Question

A kite is 100 m above a horizontal (level) field and is moving horizontally (laterally) at a speed of 10 m/s. At what rate is the angle between the string and the horizontal changing when 250 m of string has been released?

Answer 1

To find the rate at which the angle between the string of a kite and the horizontal is changing, we can use trigonometry and related rates.

Step-by-step explanation:

To solve this problem, we can use trigonometry and related rates. Let's call the length of the string released at time t as x. It is given that the kite is moving horizontally at a speed of 10 m/s, so the horizontal distance it covers is also x. We can form a right triangle with the string as the hypotenuse, and the horizontal distance as the adjacent side. The angle between the string and the horizontal is given by θ, so we can use trigonometry to write:

tan(θ) = 100 / x

Now, we need to find dx/dt, the rate at which x is changing with respect to time. It is given that 250 m of string has been released, so we can substitute x = 250 into the equation and differentiate implicitly with respect to time:

sec^2(θ) * d(θ) / dt = -100 / (x^2 * dx / dt)

Now we can solve for d(θ) / dt, which is the rate at which the angle between the string and the horizontal is changing.