# Thread: Got this interesting little problem in my test...

1. ## Got this interesting little problem in my test...

And I'd like to know the answer because it's bugging me now

A kite, 30m above the ground, moves horisontally at a speed of 5m/s.
At what rate is the angle between the string and the horisontal decreasing when 50m of string has been let out?

2. Originally Posted by janvdl
And I'd like to know the answer because it's bugging me now

A kite, 30m above the ground, moves horisontally at a speed of 5m/s.
At what rate is the angle between the string and the horisontal decreasing when 50m of string has been let out?

Let the plan distance from the string holder to the kite be $\displaystyle d$, assume zero heigh for the ground ebd of the string.

Then:

$\displaystyle \tan(\theta)=\frac{30}{x}$

differentiate wrt to time:

$\displaystyle (\sec(\theta))^2 \frac{d \theta}{dt}=-\frac{30}{x^2}\frac{dx}{dt}$

and you should be able to do it from there as $\displaystyle x'=3$ m/s.

(also the string forms a $\displaystyle 30,40,50$ m right triangle with the ground and the vertival dropped from the kite.

RonL

3. Originally Posted by CaptainBlack
Let the plan distance from the string holder to the kite be $\displaystyle d$, assume zero heigh for the ground ebd of the string.

Then:

$\displaystyle \tan(\theta)=\frac{30}{x}$

differentiate wrt to time:

$\displaystyle (\sec(\theta))^2 \frac{d \theta}{dt}=-\frac{30}{x^2}\frac{dx}{dt}$

and you should be able to do it from there as $\displaystyle x'=3$ m/s.

RonL
Thank you Captain. I did it similarly. Only with cos and i think i left out dx/dt so it will probably be wrong.