# A kite 100 ft above the ground moves horizontally at

A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?
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Lindsey Gamble
At the moment, and we want to find $\frac{d\theta }{dt}$. We can find the curren angle first
$\theta =\mathrm{arcsin}\frac{100}{200}=\frac{\pi }{6}$
Get an equation to relate the x and $\theta$. Differentiate with respect to time t
$\frac{x}{100}=\mathrm{cot}\theta$
$0.01\frac{dx}{dt}={\mathrm{csc}}^{2}\theta \frac{d\theta }{dt}$
$\frac{d\theta }{dt}=\frac{0.01\frac{dx}{dt}}{-{\mathrm{csc}}^{2}\theta }$
$=\frac{0.01\left(8\right)}{-{\mathrm{csc}}^{2}\frac{\pi }{6}}$ plug in known values
$=\frac{0.08}{-{\left(2\right)}^{2}}$

The negative means it is decreasing at 0.02 radians per second $\left(\approx \frac{{1.15}^{\circ }}{s}\right)$

Chanell Sanborn
The angle between the horizontal and the kite $\theta$
Let Height of kite be h
Horizontal distance of kite from where the string is held be x
Then the length of the string be s and will be given by the pythagoras theorem as follows:
${s}^{2}={h}^{2}+{x}^{2}$
Differentiate throughout to get
$2s\frac{ds}{dt}=2h\frac{dh}{dt}+2x\frac{dx}{dt}$
$2s\frac{ds}{dt}=2×\left(0\right)+2x×8$
$s\frac{ds}{dt}=8x\to \left(1\right)$