This problem can be solved using related rates in calculus. Here are the steps:

1. First, we need to understand the problem and draw a diagram. The kite is 100 ft above the ground and 200 ft of string have been let out. This forms a right triangle with the ground, the kite, and the point directly below the kite. The angle we're interested in is the one at the ground between the string and the ground.

2. We can use trigonometry to relate the angle θ, the height y of the kite, and the length x of the string: cos(θ) = y/x. In this case, y = 100 ft and x = 200 ft.

3. We're asked to find dθ/dt, the rate of change of the angle with respect to time. To do this, we first differentiate both sides of the equation with respect to time t: -sin(θ) * dθ/dt = -y'/x + y*x'/x^2.

4. We know that the kite is moving horizontally at a speed of 3 ft/s, so x' = 3 ft/s. The kite's height y is not changing, so y' = 0.

5. Substituting these values into the equation gives us: -sin(θ) * dθ/dt = -0 + 100*3/200^2 = 0.0075.

6. We can solve for dθ/dt by dividing both sides by -sin(θ). But first we need to find the value of sin(θ). We can use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to find that sin(θ) = sqrt(1 - cos^2(θ)) = sqrt(1 - (100/200)^2) = sqrt(3)/2.

7. Finally, we find that dθ/dt = 0.0075 / -(sqrt(3)/2) = -0.005 ft/s.

So, the angle between the string and the horizontal is decreasing at a rate of -0.005 radians per second when 200 ft of string have been let out.

Question

This problem can be solved using related rates in calculus. Here are the steps:

1. First, we need to understand the problem and draw a diagram. The kite is 100 ft above the ground and 200 ft of string have been let out. This forms a right triangle with the ground, the kite, and the point directly below the kite. The angle we're interested in is the one at the ground between the string and the ground.

2. We can use trigonometry to relate the angle θ, the height y of the kite, and the length x of the string: cos(θ) = y/x. In this case, y = 100 ft and x = 200 ft.

3. We're asked to find dθ/dt, the rate of change of the angle with respect to time. To do this, we first differentiate both sides of the equation with respect to time t: -sin(θ) * dθ/dt = -y'/x + y*x'/x^2.

4. We know that the kite is moving horizontally at a speed of 3 ft/s, so x' = 3 ft/s. The kite's height y is not changing, so y' = 0.

5. Substituting these values into the equation gives us: -sin(θ) * dθ/dt = -0 + 100*3/200^2 = 0.0075.

6. We can solve for dθ/dt by dividing both sides by -sin(θ). But first we need to find the value of sin(θ). We can use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to find that sin(θ) = sqrt(1 - cos^2(θ)) = sqrt(1 - (100/200)^2) = sqrt(3)/2.

7. Finally, we find that dθ/dt = 0.0075 / -(sqrt(3)/2) = -0.005 ft/s.

So, the angle between the string and the horizontal is decreasing at a rate of -0.005 radians per second when 200 ft of string have been let out.

Knowee AI · Accepted Answer