A balloon rises at the rate of 8 ft/sec from a point on the ground 60 ft from the observer. How do you find the rate of change of the angle of elevation when the balloon is 25 ft above the ground?

To solve this related rates (of change) problem:

Let #y# = the height of the balloon and let #theta# = the angle of elevation.
We are told that #(dy)/(dt)=8# ft/sec.
We are asked to find #(d theta)/(dt)# when #y=25# ft.

Draw a right triangle with base = 60 ft (that doesn't change), height #y# and angle opposite height #theta#.

Then # tan theta = y/60# and #y=60 tan theta#.

Differentiating with respect to #t# gives us:

#d/(dt)(y)=d/(dt)(60 tan theta)#.

#(dy)/(dt) = 60 sec^2 theta (d theta)/ (dt)#.

We are asked to find #(d theta)/(dt)# when #y=25#.

We have: #8= 60 sec^2 theta (d theta)/ (dt)#, so

#(d theta)/ (dt)=8/60 cos^2 theta = 2/15 cos^2 theta#.

We need #cos theta# when #y=25#.
With base = 60 and height = 25, we get hypotneuse #c= sqrt (60^2 + 25^2) = sqrt ((5*12)^2+(5*5)^2)=5sqrt ((12)^2+(5)^2) = 5*13 = 65#.

So, when #y=25#, we have: #cos theta = 60/65=12/13#.

So
#(d theta)/ (dt) = 2/15 cos^2 theta= 2/15 (12/13)^2 = 96/845# radians / sec
.
(Remember, in order to use #d/(d theta)(tan theta) = sec^2 theta#, we must have #theta# either a real number or the radian (not degree) measure of an angle.)