   Chapter 3.9, Problem 30E

Chapter
Section
Textbook Problem

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 200 ft of string has been let out?

To determine

To find: The rate of change of the angle between the string and the ground when 200 ft string is let out.

Explanation

Given:

Consider the height of the kite from the ground is 100 ft.

The speed is, dxdt=8ft/s.

Formula used:

Chain rule: dydx=dydududx

Calculation:

Let x be the horizontal distance travelled by the kite. And y be the length of the string from ground to kite, and θ be the angle between the string and horizontal ground, which is shown below in Figure 1.

From Figure1, A is the position of the kite and BC is the horizontal distance travelled by the kite and BA is the length of the string.

Since x, y and θ changes with time, x, y and θ are the function of the time t.

Obtain dθdt when 200 ft string is let out.

From Figure 1, consider the triangle BCA,

cotθ=BCCAcotθ=x100x=100cotθ .

Differentiate with respect to the time t,

ddt(x)=ddt(100cotθ)dxdt=100cosec2θdθdt[ddθ(cotθ)=cosec2θ]dθdt=11001cosec2θ=1100sin2θ

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