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

Answer:

-2.26×10^-4 radians

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

Answer 2

Answer:

0.11 rad/s

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

$= 2y\frac{dy}{dt} + 2x\frac{dx}{dt} \\y\frac{dy}{dt} = -x\frac{dx}{dt} \\100\frac{dy}{dt} = \sqrt{3} * 100 * \frac{dx}{dt} \\\frac{dy}{dt} = 11\sqrt{3}$

Now Lcosθ = x

diferentiate

$Lsin\theta * \frac{d\theta }{dt} = \frac{dx}{dt} \\200 * \frac{100}{200} * \frac{d\theta}{dt} = 11\\\frac{d\theta }{dt } = 0.11 rad/s$