## 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?

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

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.

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

4. anonymous

Thank you for responding.