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abhishek sinha

Done??

Now see the ultimate general method !!

I have already posted a similar but a more general problem long ago and now I a going to use it. Let me post the whole thing again for conveneience :

The problem was :

Two particles A & B start from positions ( 0, 0 ) & ( 0, -d ) and move with constant speeds v & u respectively . A moves along x - axis and B moves such that its velocity is always aimed at A . Let r be the distance between them and be the angle made by the velocity of B with X - axis , at some time t .

Prove that ,

$\frac{r}{d}=\frac{(1-\cos\theta)^{\frac{u}{v}}}{(\sin\theta)^{\frac{u}{v}+1}}$

So it is a more general case . The r for a large time is found by letting $\theta$ tends to zero ( as they now follow each other through a st. line ).

Now if we put u=v (as given here ) then taking the limit , it follows directly that r=d/2.

Now the derivation of the formula ( as done by me )is as follows:

see, the controlling differential eqns are clearly.........

dx/dt = u cos ..................... ( 1 )

dy/dt = usin .......................( 2 )

y/( x - vt) = tan ....................... ( 3 )

y = - r sin .................. ( 4 )

Now the task is to solve these coupled differential eqn .

From ( 3 ) we have

                  x - vt = y cot

dwrt 't'

                   dx/dt -v = dy/dt cot - y cosec^2() d/dt

combining ( 1 ) & ( 2 ) with this eqn we have

                     u cos   - v = u sin cos /sin + r cosec d/dt

or ,                 d/dt = -v/r sin .................. ( A )

differentiating ( 4 ) wrt 't ' , we get

                   u sin = - d/d ( r sin ) d/dt

combining with ( A ) we get

               u/v = cos + sin 1/r dr/d

or ,         u/v cosec - cot = 1/r dr/d

Now integrating it and putting the boundary condition that r= d at = /2

we get



                               $\frac{r}{d}=\frac{(1-\cos\theta)^{\frac{u}{v}}}{(\sin\theta)^{\frac{u}{v}+1}}$

What do u say about that , Kayamant?

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