a fish is reeled in at a rate of 2m/min from a point 12m above the water. There is a total of 15m of line out. At what rate is the fish moving at the same point?
let the distance between the fish and the fisher is x then we have x^2+12^2 = s^2, where s is the length of the line out
take derivative from both sides, 2x*dx/dt = = 2s*ds/dt
x*dx/dt = s*ds/dt
dx/dt = s/x*ds/dt = 15/x *2 =30/x
x = sqrt(15^2-12^2) = sqrt(225-144) = 9
dx/dt = 30/9 = 10/3 m/minutes
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