Question

homework6: problem 2
(3 points)
a kite 100ft above the ground moves horizontally at a speed of 6ft/s. at what rate is the angle between the string and the horizontal decreasing when 250ft of string has been let out?
answer (in radians per second):
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Explanation:

Step1: Establish the relationship

Let \(y = 100\) (height of the kite, constant), \(x\) be the horizontal distance of the kite from the person flying it, \(l\) be the length of the string, and \(\theta\) be the angle between the string and the horizontal. We know that \(\sin\theta=\frac{y}{l}\) (since \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) in a right - triangle, here opposite side \(y = 100\) and hypotenuse \(l\)). Also, by the Pythagorean theorem \(l=\sqrt{x^{2}+y^{2}}\). But since \(y = 100\), \(l=\sqrt{x^{2}+10000}\). And \(\sin\theta=\frac{100}{l}\), so \(l=\frac{100}{\sin\theta}\).

Differentiating both sides of \(\sin\theta=\frac{100}{l}\) with respect to time \(t\) using the chain rule. The derivative of \(\sin\theta\) with respect to \(t\) is \(\cos\theta\frac{d\theta}{dt}\), and the derivative of \(\frac{100}{l}\) with respect to \(t\) is \(-\frac{100}{l^{2}}\frac{dl}{dt}\).

We also know that \(x\) and \(l\) are related by \(l^{2}=x^{2}+100^{2}\). Differentiating both sides with respect to \(t\): \(2l\frac{dl}{dt}=2x\frac{dx}{dt}\), so \(\frac{dl}{dt}=\frac{x}{l}\frac{dx}{dt}\).

Since \(\sin\theta=\frac{100}{l}\), when \(l = 250\), \(\sin\theta=\frac{100}{250}=\frac{2}{5}\), and \(\cos\theta=\sqrt{1-\sin^{2}\theta}=\sqrt{1 - (\frac{2}{5})^{2}}=\frac{\sqrt{21}}{5}\).

From \(l^{2}=x^{2}+100^{2}\), when \(l = 250\), \(x=\sqrt{l^{2}-100^{2}}=\sqrt{250^{2}-100^{2}}=\sqrt{(250 + 100)(250-100)}=\sqrt{350\times150}=\sqrt{52500}=50\sqrt{21}\).

We are given that \(\frac{dx}{dt}=6\) ft/s.

Step2: Substitute into the derivative equation

From \(\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\frac{dl}{dt}\) and \(\frac{dl}{dt}=\frac{x}{l}\frac{dx}{dt}\), we substitute \(\frac{dl}{dt}\) into the first equation:

\(\cos\theta\frac{d\theta}{dt}=-\frac{100}{l^{2}}\cdot\frac{x}{l}\frac{dx}{dt}\)

Substitute \(x = 50\sqrt{21}\), \(l = 250\), \(\cos\theta=\frac{\sqrt{21}}{5}\), and \(\frac{dx}{dt}=6\)

\(\frac{\sqrt{21}}{5}\frac{d\theta}{dt}=-\frac{100}{250^{2}}\cdot\frac{50\sqrt{21}}{250}\times6\)

First, simplify the right - hand side:

\(-\frac{100\times50\sqrt{21}\times6}{250^{3}}=-\frac{100\times50\sqrt{21}\times6}{15625000}=-\frac{30000\sqrt{21}}{15625000}=-\frac{6\sqrt{21}}{3125}\)

Then solve for \(\frac{d\theta}{dt}\):

\(\frac{d\theta}{dt}=-\frac{6\sqrt{21}}{3125}\times\frac{5}{\sqrt{21}}=-\frac{6}{625}=- 0.0096\)

Answer:

\(\frac{6}{625}\) radians per second.