Question

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 changing when 200 ft of string have been let out? is the angle increasing or decreasing? rad/s

Explanation:

Step1: Set up the relationship

Let \( \theta \) 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.

Step2: Differentiate with respect to time

Differentiate \(\sin\theta=\frac{100}{l}\) with respect to \(t\). Using the chain - rule, \(\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}\).

Step3: Find \(\cos\theta\)

Since \(\sin\theta=\frac{100}{200}=\frac{1}{2}\), then \(\theta = 30^{\circ}\) or \(\frac{\pi}{6}\) radians. And \(\cos\theta=\frac{\sqrt{3}}{2}\).

Step4: Substitute values

We are given \(\frac{dl}{dt}\) (related to the horizontal movement. Let \(x\) be the horizontal distance. By Pythagoras \(l^{2}=x^{2}+100^{2}\). Differentiating \(2l\frac{dl}{dt}=2x\frac{dx}{dt}\). When \(l = 200\), \(x=\sqrt{200^{2}-100^{2}} = 100\sqrt{3}\). And \(\frac{dx}{dt}=8\) ft/s. From \(l\frac{dl}{dt}=x\frac{dx}{dt}\), \(\frac{dl}{dt}=\frac{x}{l}\frac{dx}{dt}=\frac{100\sqrt{3}}{200}\times8 = 4\sqrt{3}\) ft/s.
Substitute into \(\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}\). \(\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=-\frac{100}{200^{2}}\times4\sqrt{3}\)

Step5: Solve for \(\frac{d\theta}{dt}\)

\(\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=-\frac{100\times4\sqrt{3}}{40000}\)
\(\frac{d\theta}{dt}=-\frac{1}{50}=- 0.02\) rad/s

Answer:

\(\frac{d\theta}{dt}=-0.02\) rad/s. The negative sign indicates that the angle is decreasing.