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anonymous
 5 years ago
A kite 100 ft above the ground moves horizontally at a speed of 9 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
anonymous
 5 years ago
A kite 100 ft above the ground moves horizontally at a speed of 9 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

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sgadi
 5 years ago
Best ResponseYou've already chosen the best response.0let h be height of kite let d be distance from the standing position and l be length of string and Theta be angle between string and ground \[l^2=h^2+d^2\] \[\tan{\theta}=\frac{h}{d}\] by diffrentiating, we get \[\sec^2{\theta}d\theta=\frac{h}{d^2}dd\] \[d\theta=\cos^2{\theta}\frac{h}{d^2}dd\] \[d\theta=\frac{h^2}{h^2+d^2}dd\] given that dd = 9 d=200 h=100 so we get \[d\theta=1.9 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sqadi: \[d\theta=\cos^2{\theta}\frac{h}{d^2}dd\] If you replace the d in d^2 with \[l \cos(\theta)\] then \[d\theta={}\frac{h}{l^2}dd\] plug in dd=9, l=200 and h=100, then \[d \theta ={9\over400} = 0.0225\] Your dd value of 1.9 radians, close to 109 degrees, looks a little on the high end, although I could be incorrect.

sgadi
 5 years ago
Best ResponseYou've already chosen the best response.0Hi Robtobey, you are right. I did mistake. I have taken d=200, actually it is l=200 Thanks for correcting :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you for responding.
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