a fisherman hooks a trout and reels in his line at speed of 4 inches/second. assume tip of fishing rod is 12 ft above the water directly above the fisherman and the fish is pulled horizontally directly towards the fisherman. Find the horizontal speed of the fish when it is 20 ft from the fisherman.

Question

anonymous · Answer

OK, I'm not sure this is exactly right, but hopefully you have the answer to check against.
cos (theta) = x/r, so
x = r cos (theta)
We want dx/dt, so
dx/dt = (dr/dt) cos (theta)
dr/dt is the rate of reeling in the line.  Converting inches to feet we use 0.25 ft/sec. So,
dr/dt = 0.25 cos (theta)
cos (theta) is adjacent over hypotenuse.  Use the conditions of the triangle to solve for the hypotenuse.  When the fish is 20 ft from the boat you have hyp^2 = 20^2 + 12^2.

Solving give hyp = sqrt(544).  So,

dx/dt = 0.25 (20/sqrt(544))
dx/dt = 0.214 ft/sec.  Then, converting to inches gives
dx/dt = 2.57 in/sec.

So, if I haven't screwed up, the horizontal speed of the fish when it's 20 ft from the boat is 2.57 ft/sec.