Help with Extrema problem Calculus?

Question

Some hotels have glass elevators on the outside of the building. One such hotel is 300 feet high. You are looking out a window 100 feet above the ground in a building 150 feet away from the hotel, as the elevator descends at a constant speed of 30ft/sec, starting at the top of the hotel at time t = 0 sec. Let Ө be the angle between your horizontal line of vision and your line of sight to the elevator.

The rate of change of Ө is a measure of how fast the elevator appears to you to be moving. At what height above the ground does the elevator appear to be moving fastest?

you can see a drawing of the problem on PROBLEM E at
http://www.flickr.com/photo_zoom.gne?id=296714838&size=l

Answer 1

I am taking College Algebra with Trigonometry and that is hard, so I know Calculus is real hard. That is a hard question to figure out.

Answer 2

Ok, here's how I think you do it.

First you must draw a triangle where 150 ft is the adyascent side (as) and os is the opposite side according to the angle.

since you know the speed of the elevator you can get the os, which varies with time, thing that doesn't happen to the as so the 150 is going to be a constant

os = 200 - 30t

200 because 300-100 which are the heights of the buildings.

Remember tu use allways dimensinals, i'm not using them right now because i'm lazy.

so, now you can get an equation involving the angle

tan (θ) = (os/as)

so you substitute:

θ = tan^-1(os/as)

θ = tan^-1((200-30*t)/150)

θ = tan^-1(4/3-t/5)

and you derive

dθ/dt=-45/(9*t^2-120*t+625)

That's going to be the rate of the change of the angle with the time, which is the measure of how fast the elevator appears to you to be moving.

in order to get the time where the velocity seems to be the fastest for you, you derive again and equal it to 0

d^2θ/dt^2=270*(3*t-20)/(9*t^2-120*t+625)^2 = 0

now, solving for t

t = 20/3s

so finally you get the height

height = 300-30*t

height = 300 - 30*20/3 = 300-200 = 100 ft

So the height where you are going to see it moving faster is going to be 100 ft from the ground.

Hope this helps.

PS. I didn't put the procedure to derive because i'm lazy as i said before, but if you need help with that you can tell me ;o)