A Kite 60 feet above the ground moves horizontally at a speed of 7 ft/s. At what rate is the angle between the string and the ground decreasing when 120 feet of string is out?

Answer 1

at r=120, θ=π/6,,,t=((√3)/2)*120/7,,,y=60,,,,x=[(√3)/2]*120

have tan(θ)=60/(((√3)/2)*120+7t)

sec²(θ)θ'=(-60*7)/(((√3)/2)*120+7t)²,,,derivative wrt to t

at r=120

θ'=(3/4)(-60*7)/[(120²(√3)²]

=-1260/43200

=(-.03) radian/s approximately

=(-1.7) degrees/s approximately

3/4=1/sec²(π/6)

Answer 2

Maria,

Let A be the point where the person holding the cl the kite is standing. Let B be the point 60 ft above point A. Let point C at the point where the kite currently is. 

In the right triangle ABC the leg BC = 7 x t why the leg AB is 60 ft and independent on time. No according to Pythagorean theorem the hypotenuse 

1. AC=Sqrt(AB^2 + BC^2 ) (1)

On the other hand the angle between the line AC and the ground is equal to the angle BCA which is equal to 

1.  θ(t)=arctan(60/7t) (2)    

from (2) find derivative of over time

1. dθ(t)/dt=1/(1+(60/7t)^2 x (-60/7t^2) (3)

, from 

AC=120=Sqrt(60^2+7^2 x t^2) (4)

find the time t to be then used in (3) to get the result

Hope it is helpful

Dr Ariel B.