4. a kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. at what rate is the angle between…

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, $\sin\theta=\frac{100}{200}=\frac{1}{2}$, so $\theta=\frac{\pi}{6}$. 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 $\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}$.

Step2: Find $\frac{dl}{dt}$ in terms of $\frac{dx}{dt}$

Since $x^{2}+100^{2}=l^{2}$, differentiating both sides with respect to $t$ gives $2x\frac{dx}{dt}=2l\frac{dl}{dt}$, so $\frac{dl}{dt}=\frac{x}{l}\frac{dx}{dt}$. When $l = 200$ ft, $x=\sqrt{200^{2}-100^{2}}=\sqrt{40000 - 10000}=\sqrt{30000}=100\sqrt{3}$ ft. Given $\frac{dx}{dt}=8$ ft/s, then $\frac{dl}{dt}=\frac{100\sqrt{3}}{200}\times8 = 4\sqrt{3}$ ft/s.

Step3: Solve for $\frac{d\theta}{dt}$

We know that $\cos\theta=\frac{\sqrt{3}}{2}$ (since $\theta=\frac{\pi}{6}$), and from $\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}$, substituting $l = 200$ ft, $\cos\theta=\frac{\sqrt{3}}{2}$, and $\frac{dl}{dt}=4\sqrt{3}$ ft/s. We have $\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=-\frac{100}{200^{2}}\times4\sqrt{3}$. Simplify the right - hand side: $-\frac{100}{200^{2}}\times4\sqrt{3}=-\frac{400\sqrt{3}}{40000}=-\frac{\sqrt{3}}{100}$. Then $\frac{d\theta}{dt}=-\frac{1}{50}$ rad/s. The negative sign indicates that the angle $\theta$ is decreasing.

Answer:

$\frac{1}{50}$ rad/s