Question

A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 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

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

$y=150-\frac{1}{40}(x-50)^2$

Differentiating with respect to x,

$y'=-\frac{1}{20}(x-50)$

∵ arc length of a curve is,

$L=\int_{a}^{b} \sqrt{1+y'^2}dx$

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

$L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx$

Put $-\frac{1}{20}(x-50)=tan\theta$

$\implies -dx=-20 sec^2\theta d\theta$

$\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

By integration by parts,

$=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}$

If x = 80, $tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}$

$sec \theta = \frac{\sqrt{13}}{2}$

$\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}$

$sec \theta = \frac{\sqrt{29}}{2}$

Thus, the length of the curve is,

$=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)$

$\approx 122.8\text{ feet}$