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Use the model to compare

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is given.
    Use this model to compare how the number of hours of daylight is increasing in Philadelphia on June 5 and June 15. (Round your answers to four decimal places).

    2. Relevant equations

    L(t)=12+2.8sin[(2pi/365)(t-80)]

    3. The attempt at a solution

    i have no clue were to start. or what to do, please help.

    i tried
    L(t)=12+2.8sin[(2pi/365)(5-80)]=9.3088
    L(t)=12+2.8sin[(2pi/365)(15-80)]= 9.481

    it seemed to easy to be right, and of course it was. please help?
     
  2. jcsd
  3. Nov 30, 2011 #2

    Mark44

    Staff: Mentor

    This is L(5). L(t) is what you have in Relevant equations.
    This is L(15).
    What they're asking about is the slope of the tangent line at a) (5, L(5)) and b) (15, L(15)).

    For the first one, calculate the slope of a secant line for two points on the graph of your function. The two points should straddle (5, L(5)). The closer together the two points are, the better your approximation will be.

    Do the same for the other point at (15, L(15)).
     
  4. Nov 30, 2011 #3
    okay, so
    a) (5, 9.3088)
    for the slope, should i use a point to the left and right of 5, such as 4.99 and 5.01?

    b) (15, 9.481)
    and the same for this part, 14.99 and 15.01?
     
  5. Nov 30, 2011 #4

    Mark44

    Staff: Mentor

    Right, those would be good choices.
     
  6. Nov 30, 2011 #5
    a) (4.99,9.3087) (5.01,9.309)=(9.309-9.3087)/(5.01-4.99)= .015

    b) (14.99,9.4808)(15.01,9.4812)= (9.4812-9.4808)/(15.01-14.99)= .02

    so thats my answer?
     
  7. Nov 30, 2011 #6

    Mark44

    Staff: Mentor

    Let's clean this up.
    The slope of the secant line between (4.99,9.3087) and (5.01,9.309) is
    (9.309-9.3087)/(5.01-4.99)= .015

    Unfortunately, you aren't evaluating your function correctly, something I didn't notice before.
    L(5) represents the number of hours of daylight on Jan 5. You're going to have to count days to see which days of the year June 5 and June 15 are.



     
  8. Nov 30, 2011 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Since this is the "Calculus and Beyond" Forum, my first suggestion would be to use exact derivatives. If you don't know what derivatives are or how to compute them, then we have a different discussion!

    RGV
     
  9. Nov 30, 2011 #8

    Mark44

    Staff: Mentor

    That's probably good advice. I was operating under the assumption (probably unwarranted) that the OP had not yet been exposed to differentiation.
     
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