007A Sample Midterm 2, Problem 5

A kite 30 (meters) above the ground moves horizontally at a speed of 6 (m/s). At what rate is the length of the string increasing when 50 (meters) of the string has been let out?

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Solution:
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$30^{2}+h^{2}=s^{2}$
$s=50$
${\begin{array}{rcl}\displaystyle {h}&=&\displaystyle {\sqrt {s^{2}-30^{2}}}\\&&\\&=&\displaystyle {\sqrt {50^{2}-30^{2}}}\\&&\\&=&\displaystyle {40}\end{array}}$
$2hh'=2ss'$
$2(40)6=2(50)s'$
${\begin{array}{rcl}\displaystyle {s'}&=&\displaystyle {\frac {2(40)(6)}{2(50)}}\\&&\\&=&\displaystyle {{\frac {24}{5}}{\text{ m/s}}}\end{array}}$

Detailed Solution

Foundations:
The Pythagorean Theorem
For a right triangle with side lengths  $a,b,c$  where  $c$  is the length of the
hypotenuse, we have  $a^{2}+b^{2}=c^{2}.$

Step 1:
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From the diagram, we have  $30^{2}+h^{2}=s^{2}$  by the Pythagorean Theorem.
Taking derivatives, we get
$2hh'=2ss'.$

Step 2:
If   $s=50,$  then
$h={\sqrt {50^{2}-30^{2}}}=40.$
So, we have
$2(40)6=2(50)s'.$
Solving for   $s',$  we get   $s'={\frac {24}{5}}{\text{ m/s.}}$

Final Answer:
$s'={\frac {24}{5}}{\text{ m/s}}$

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