Question

A kite 100 ft above the ground moves horizontally at a speed of 11 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

0.11 rad/s

Step-by-step explanation:

Given that,

kite is 100 ft high

and moves horizontally at 7 ft/s

Total string let out =200 ft

String length(l), vertical(y) & Horizontal(x) distance of kite will form a right angle triangle

`L^2 = y^2 + x^2`

differentiate both side

`= 2y(dy)/(dt) + 2x(dx)/(dt) \\y(dy)/(dt) = -x(dx)/(dt) \\100(dy)/(dt) = √(3) * 100 * (dx)/(dt) \\(dy)/(dt) = 11√(3)`

Now Lcosθ = x

diferentiate

`Lsin\theta * (d\theta )/(dt) = (dx)/(dt) \\200 * (100)/(200) * (d\theta)/(dt) = 11\\(d\theta )/(dt ) = 0.11 rad/s`

Answer 2

-2.26×10^-4 radians

Step-by-step explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians