Question

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 h be height of kite
let d be distance from the standing position
and l be length of string
and Theta be angle between string and ground
\[l^2=h^2+d^2\]
\[\tan{\theta}=\frac{h}{d}\]
by diffrentiating, we get
\[\sec^2{\theta}d\theta=-\frac{h}{d^2}dd\]
\[d\theta=-\cos^2{\theta}\frac{h}{d^2}dd\]
\[d\theta=-\frac{h^2}{h^2+d^2}dd\]
given that
dd = 9
d=200
h=100

so we get

\[d\theta=-1.9 \]

Answer 2

sqadi:

\[d\theta=-\cos^2{\theta}\frac{h}{d^2}dd\]
If you replace the d in d^2 with
\[l \cos(\theta)\]
then
\[d\theta=-{}\frac{h}{l^2}dd\]
plug in dd=9, l=200 and h=100, then
\[d \theta =-{9\over400} = -0.0225\]
Your dd value of -1.9 radians, close to -109 degrees, looks a little on the high end, although I could be incorrect.

Answer 3

Hi Robtobey,
you are right. I did mistake.
I have taken d=200, actually it is l=200


Thanks for correcting :-)

Answer 4

Thank you for responding.