More Related Rates: Kite flying horiz. at 8 feet per second

[scrum]

So I have that. and I make this diagram, getting in rads the angle with arcsin(100/200)

This is my diagram of what I think is going on.

I also get 100^2 + x^2 = 200^2 as the forumla for the base of the triangle but I don't quite get how to find the rate of change. These are really weirding me out as each word problem is about something else and I don't know how to apply the concept to each one.

 

[galactus]

We know y=100, dx/dt=8

Let the horizontal distance from the flyer to the kite be x.

\(\displaystyle \L\\cos{\theta}=\frac{x}{200}\)

Differentiate wrt time:

\(\displaystyle \L\\-sin{\theta}\frac{d{\theta}}{dt}=\frac{1}{200}\frac{dx}{dt}\)

You found theta when z=200. It is \(\displaystyle {\theta}=\frac{\pi}{6}\)

Now, sub in your knowns and solve for \(\displaystyle \frac{d{\theta}}{dt}\)

 

[arthur ohlsten]

let us define the base as x, and the hypoteneuse as h.
then:
h^2=100^2+x^2
take the derivative with respect to time.in seconds
2h dh/dt =2x dx/dt
dh/dt =[ x dx/dt ]/200

x=[200^2-100^2]^1/2
x=[30000]^1/2
x=100 [3^1/2] ft
dx/dt= 8ft/sec

dh/dt = 100sqrt3[8] / 200
dh/dt= 4 sqrt3 ft/sec answer

Arthur

 

[arthur ohlsten]

I am sorry , I solved the wrong problem. I thought you asked how the hypoteneuse was increasing. I didn't read the first line o your message
Arthur

 

[scrum]

That's cool.

I sorta see what you guys are doing but I still can't do it myself on the other problems involving similar things. I am so confused by these.

I'll look over your method and what galactus did and try to do my own.