A person whose eyes are 6 feet from the ground stands 35 feet from a bottle rocket on the ground and watches it as it takes off vertically into the air at a rate of 17 ft/sec. Find the rate at which the angle of elevation between the person’s eyes and the rocket changes when the rocket is 41 feet in the air.
Wayne DeguMan
tanθ = y/x => y(1/x)
so, sec²θ.dθ/dt = (-1/x²)y.dx/dt + (1/x)dy/dθ
Now, as x is constant, dx/dt = 0 so,
sec²θ.dθ/dt = (1/x).dy/dt
Then, (1 + tan²θ).dθ/dt = (1/x).dy/dt
Now, when the rocket is at 41 feet, y = 35, x = 35 and dy/dt = 17 we have:
(1 + (35/35)²).dθ/dt = (1/35)(17)
Note: In your working, the dθ/dt on the left appears as dy/dt..??
so, 2.dθ/dt = 17/35
=> dθ/dt = 17/70
i.e. 0.243 radians/sec....all looks good.
P.S. It's always good to see someone's workings as it helps to highlight any errors and misconceptions..well done!!
:)>