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 60-80-100 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

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