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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 - \frac{1}{40} (x - 50)^2 $. Find the distance traveled by the kite.

$L=10\left[\frac{3 \sqrt{13}}{4}+\frac{5 \sqrt{29}}{4}-\ln \left|\frac{\sqrt{13}-3}{\sqrt{29}+5}\right|\right] \approx 122.76$

Applications of Integration

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Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Idaho State University

Hey, it's Clara. So when you raid here, so we have a deal, I over DX, that's what we're first going to find is equal to negative one over 20 times X minus 50. So when we plug this into the arc length formula gives us l is equal to from 0 to 80 square of one plus negative one over 20 X minus 50 square the ex So are we going to substitute negative won over 20? That's minus 50. I was 10. So negative one over 20 D X is he could equal to seek in square Think vato. So we have l this equal to 0 to 80 square root of one plus 10 square negative 20 sequence Square de Fada which gives us from 0 to 80 for X negative 20 Seeking cute The data. We just have to do integration by parts This gives us l is equal to negative 20 over too. I'm seeking plus tangent waas Ln a second plus tangent from this is from X is equal to cereal and X is equal to a T. So when we look at the limits, if X is equal to 80 than tangent and equal to negative three house. So that means that seat int is equal to square of 13 over too and access equal to zero. Then tangent is equal to five halves and seeking ISS square root of 29 over, too. And when we put these values inside, we get negative 10 times square root of 13 over two times negative three house plus l and the square root of 13 over too, minus three homes minus five. It's where we're 29 over four, minus L on a square root of 29 over too, plus my hands. This is approximately 122.76