Painevg

2021-12-18

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?

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
$2s\frac{ds}{dt}=2h\frac{dh}{dt}+2x\frac{dx}{dt}$
$2s\frac{ds}{dt}=2×\left(0\right)+2x×8$
$s\frac{ds}{dt}=8x\to \left(1\right)$

Do you have a similar question?