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An angler hooks a trout and reels in his line at 3 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 16 ft from the angler ... 11 ft Decreasing at 3 in./s Let x be the horizontal distance from the angler to the fish and z be the length of the fishing line, where both x and z are measured in inches. Write an equation relating x and z​

User Ser
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Equation relating x and z is z² = x² + 121.

The horizontal speed of the fish when it is 16 ft from the angler is 4.14 in/s.

How the horizontal speed of the fish is calculated.

Let x be the horizontal distance from the angler to the fish and z be the length of the fishing line, where both x and z are measured in inches.

This arrangement form right triangle

using Pythagorean theorem

z² = x² + 11²

z² = x² + 121 -----------1

Let differentiate the equation with respect to time t.

d/dt(z²) = d(x² + 121)/dt

2zdz/dt = 2xdx/dt + 0

Make dx/dt subject of formula

dx/dt = (z/x)*dz/dt

Given that

The angler hooks a trout and reels in his line at 3 in /s.

The distance between the fish and angler is 16 ft

Substitute into z² = x² + 121

16² = x² + 121

256 = x² + 121

x² = 256 -121

= 135

x = √135 = 11.6 ft

Let us convert to same unit

1ft = 12 inches

z = 16 ft = 192 inches

x = 11.6 ft = 139.2 inches

dz/dt = 3 in/s (given)

Substitute into dx/dt = (z/x)*dz/dt

dx/dt = (192/139.2)*3

= 4.14 in/s

Therefore, the horizontal speed of the fish when it is 16 ft from the angler is 4.14 in/s.

An angler hooks a trout and reels in his line at 3 in /s. Assume the tip of the fishing-example-1
An angler hooks a trout and reels in his line at 3 in /s. Assume the tip of the fishing-example-2
User Cschol
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