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?

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• 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