A kite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

Answer: So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z. So we have x2 + 1002 = z2. Now take its derivative in terms of time to get 2x (dx/dt) = 2z (dz/dt) So at your specific moment z = 200, x = 100âš3 and dx/dt = + 8 substituting, that makes dz/dt = 800âš3 / 200 or 4âš3. Part 2 sin a = 100/z = 100 z-1. Now take the derivative in terms of t to get cos a (da./dt) = - 100 / z2 (dz/dt) So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or Ď€/6 radians. Substitute to get cos (Ď€/6) (da/dt) = (-100 / 40000) (4âš3) âš3 / 2 (da/dt) = - âš3 / 100 da/dt = - 1/50 radians