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Word problem involving sinusoidal model

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data

    O4VHjHQ.png


    2. Relevant equations

    [tex]y=30\sin(\dfrac{2\pi}{20}t)+270[/tex]

    General principal solutions:

    [tex]t=\left(\dfrac{\arcsin(\dfrac{1}{3}) 20}{2\pi}\right)+20k, k\in \mathbb{Z}[/tex]

    [tex]t=1.08173+20k[/tex]


    General symmetry solutions:

    [tex]t=\left(-\dfrac{(\arcsin(\dfrac{1}{3})-\pi)20}{2\pi}\right)+20k, k\in \mathbb{Z}[/tex]

    [tex]t=8.91827+20k[/tex]

    3. The attempt at a solution

    What I did was I divived 30 by the difference between these two values of time to find how many such segments I need to accumulate 30 min of 280 degrees:

    [tex]\dfrac{30}{8.91827-1.08173}=3.828[/tex]

    And then I can find the answer with a bunch of manual work, involving referring to the graph, but I just feel like this is a very ineffective approach.
     
    Last edited: Jul 14, 2014
  2. jcsd
  3. Jul 14, 2014 #2

    Simon Bridge

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    The problem statement did not come through.
     
  4. Jul 14, 2014 #3
    Thanks. Fixed it.
     
  5. Jul 14, 2014 #4

    Simon Bridge

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    ... this is the equation for the temperature of the oven right?
    Did you try sketching this and drawing a horizontal line for the desired temperature?

    So your strategy is to find out how much time T1 in one cycle y>280F, divide the cooking time by T1 to find out how many cycles you need, then multiply that by the period.

    I think the main thing you have to watch for is the last fraction of a period... that what you wanted to refer to the graph for?

    Anyway - that's the approach all right.
     
  6. Jul 14, 2014 #5
    5qyxD9x.png

    Yes, well, I find where the 4th segment ends and then subtract (4-3.828) times segment from it.

    [tex] t=8.91827+20*3 [/tex] gives me coordinate where 4th segment ends and then I do the following:
    [tex] t=(8.91827+20*3)-((4-3.828)*8.91827) [/tex]

    So, yes, I just wanted to know if there is a better way to solve it or maybe it is supposed to be such a graph-involving problem. Well, maybe with calculus though, but I'm not yet at that level.
     
  7. Jul 14, 2014 #6
    Well, actually not that much of "manual" work, just a less algebraic approach.
     
  8. Jul 14, 2014 #7

    Simon Bridge

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    Well you could do it algebraically - but the graph approach is easier to think about.
    AFAIK there is no fancy magic-wand approach - pretty much all alternatives involve headaches.
     
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