Related Rates: Rising Balloon Angle?

Question

A Ballons rises at the rate of 8 ft/sec from a point on the ground 60 ft from the observer. Find the rate of change of the angle of elevation when the balloon is 25 ft above the ground.

Let a= 60 ft
Let b= 25 ft
Let db/dt= 8
Find d(theta)/dt

Formula relating angle of elevation to height of balloon and distance from observer.

tan(theta)=b/60

Differentiate implicitly, and solve for d(theta)/dt:

d(theta)/dt = (cos^2(theta)/60) * (db/dt)

Need to find what theta is when balloon is 25 ft above ground:
USE:

arctan(5/12)= theta

Now plug theta and db/dt into the equation and I got:

d(theta)/dt= 1152/169 radians/second

Something doesn't seem right, I either missed up on a step, am paranoid, or both.

Answer 1

You need to make a correction.

Let
h = height balloon
θ = angle of elevation

Given dh/dt = 8 ft/sec

Find dθ/dt when h = 25 ft.
_____________

tanθ = h/60 = 25/60 = 5/12
when h = 25 ft

The hypotenuse of the triangle is:
√(60² + h²) = √(60² + 25²) = √(3600 + 625) = √4225 = 65

cosθ = 60/65 = 12/13
when h = 25 ft
_______________

tanθ = h/60

Differentiate implicitly.

(sec²θ)(dθ/dh) = 1/60
dθ/dh = cos²θ/60 = (12/13)² / 60 = (144/169) / 60
dθ/dh = (12/169) / 5 = 12/(5*169) = 12/845

dθ/dt = (dθ/dh)(dh/dt) = (12/845)(8)
dθ/dt = 96/845 radians/sec

Answer 2

Related Rates Balloon