Question

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. ...?

Answer 1

Answer:

v = 4.66 inch/s

Explanation:

As we know that the height of the rod and the distance from the fisherman is related to each other as

$x^2 + y^2 = L^2$

now differentiate it with respect to time

$2x\frac{dx}{dt} + 0 = 2L\frac{dL}{dt}$

here we know that

$\frac{dL}{dt} = 4 inch/s$

now when

x = 20 ft  and y = 12 ft then the length of the line is given as

$L^2 = 12^2 + 20^2$

$L = 23.3 ft$

now from above relation

$\frac{dx}{dt} = \frac{L}{x}(\frac{dL}{dt})$

now we have

$v_x = \frac{23.3}{20} (4 inch/s)$

$v_x = 4.66 inch/s$

Answer 2

 Let x be the distance of the fish to the fisherman and h be the height of the tip of fishing rod then we want to fiind dx/dt when x = 20. By property of similar triangle we have 12/x = h/20 --> h = 240/x 


Differentiating w.r.t. we obtain dh/dt = (-240/x^2)*dx/dt --> dx/dt = [(x^2)/240]*dh/dt 


At x = 20, dh/dt = 4 --> dx/dt = (-400/240)*4 = 20/3


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