#### To determine

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

#### Answer

The rate change of the angle between the ground and the string is dθdt=−150 rad/s.

#### Explanation

**Given:**

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

The speed is, dxdt=8 ft/s.

**Formula used:**

Chain rule: dydx=dydu⋅dudx

**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,

cot θ=BCCAcot θ=x100x=100cot θ .

Differentiate with respect to the time t,

ddt(x)=ddt(100cot θ)dxdt=−100cosec2θdθdt [Qddθ(cot θ)=−cosec2θ]dθdt=−1100⋅1cosec2θ=−1100sin2 θ [Q1cosec2θ=sin2θ]

From the ,

sin θ=CABA=100y=100200 [Qy=200]=12

Substitute sin θ=12 and dxdt=8 in dθdt,

dθdt=−1100(12)2(8)=−8100×4=−150 rad/s

Therefore, the rate change of the angle between the ground and the string is dθdt=−150 rad/s .

That is, the angle decreases at a rate of 150 rad/s.