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A Rate of change and a differentials problem

  1. Nov 19, 2007 #1
    First off i'd like to say Hi to the forums, hehe. I didn't really see a new member area but I suppose this will do. Right now I'm kind of struggling with these two problems that I recently took a quiz on and didn't do so well. I've been trying to figure out how to work them out but i just dont really get it. if anyone could help out it'd be awesome.

    (1)

    The position (in cm) of a particle along the x-axis at time t(in s) is given by X(t) = t^3-6t^2+8t, t is greater than or equal to 0.

    the question:
    When is the particle moving to the left? the right?
    Acceleration of the particle when the speed is zero.
    Total distance traveled from t=0 to t=2.


    (2)

    ( and the one that REALLY gets me)
    A person standing 30ft from the base of a building measures the angle of elevation to the top of the building as being 75 degrees with an error of +or- 1.2 degrees.

    What is the max error in estimating the height of the building, and the percentage error?

    This one has taken me quite a while, I am thinking it has something to do with the law of sines, but when i try it with that method i'm getting a completely unreasonable answer. I kind of have the right idea, i hope, but for some reason i'm just not seeing it.

    Thank you in advance for those who wish to take on these.
     
  2. jcsd
  3. Nov 19, 2007 #2
    (1.) Let x' be the first derivative with respect to t, x'' the second.
    a.) Right when x'>0, left when x'<0.
    b.) Find at what time t the velocity x'=0, and then substitute this into x''.
    c.) compute x(2)-x(0)

    (2.) If you know the angle of elevation, a, the height of the building is given by
    h = 30ft * tg(a)
    Now the error in your case is given by Delta a = +-1.2 degrees = +- 0.02 radians, and the resulting error in h is:
    [tex]\Delta h = (1+\tan(a)^2) \Delta a [/tex]
    (If you need further explanations on that one, ask.)
    Now you merely need to substitute.
    Good luck!



    --------
    Assaf
    Physically Incorrect
     
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