A kite 100 ft above the ground moves horizontally at a speed of 9 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

let the angle be Ø, and the horizontal distance be x
given : dx/dt = 9
find dØ/dt

we have a right angled triangle, and
tanØ =100/x
xtanØ = 100

x sec^2 Ø dØ/dt + tanØ (dx/dt) = 0 ***

when 200 ft are let out,
x^2 + 100^2 = 200^2
x^2 = 300
x = 10√3
secØ = 10√3/100 = √3/10
sec^2 Ø = 3/100
tanØ = 100/10√3 = 10/√3

in ***
(10/√3)(3/100) dØ/dt + (10/√3)(9) = 0

solve for dØ/dt

Answer 2

sorry I didnt get whats the answer

Answer 3

oh, please -- do some of the work, okay?

(10/√3)(3/100) dØ/dt + (10/√3)(9) = 0

solve for dØ/dt

Now it's just Algebra I ...

Answer 4

Just noticed my "late-night" calculations have an errors and typos, sorry about that.

corrected version:

when 200 ft are let out,
x^2 + 100^2 = 200^2
x^2 = 30,000
x = 100√3
secØ = 200/ (100√3) = 2/√3
sec^2 Ø = 4/3
tanØ = 100/100√3 = 1/√3

then my equation should be

(100√3)(4/3) dØ/dt + (1/√3)(9) = 0

I get dØ/dt = -.0225 radians/s

check my work again carefully, still on my first coffee.