"Flying a Kite"

[kaebun]

Hi! Can anyone tell me if this is right or am I way off? I have a feeling I didn't start out right, because I feel like I should have to plug in S=500 to the final equation because thats how its been in previous problems

The problem: Inge flies a kite at a height of 300ft , the wind carries the kite horizontaly at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her?

s^2= x^2+y^2
(500)^2=(300)^2+y^2
250000-90000=y^2
y=400


s^2=300^2+y^2
s=sqrt(90000+y^2)

ds/dt=(9000+y^2)^(-1/2) (2y(dy/dt))
ds/dt=2(400)(25)/sqrt(90000+400^2)

ds/dt=0.8

thanks

 

[arthur ohlsten]

S=500ft
x=300 ft
dx/dt=0, constant altitude
y=400 ft as calculated
dy/dt = 25 ft/sec

find ds/dt

S^2=x^2+y^2 take derivative with respect to t
2S ds/dt = 2x dx/dt +2y dy/dt substitute
2[500] ds/dt = 2[300][0] +2[400][25]
1000 ds/dt = 20000
ds.dt = 20 ft/ sec

this is .8 times 25 ft/sec

Arthur

 

[soroban]

Hello, kaebun!

Arthur is absolutely correct.

A suggestion: do not plug in the constants until the end.

Inge flies a kite at a height of 300 ft.
The wind carries the kite horizontaly at a rate of 25 ft/sec.
How fast must she let out the string when the kite is 500 ft away from her?

              y
      + - - - - - - - *
      :             /
      :           /
      :         /
  300 :       / s
      :     /
      :   /
      : /
  - - * - - - - - - - - -

From Pythagorus, we have: \(\displaystyle \,s^2\:=\:y^2\,+\,300^2\;\) [1]

Differentiate with respect to time: \(\displaystyle \L\:2s\left(\frac{ds}{dt}\right) \:=\:2y\left(\frac{dy}{dt}\right)\)
. . and we have: \(\displaystyle \L\:\frac{ds}{dt} \:=\:\frac{y}{s}\left(\frac{dy}{dt}\right)\;\) [2]


Now we work out the values of the right side.

We are told that: \(\displaystyle \frac{dy}{dt} \:=\:25\) ft/sec

"The kite is 500 feet from her" means \(\displaystyle s \,=\,500\)
. . From [1], we have: .\(\displaystyle 500^2\:=\:y^2\,+\,300^2\;\;\Rightarrow\;\;y\,=\,400\)

Substitute into [2]: \(\displaystyle \L\:\frac{ds}{dt} \:=\:\frac{400}{500}(25) \:=\:20\) ft/sec