A river is flowing with the velocity of 5Km/hr as shown in the figure. A boat starts from A and reaches the other bank by covering the shortest possible distance. If the velocity of the boat is 3km/hr in still water then distance boat covers is?

[Delete this answer]

• Let us assume that the distance covered by the boat to cross the river will be minimum, if it directs its velocity making an angle #theta # with the vertical line AB.

• Then the velocity of boat along AB will be #v_(AB)=vcostheta=3costheta # km/h.

• So time to cross the breadth #AB =300m=0.3km# of the river will be

• #t=(AB)/ v_(AB)=0.3/(3costheta)=0.1sectheta# hr

• The actual velocity of the boat will be
  #v=sqrt(3^3+5^2+2*3*5*cos(90+theta))#
  #=>v=sqrt(34-30sintheta)# km/h

• So distance traveled to cross the river
  #s=vxxt#
  #=>s=sqrt(34-30sintheta)xx0.1sectheta......[1]# km
  #=>100s^2=(34-30sintheta)xxsec^2theta#

Differentiating w r to #theta#

we get

#200s(ds)/(d theta)=(34-30sintheta)xx2sec^2thetatantheta-30costheta xxsec^2theta#

Imposing the condition of minimization #(ds)/(d theta)=0# we get

#(34-30sintheta)xx2sec^2thetatantheta-30costheta xxsec^2theta=0#

Multiplying both sides by #cos^3theta# we have

#(34-30sintheta)xx2sintheta-30cos^2theta=0#

#=>(34-30sintheta)xxsintheta-15cos^2theta=0#

#=>(34-30sintheta)xxsintheta-15(1-sin^2theta)=0#

#=>(34sintheta-30sin^2theta-15+15sin^2theta)=0#

#=>15sin^2theta-34sintheta+15=0#

So

#sintheta=(34-sqrt(34^2-4xx15xx15))/30=18/30=3/5#

This gives # sectheta=1/costheta=1/sqrt(1-sin^2theta)#

#=1/sqrt(1-(3/5)^2)=5/4#

Inserting #sintheta and sectheta # in equation [1] we get the
distance traveled by the boat to cross the river

#=>s=sqrt(34-30sintheta)xx0.1sectheta# km

#=>s=sqrt(34-30xx3/5)xx0.1xx5/4# km

#=>s=sqrt(34-18)xx0.1xx5/4# km

#=>s=4xx0.1xx5/4=0.5# km