. The change in the direction of the fly now is $d\theta = \frac{v dt}{ r\sqrt{2} }$  and it is in the direction of origin.  so $\sqrt{2} r\ d\theta\ =\ dr\ =\ -v\ dt\ or\ \frac{dr}{dt}\ = -v$

$\frac{-dr}{dt} = r \sqrt2 \frac{d\theta}{dt} \\ \\ \frac{dr}{r} = -\sqrt2 * d\theta \\ \\ Ln\ \frac{r}{r_0} = -\sqrt2 (\theta - \theta_{0}) \\ \\ r_0 = \frac{a}{\sqrt2}\ \ and\ \ \ \theta_0\ = \ 0^0, Say\ so \\ \\ r = \sqrt2 \ a\ e^{-\sqrt2\ \theta}$