findind the rate of change of distance

Question

A helicopter is taking at an angle of 60 degree,flying at 20 m/sec,in a straight line away from an observer at ground level 100m away from take off point.Find the rate of change of the distance between the helicopter and the observer 5th second after take off.

Answer 1

pythagorean theorem for 30-60-90 right triangle

d^2= 100^2 + (100sqr3)2= 40000

d= sqr40000= 200m in 5 sec

d^2=a^2+b^2

take derivatives, divide by 2

dd'=aa'+bb',

b'= 20m/s, b=100m, a=100sqr3, d=200, a'=0

plug in known values

d'= bb'/d= 10 m/s