A kite 100ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 150 ft of string has been let out?

Question

A kite 100ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 150 ft of string has been let out?

Answer 1

From the problem description, we have a triangle with a constant height, 100ft, and variable hypotenuse and width, which I call s(t) (for "string") and x(t) respectively. Using our trig identities, we can relate the angle, theta, to the information that we have:

sin(theta) = 100ft / s(t)

Since we care about the rate of change, let's differentiate on both sides:

cos(theta) d(theta)/dt = 100ft * (-1/s(t)²) * ds/dt

Note that d(theta)/dt and ds/dt come from the chain rule. The good news is that we have d(theta)/dt, which is exactly what we're looking for. On the downside, we still need to find ds/dt and cos(theta). Let's start with ds/dt by using the pythagorean theorem:

s(t)² = (100ft)² + x(t)²

2s(t)*ds/dt = 2x(t)*dx/dt

ds/dt = x(t)/s(t) * dx/dt

You may notice that x(t)/s(t) represents cos(theta), since x(t) adjacent to theta, and s(t) is still our hypotenuse. Additionally, we know dx/dt = 8ft/sec from the problem statement. We can therefore rewrite as:

ds/dt = cos(theta) * 8 ft/sec

Substituting back in our original equation involving cos(theta), we have:

cos(theta) d(theta)/dt = 100ft * (-1/s(t)²) * ds/dt

cos(theta) d(theta)/dt = 100ft * (-1/s(t)²) * cos(theta) * 8 ft/sec

d(theta)/dt = 100ft * (-1/s(t)²) * 8 ft/sec

And finally, since we're evaluating when the length of the string is 150ft, we can plug in that piece of information:

d(theta)/dt = 100ft * (-1/(150ft)²) * 8 ft/sec

d(theta)/dt = -8/225 rad/sec