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Homework Help: Rate of Change Question

  1. Dec 4, 2017 #1
    1. The problem statement, all variables and given/known data
    A building has an external elevator. The elevator is rising at a constant rate of ##2 \; \text{ms}^{-1}##.
    Sarah is stationary, watching the elevator from a point 30m away from the base of the elevator shaft.
    Let the angle of elevation of the elevator floor from Sarah's eye level be ##\theta##.

    Find the rate at which the angle of elevation is increasing when the elevator floor is 20m above Sarah’s eye level.

    2. Relevant equations


    3. The attempt at a solution
    Am I trying to find ##\frac{d\theta}{dt} = \frac{d\theta}{dh}\times \frac{dh}{dt}##. We know ##\frac{dh}{dt} = 2##. Am I on the right track???
     
  2. jcsd
  3. Dec 4, 2017 #2

    Ray Vickson

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    Why not just keep going and see what you get?
     
  4. Dec 4, 2017 #3
    Ok then. ##h = 30 \tan{(\theta)}## so ##\frac{dh}{d\theta} = 30\sec^2{(\theta)}##. Hence, ##\frac{d\theta}{dt} = \frac{\cos^2{(\theta)}}{15}##.
     
  5. Dec 4, 2017 #4

    Dick

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    Fine so far. But you should be able to express ##\frac{d\theta}{dt}## as a numerical value. What is ##\cos(\theta)## when ##h=20##?
     
  6. Dec 4, 2017 #5

    Ray Vickson

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    Good. That is exactly what I was hoping you would do.

    I was basically encouraging you to do more on your own, by taking a chance and trying out something---win or lose. That is how we all learned.
     
  7. Dec 5, 2017 #6
    Thanks for all your help.
    I found the length of the hypotenuse then calculated ##\theta = \cos^{-1}\left(\frac{30}{\sqrt{1300}}\right)## then threw that solution into ##\frac{d\theta}{dt}##.
     
  8. Dec 6, 2017 #7

    Dick

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    Right. Though you actually didn't need to find ##\theta##, right? All you need is ##\cos(\theta)=\frac{30}{\sqrt{1300}}##.
     
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