Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 28

How fast is the angle between the string and the horizontal decreasing when 200ft of string has been let out?

Given:

$\qquad $ height of the kite = $100 ft$

$\qquad $ horizontal speed of the kite = $8 ft/s$

$\qquad \displaystyle \frac{dx}{dt} = 3 cm/s $

Required: The angle between the string and the horizontal when $200 ft$ of string has been let out.

Solution:







We use the sine function to get the unknown

$\displaystyle \sin \theta = \frac{100}{s};$ when $s = 200, \theta = \sin^{-1} = 30^0$

We also use the tan function to solve the required answer

$\displaystyle \tan \theta = \frac{100}{x}; x = \frac{100}{\tan (30)} = 173.21 ft$

Taking the derivative with respect to time,


$
\begin{equation}
\begin{aligned}

\sec ^2 \theta \frac{d \theta}{dt} =& \frac{\displaystyle -100 \frac{dx}{dt}}{x^2}
&& \text{Solving for $\large \frac{d \theta}{dt}$}
\\
\\
\frac{d \theta}{dt} =& \frac{\displaystyle -100 \cos ^2 \theta \frac{dx}{dt}}{x^2}; \qquad \cos \theta = \frac{ 1 }{\sec \theta}
&&
\\
\\
\frac{d \theta}{dt} =& \frac{-100 [\cos (30)]^2 (8)}{(173.21)^2} = \frac{-0.02^0}{s}
&& \text{negative value for decreasing rate}

\end{aligned}
\end{equation}
$


The final answer is $\displaystyle \frac{d \theta}{dt} = \frac{0.02^0}{s}$ because we are asked to find the decreasing rate of the angle.