anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sgadi
  • sgadi
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 \[d\theta=-1.9 \]
anonymous
  • 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.
sgadi
  • sgadi
Hi Robtobey, you are right. I did mistake. I have taken d=200, actually it is l=200 Thanks for correcting :-)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Thank you for responding.

Looking for something else?

Not the answer you are looking for? Search for more explanations.