Posted by need help on Friday, December 16, 2011 at 4:41pm.
A kite 80 feet above the ground moves horizontally at a speed of 8 ft/sec. At what rate is the angle between the string and the horizontal decreasing when 100 ft of string have been let out?

calc  Steve, Friday, December 16, 2011 at 6:24pm
Draw your little triangle.
cos a = x/s
where s is the string length (the hypotenuse) and x is the horizontal distance from the flyer.
x = s*cos a
dx/dt = ds/dt cos a  s*sin a da/dt
At our moment, s=100, so
8 = ds/dt cos a  100 * sin a da/dt
Still not home free. What are ds/dt and a?
a little examination shows that we have a 6080100 triangle, so
8 = ds/dt * .6  100 * .8 da/dt
8 = .6 ds/dt  80 da/dt
What's ds/dt?
well, s^2 = x^2 + 80^2
2s ds dt = 2x dx/dt
200 ds/dt = 120 * 8 = 160
ds/dt = .8
8 = .6*.8  80 da/dt
8 = .48  80 da/dt
7.52/80 = da/dt
.094 = da/dt
That's in radians, so at the moment in question,
a = 53.1° and
da/dt = 5.38°/sec