a kite 100 ft above the ground moves horizontally at a speed of 4 ft/s. at what rate (in rad/s) is the angle…

Explanation:

Step1: Establish a trigonometric relationship

Let $x$ be the horizontal distance of the kite from the person flying it and $\theta$ be the angle between the string and the horizontal. We know that $\sin\theta=\frac{100}{l}$, where $l$ is the length of the string. Given $l = 200$ ft, at this moment $\sin\theta=\frac{100}{200}=\frac{1}{2}$, so $\theta=\frac{\pi}{6}$ radians. Also, by the Pythagorean theorem, $x=\sqrt{l^{2}-100^{2}}$. Differentiating $\sin\theta=\frac{100}{l}$ with respect to time $t$ using the chain - rule, we get $\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}$. We can also use the relationship $\tan\theta=\frac{100}{x}$. Differentiating both sides with respect to $t$ gives $\sec^{2}\theta\frac{d\theta}{dt}=-\frac{100}{x^{2}}\frac{dx}{dt}$.

Since $\frac{dx}{dt} = 4$ ft/s, and $\tan\theta=\frac{100}{x}$, when $l = 200$, $x=\sqrt{200^{2}-100^{2}}=\sqrt{40000 - 10000}=\sqrt{30000}=100\sqrt{3}$ ft. And $\sec\theta=\frac{l}{x}=\frac{200}{100\sqrt{3}}=\frac{2}{\sqrt{3}}$.

Step2: Solve for $\frac{d\theta}{dt}$

From $\sec^{2}\theta\frac{d\theta}{dt}=-\frac{100}{x^{2}}\frac{dx}{dt}$, substituting $\sec\theta=\frac{2}{\sqrt{3}}$, $x = 100\sqrt{3}$ ft and $\frac{dx}{dt}=4$ ft/s.

$\left(\frac{4}{3}\right)\frac{d\theta}{dt}=-\frac{100}{(100\sqrt{3})^{2}}\times4$

$\left(\frac{4}{3}\right)\frac{d\theta}{dt}=-\frac{100}{30000}\times4$

$\left(\frac{4}{3}\right)\frac{d\theta}{dt}=-\frac{4}{300}$

$\frac{d\theta}{dt}=-\frac{4}{300}\times\frac{3}{4}=-\frac{1}{100}$ rad/s. The negative sign indicates that the angle $\theta$ is decreasing.

Answer:

$\frac{1}{100}$