Question

A balloon, 50 feet from an observer, is rising at 20 ft/sec. At 5 seconds after lift off

1. How fast is the distance between the observer and the balloon changing?
2. How fast is the angle of elevation changing?

I need help on both questions. Thanks in advance smart people!!

Answer 1

At a time of t sec, let the height be h ft.
let the angle of elevation be k
let the distance between them be y

1.
h^2 + 50^2 = y^2
2h dh/dt = 2y dy/dt
dy/dt = (h dh/dt)/y

when t = 5, h = 100
y^2 = 50^2 + 100^2 = 12500
y= 50sqrt(5)
dy/dt = 100(20)/sqrt(5)
= 50/sqrt(5) = appr 22.36 ft/sec

2.
then tank = h/50
h = 50tank
dh/dt = 50sec^2 k dk/dt
given: dk/dt = 20 ft/sec
when t =5, h = 100 ft
tank = 200/50 = 2
I sketched a triangle, and
cosk = 1/squr(5)
seck = squr(5)
sec^2 k = 5

dh/dt = 50sec^2 k dk/dt
20 = 50(5) dk/dt
dk/dt = 20/(250) rad/sec
= .08 rads/sec