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

Beginner2021-12-19Added 38 answers

At the moment, $\frac{dx}{dt}=8\text{}f\frac{t}{s}$ 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

$=\frac{0.08}{-{\left(2\right)}^{2}}$

$=-0.02\text{}ra\frac{d}{s}$

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

Get an equation to relate the x and

The negative means it is decreasing at 0.02 radians per second

Chanell Sanborn

Beginner2021-12-20Added 41 answers

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

$2s\frac{ds}{dt}=2h\frac{dh}{dt}+2x\frac{dx}{dt}$

$2s\frac{ds}{dt}=2\times \left(0\right)+2x\times 8$

$s\frac{ds}{dt}=8x\to \left(1\right)$

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:

Differentiate throughout to get

22+64

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