Related Rate: Kite Problem

Question

A kite 100 ft above the ground moves horizontally at a speed of 14 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

Answer 1

This Desmos graph gives a complete picture:

desmos.com/calculator/wuvkqrck6m

Be sure to go to full screen by collapsing the lefthand panel (click the << at top of that panel), when visiting the graph.

Let's say that Joe is holding the apparatus that let's out the string for the kite, and that initially the kite somehow is directly above Joe at a height of 100 feet. At that point in time the string is making a 90 degree angle with the ground. Now the wind starts moving the kite horizontally (so vertical height remains 100 feet)., at 14 ft/sec). In order to accommodate the movement Joe has to let out extra length of string (call its length s).

Picture the string, the horizontal distance the kite has travelled (but picture along the ground), and a vertical leg opposite the angle formed by the string and the ground forming a right triangle. The angle between the ground and string is θ.

cot θ = x/100 (x is the horizontal distance the kite has moved)

θ = arccot(x/100)

dθ/dt = [-1/(1+(x/100)²)] (dx/dt)

When s = 200, x² = 200² - 100² = 30000

Substitute 14 (ft/sec) for (dx/dt) and 30000 for x² and simplify. See the graph for more details.

There are a few other ways to do this problem (including using tangent instead of cotangent), but this seems most straightforward. Another way would be to take the derivative implicitly with respect to t.