Question

Need help please!
A Kite 70 feet above the ground moves horizontally at a speed of 4 ft/s. At what rate is the angle between the string and the ground decreasing when 150 feet of string is out?

Answer 1

`150 feet of string is out` so the hypotenuse is 150

Answer 2

use this to find theta

Answer 3

which I get to be 27.82 degrees

Answer 4

me too

Answer 5

\[\Large \tan(\theta) = \frac{x}{70}\]
derive both sides with respect to t (apply the chain rule)
\[\Large \frac{d}{dt}\left[\tan(\theta)\right] = \frac{d}{dt}\left[\frac{x}{70}\right]\]
\[\Large \sec^2(\theta)\frac{d \theta}{dt} = \frac{1}{70}*\frac{dx}{dt}\]

Answer 6

since the kite moves `horizontally at a speed of 4 ft/s` this means dx/dt = 4

Answer 7

the goal is to find dtheta/dt

Answer 8

I still can't get the answer

Answer 9

oh my bad, I messed up when I set up tangent

Answer 10

let me fix

Answer 11

\[\Large \tan(\theta) = \frac{70}{x}\]
\[\Large \tan(\theta) = 70x^{-1}\]
derive both sides with respect to t (apply the chain rule)
\[\Large \frac{d}{dt}\left[\tan(\theta)\right] = \frac{d}{dt}\left[70x^{-1}\right]\]
\[\Large \sec^2(\theta)\frac{d \theta}{dt} = -70x^{-2}*\frac{dx}{dt}\]
\[\Large \sec^2(\theta)\frac{d \theta}{dt} = -\frac{70}{x^2}*\frac{dx}{dt}\]

Answer 12

so because x is in the \(\Large \frac{d \theta}{dt}\) equation, you'll need to find the value of x in the drawing