more derivatives

[mlane]

A hotel is 300 ft high has an outside elevator. You are standing by a window 100ft above the ground and 150ft away from the hotel, and the elevator descends at a constant speed of 30 ft/sec, starting a time t=0, where t is time in secons. Le theta be the angle between the line of your horizon and your line of sight to the elevator.
a) Find formula h(t), the elevator's height above the ground as it descends from the top of the hotel.
b) Using your answer to part (a) express theta as a function of time t and find the rate of change of theta with respect to t.
c) The rate of change of theta is a measure of how fast the elevator appears to yhou to be moving. At what height is the elevator when it appears to be moving fastest?

 

[Unco]

G'day!

a) Find formula h(t), the elevator's height above the ground as it descends from the top of the hotel.

I'm sure you drew a diagram. If h is the vertical distance from the elevator to you, you might have something like this:

        |\
        |  \  
        |    \
    h   |      \
        |   theta\
        |__________\
            150

Let theta = Q.

tan(Q) = h/150
h = 150tan(Q)

The height of the elevator, H, will be
H = 100 + h
H = 100 + 150tan(Q)

In terms of time:
H(t) = 100 + 150tan(Q(t))

b) Using your answer to part (a) express theta as a function of time t and find the rate of change of theta with respect to t.

dQ/dt = dQ/dH * dH/dt

It doesn't seem necessary to rearrange the equation above from H(t) as you can just take the reciprocal of dH/dQ.

c) The rate of change of theta is a measure of how fast the elevator appears to yhou to be moving. At what height is the elevator when it appears to be moving fastest?

If dQ/dt is a measure of speed, then speed will be maximised when its derivative, (d^2Q)/(dt^2) = 0.

 

[mlane]

thanks. I am not quite sure what you did but we went over this one in class today and did it a little differently. we used 300-30t for our original equation since the elevator we are told is decending.