Aballoon is rising at a rate of 6 meters per second from a point on the ground 53 meters from an observer. find the rate of change of the angle of elevation from the observer to the balloon when the balloon is 32 meters above the ground.

0.0830 radians per second

Step-by-step explanation:

If θ is the angle of elevation and h is the height of the balloon, then:

tan θ = h / 53

Take the derivative of both sides with respect to time:

sec² θ dθ/dt = 1/53 dh/dt

Use Pythagorean identity:

(1 + tan² θ) dθ/dt = 1/53 dh/dt

Substitute:

(1 + (h/53)²) dθ/dt = 1/53 dh/dt

Multiply both sides by 53²:

(53² + h²) dθ/dt = 53 dh/dt

Given dh/dt = 6 and h = 32:

(53² + 32²) dθ/dt = 53 (6)

dθ/dt = 318 / 3833

dθ/dt ≈ 0.0830

The angle of elevation is increasing at a rate of 0.0830 radians per second.

idk

Step-by-step explanation: