Question

A balloon leaves the ground 500 feet away from an observer and rises vertically at the rate of 140 feet per minute. At what rate is the angle of inclination of the observer's line of sight increasing at the instant when the balloon is exactly 500 feet above the ground?

Please show me how you got the answer!!

Answer 1

tan A = h/500

d tan A/ dh = 500(-1)/500^2 = -1/500

d tan A/dA = sec^2 A
so d tan A = sec^2 A dA

sec^2 A dA/dh = -1/500
dA/dh = cos^2 A /500
dA/dt = dh/dt * dA/dh
dA/dt = 140 cos^2 A / 500
at 500 ft. cos A = cos pi/4 = .707
so cos^2 A = .5
dA/dt = 140 * .5 /500
= 0.14 rad/min
or 8 degrees/minute

Answer 2

tan A = h/500

d tan A/ dh = 500(1)/500^2 = 1/500

d tan A/dA = sec^2 A
so d tan A = sec^2 A dA

sec^2 A dA/dh = 1/500
dA/dh = cos^2 A /500
dA/dt = dh/dt * dA/dh
dA/dt = 140 cos^2 A / 500
at 500 ft. cos A = cos pi/4 = .707
so cos^2 A = .5
dA/dt = 140 * .5 /500
= 0.14 rad/min
or 8 degrees/minute