Please explain how to solve this question?

This is an algebra problem. Let b be the angle in radians that the boat is facing. If b = 0 then it is pointed straight across the river. It will take 166 2/3 seconds to cross the river. During this time he will have drifted down 333 1/3 meters. It will take him 66 2/3 seconds to walk back. Total time is 233 1/3 seconds.

His 3 m/s velocity is composed of 3cos(b) in the direction straight across the river and 3sin(b) in the direction against the current of the river.

Lets look at another example. b = .7297 radians so the boat is aimed upstream some. Now his across velocity is 3cos(.7927) = 2.236 m/s. It will take 223.6 seconds to cross the stream. Now his against stream velocity is 3sin(.7926) = 2 m/s. This adds to 0 velocity with the river current. So he does not have to walk.

His time is better now than before. The trick is to express the time in terms of b as an equation. Then take the first derivative of this equation. So you will have dT / db = blah blah blah.

Set blah blah blah to 0 and solve. This will give a value for b that gives an extreme value for T. Just verify that this is an extreme low and not high.

♠ conventions: x-axis is along river flow; his destiny point p=j*0.500 km;

his path s=s1+s2, where s1=-i*5*t1 he walked up the river, s2=i*2*t2 +j*3*t2 he was carried down the river and rowing across;

♣ thus s=p; j*0.500 = -i*5*t1 + i*2*t2 +j*3*t2,

hence -i*5*t1 + i*2*t2=0, and j*0.500 = j*3*t2;

thus t2= 0.5/3, and 5*t1=2*0.5/3, or t1=0.2/3;

(a) he walked = 5*t1= 1/3 km;

(b) total time =t1+t2= 0.7/3 hours = (0.7/3)*60 = 14 min;

i dont know.