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how to find the shortest path travelled by a boatman ,when,VELOCITY OF THE RIVER "vR" IS GREATER THAN THE VELOCITY OF THE BOATMAN W.R.T THE RIVER "vBR".(vR>vBR).
MY BOOK SAYS TO TAKE "dx/dΦ=0" {x=min displ.,Φ=angle w.r.t y-axis}.BUT I DO NOT UNDERSTAND THE PURPOSE OF IT.IS IT DUE TO MAXIMA AND MINIMA????
Dear aruka
consider the situation as given in diagram
time taken to cross the river is t=d/vcosΘ
so distance AC =(u-vcosΘ)t
= (u-vcosΘ)d/vcosΘ
so distance travel OC =x =[d2 + {(u-vcosΘ)d/vcosΘ}2]1/2
so distance travel is dependent upon the angel Θ ,so for minimum value of x differentiate x w.r.t Θ
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
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