Question

A steady wind blows a kite due west. the kite's height above ground from horizontal position x = 0 to x = 120 ft is given by y = 150 − 1 40 (x − 50)2. find the distance traveled by the kite. (round your answer to one decimal place.)

Answer 1

The derivative of the vertical position is
.. y' = 2/40*(x -50)
so the integral that gives the length of the curve the kite traveled is
$distance= \int\limits^{120}_{0} {\sqrt{1+(\frac{1}{20}(x-50))^{2}} \, dx$

It is convenient to let a graphing calculator compute this integral numerically. If you want to do it analytically, a table of integrals can be handy. The indefinite integral is
$\frac{x-50}{40} \sqrt{(x-50)^{2}+400} +10 arcsinh(\frac{x-50}{20})$
This evaluates to
$\frac{1}{2}(25 \sqrt{29} +35 \sqrt{53}) +10(arcsinh( \frac{5}{2}) +arcsinh( \frac{7}{2})) \approx 230.8$ feet