A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle of elevation between the string and the horizontal decreasing when 200 ft of string have been let out?
To find the rate at which the angle of elevation is decreasing, we need to find an expression for the rate of change of the angle with respect to time. Let's call this rate of change dθ/dt.
To do this, we'll use the concept of similar triangles. Let's consider the triangle formed by the kite, the string, and the ground. At any given time, this triangle will have a constant shape, regardless of the length of the string.
Let's denote the height of the kite above the ground as h, and the length of the string let out as s. We're given that h = 100 ft and ds/dt = 8 ft/s.
We can use the Pythagorean theorem to relate h, s, and the distance x along the ground between the base of the kite and the point directly below it:
x^2 + h^2 = s^2
Differentiating both sides of this equation with respect to time t, we get:
2x(dx/dt) + 0 = 2s(ds/dt)
Since we're interested in dx/dt, the rate at which x is changing with respect to time, let's solve this equation for dx/dt:
dx/dt = (s(ds/dt)) / x
To find dx/dt when 200 ft of string have been let out (s = 200 ft), we need to find the value of x. This can be done using the Pythagorean theorem:
x = sqrt(s^2 - h^2)
= sqrt(200^2 - 100^2)
x = sqrt(40000 - 10000)
x = sqrt(30000) ≈ 173.21 ft
Now we can substitute the values of s, ds/dt, and x into the equation for dx/dt:
dx/dt = (200 ft * (8 ft/s)) / (173.21 ft)
≈ 9.2 ft/s
Therefore, the rate at which the angle of elevation between the string and the horizontal is decreasing when 200 ft of string have been let out is approximately 9.2 ft/s.