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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?

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  1. sgadi
    • 5 years ago
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    let 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 \[d\theta=-1.9 \]

  2. anonymous
    • 5 years ago
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    sqadi: \[d\theta=-\cos^2{\theta}\frac{h}{d^2}dd\] 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.

  3. sgadi
    • 5 years ago
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    Hi Robtobey, you are right. I did mistake. I have taken d=200, actually it is l=200 Thanks for correcting :-)

  4. anonymous
    • 5 years ago
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    Thank you for responding.

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