- #1

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P.S Im still looking for number 2 on my other post. I thought I had the answer but it was wrong.

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- Thread starter Matthew
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- #1

- 7

- 0

P.S Im still looking for number 2 on my other post. I thought I had the answer but it was wrong.

- #2

HallsofIvy

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I am going to take it that Jill will remain where she is and Jack wants to get to her position in the shortest time.

Let theta be the angle at which he rows relative to the straight line across the river (directly across would be theta= 0, aiming downstream positive, upstream negative). Since his speed in still water is 6 km/hr, his "velocity vector" would be (6 sin(theta), 6 cos(theta)). Since the river is flowing at 3 km/, its velocity vector is (3, 0) and Jack's "velocity made good" is (6 sin(theta)+ 3, 6 cos(theta)). He will cross the river in time t

6 cos(theta)t

At that time, his position on the far shore will be (6 sin(theta)+ 3)sin(theta), positive if downstream from Jill, negative if upstream.

In any case, he now has that distance to run at 10 km/h

That will require time t

You want to minimize t

1/(6 cos(theta))+10(6 sin(theta)+ 3)sin(theta).

Differentiate with respect to theta and set equal to 0.

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