I'm having trouble with this math problem, can someone please help, its urgent!?

T = angle with respect to the perpendicular that they aim the boat

D = width of the river (stream, whatever)

U = velocity perpendicular to the shore = Vm*cos(T)

W = velocity parallel to the shore = Vc - Vm*sin(T)

t = time to cross = D/U = D/[Vm*cos(T)]

s = distance moved along the shore during the time t = t*W = t*[Vc - Vm*sin(T)]

s = D*[Vc - Vm*sin(T)]/[Vm*cos(T)]

(s/D) = [Vc - Vm*sin(T)]/[Vm*cos(T)] = (Vc/Vm)/cos(T) - tan(T)

We want to minimize s which is the same as minimizing s/D so take the derivative.

d(s/D)/dT = (Vc/Vm)sin(T)/cos^2(T) - sec^2(T)

And set equal to 0.

0 = (Vc/Vm)sin(T)/cos^2(T) - sec^2(T)

(Vc/Vm)sin(T)/cos^2(T) = sec^2(T) = 1/cos^2(T)

(Vc/Vm)sin(T) = 1

sin(T) = Vm/Vc and T = arcsin(Vm/Vc)

So the answer is A

Check. First of all we can eliminate c since this does not give an angle. All I can offer are qualitative arguments. If Vc is very, very large then it would be best to aim the boat almost directly across the river so the time to cross will be minimized. So T -> 0 as Vc gets larger and the arcsin(Vm/Vc) is the only answer that meets this criteria since this goes to arcsin(0) which is an angle of 0. Answers B and D, on the other hand, go in the opposite direction and approach 90 degrees. In the opposite case, when the speed of the current approaches that of the boat, you want to aim the boat almost directly against the current to minimize distance parallel to the shore that the boat moves.

section a, i think of you have have been given a challenge. Your deliver is moving east and south (2i - 3j), and you're announcing that is moving alongside a bearing of 70°, it particularly is someplace between east and north. so as that would no longer be precise. ....2 ------->| ...?...| 3 .........| ......?.V tan ? = 3/2 ? = arctan(3/2) = fifty six° and because ? is measured from east, we'd desire to function yet another ninety° to it, and we get the bearing 146°. section b is solid, different than i might verify that if my time is capital T, i might use capital T in my place vector equations. i might take "due north-east" to point precisely between north and east, it particularly is, from deliver R's place, the path to deliver S is a 40 5° bearing. Does that help? are you able to take it from right here?

tan x= ( vm/vc) [no match]

arctanx= 1( vm/ vc )= vc/vm

It's B