Question

A kite 100 ft above the ground moves horizontally at a speed of 7 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

The rate at which the angle is changing is determined as - 0.0175 rad/s.

How to calculate the rate at which the angle is changing?

The rate at which the angle is changing is calculated by applying the following formula as shown below.

From the right triangle attached;

tan θ = y / x

So, cot θ = x / y

cot θ = x / 100

we will take the derivative of both sides of the equation;

- csc²θ (dθ/dt) = 1/100 (dx / dt)

Since we are looking for the rate of change of the angle, we will divide both sides by "- csc²θ".

`(d\theta )/(dt) = ((1)/(100) * (dx)/(dt) )/(-(csc \theta) ^2)`

But cscθ = 1/sin θ

sin θ = 100/200

sin θ = 1/2

cscθ = 2

Also, we are given dx/dt = 7 ft/s

Now, we will calculate the rate at which the angle is decreasing;

`(d\theta )/(dt) = ((1)/(100) * (dx)/(dt) )/(-(csc \theta) ^2)\\\\(d\theta )/(dt) = ((1)/(100) * 7 )/(-(2) ^2)\\\\(d\theta )/(dt) = (7)/(-100 * 4) \\\\(d\theta )/(dt) = - (7)/(400) \ rad/s\\\\(d\theta )/(dt) = - 0.0175 \ rad/s`

Answer 2

Answer:0.07 rad/s

Step-by-step explanation:

Given

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`

Differentiating both side

`0=2y\frac{\mathrm{d} y}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}`

`y\frac{\mathrm{d} y}{\mathrm{d} t}=-x\frac{\mathrm{d} x}{\mathrm{d} t}`

`100* \frac{\mathrm{d} y}{\mathrm{d} t}=-√(3)* 100* \frac{\mathrm{d} x}{\mathrm{d} t}`

`\frac{\mathrm{d} y}{\mathrm{d} t}=7√(3)`

Now
`Lcos\theta =x`

Differentiating

`Lsin\theta * \frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{\mathrm{d} x}{\mathrm{d} t}`

`200* (100)/(200)* \frac{\mathrm{d} \theta }{\mathrm{d} t}=7`

`\frac{\mathrm{d} \theta }{\mathrm{d} t}=0.07 rad/s`