an angler hooks a trout and reels in his line at 4 in/s. assume the tip of the fishing rod is 11 ft above…

an angler hooks a trout and reels in his line at 4 in/s. assume the tip of the fishing rod is 11 ft above the water and directly above the angler, and the fish is pulled horizontally directly toward the angler (see figure). find the horizontal speed of the fish when it is 20 ft from the angler. when the fish is 20 ft from the angler, its horizontal speed is about (round to two decimal places as needed.)

an angler hooks a trout and reels in his line at 4 in/s. assume the tip of the fishing rod is 11 ft above the water and directly above the angler, and the fish is pulled horizontally directly toward the angler (see figure). find the horizontal speed of the fish when it is 20 ft from the angler. when the fish is 20 ft from the angler, its horizontal speed is about (round to two decimal places as needed.)

Answer

Explanation:

Step1: Establish the relationship

Let (x) be the horizontal distance of the fish from the angler and (z) be the length of the fishing - line. By the Pythagorean theorem, (z^{2}=x^{2}+11^{2}) (since the vertical height (y = 11) ft).

Step2: Differentiate with respect to time (t)

Differentiating both sides of (z^{2}=x^{2}+121) with respect to (t): [ \begin{align*} 2z\frac{dz}{dt}&=2x\frac{dx}{dt}\ z\frac{dz}{dt}&=x\frac{dx}{dt} \end{align*} ] We know that (\frac{dz}{dt}=- 4) in/s (negative because (z) is decreasing). First, convert units: (x = 20) ft (=20\times12 = 240) in, (y = 11) ft (=11\times12=132) in. Then find (z) using (z=\sqrt{x^{2}+y^{2}}=\sqrt{240^{2}+132^{2}}=\sqrt{57600 + 17424}=\sqrt{75024}=274) in.

Step3: Solve for (\frac{dx}{dt})

Substitute (z = 274), (\frac{dz}{dt}=-4), and (x = 240) into (z\frac{dz}{dt}=x\frac{dx}{dt}): [ \begin{align*} 274\times(-4)&=240\times\frac{dx}{dt}\ \frac{dx}{dt}&=\frac{-274\times4}{240}\ \frac{dx}{dt}&=\frac{-1096}{240}\approx - 4.57 \end{align*} ] The negative sign indicates the direction (towards the angler). The speed is the magnitude.

Answer:

(4.57) in/s