Question

a kite 100 ft above the ground moves horizontally at a speed of 6 ft/s. at what rate (in rad/s) is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
rad/s

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the top of a ladder slides down a vertical wall at a rate of 0.125 m/s. at the moment when the bottom of the ladder is 5 m from the wall, it slides away from the wall at a rate of 0.3 m/s. how long is the ladder?
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Explanation:

Step1: Set up the relationship for the kite problem

Let $x$ be the horizontal distance of the kite from the person flying it, $y = 100$ ft (constant height of the kite) and $l$ be the length of the string. Let $\theta$ be the angle between the string and the horizontal. We know that $\sin\theta=\frac{y}{l}=\frac{100}{l}$. Differentiating both sides with respect to time $t$ using the chain - rule, we get $\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}$.

When $l = 200$ ft, $\sin\theta=\frac{100}{200}=\frac{1}{2}$, so $\theta=\frac{\pi}{6}$ and $\cos\theta=\frac{\sqrt{3}}{2}$. Also, we are given that $\frac{dx}{dt}=6$ ft/s. By the Pythagorean theorem, $x=\sqrt{l^{2}-y^{2}}$. Differentiating $x^{2}+y^{2}=l^{2}$ 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 and $y = 100$ ft, $x=\sqrt{200^{2}-100^{2}} = 100\sqrt{3}$ ft. Then $\frac{dl}{dt}=\frac{100\sqrt{3}}{200}\times6 = 3\sqrt{3}$ ft/s.

Substituting $\cos\theta=\frac{\sqrt{3}}{2}$, $l = 200$ ft and $\frac{dl}{dt}=3\sqrt{3}$ ft/s into $\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}$, we have $\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=-\frac{100}{200^{2}}\times3\sqrt{3}$.

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

First, simplify the right - hand side of the equation: $-\frac{100}{200^{2}}\times3\sqrt{3}=-\frac{3\sqrt{3}}{400}$. Then, solve for $\frac{d\theta}{dt}$ from $\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=-\frac{3\sqrt{3}}{400}$. Cross - multiply to get $\frac{d\theta}{dt}=-\frac{3\sqrt{3}}{400}\times\frac{2}{\sqrt{3}}=-\frac{3}{200}$ rad/s.

For the ladder problem:
Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the distance of the top of the ladder from the ground. Let the length of the ladder be $L$. By the Pythagorean theorem, $x^{2}+y^{2}=L^{2}$. Differentiating both sides with respect to time $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.

We are given that $\frac{dy}{dt}=- 0.125$ m/s and $\frac{dx}{dt}=0.3$ m/s when $x = 5$ m. Substitute into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$, we get $5\times0.3+y\times(-0.125)=0$. Then $1.5 - 0.125y = 0$, so $y = 12$ m.

Since $x^{2}+y^{2}=L^{2}$, when $x = 5$ m and $y = 12$ m, $L=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$ m.

Answer:

For the kite problem: $\frac{3}{200}$
For the ladder problem: $13$