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Homework Help: Possible Integration Problem

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data
    The mean daily temperature in degrees fahrenheit in Athens, GA, t months after July 15 is closely approximated by T=61+18cos(πt/6)
    Find the average temp. between September 15 (t=2) and December 25 (t=5).

    2. Relevant equations
    T=61+18cos(πt/6)

    3. The attempt at a solution
    Can I just plug in t=2 and t=5 into the equation, add the answers, then divide by two to get the average temp. Or do I have to use integration. I'm not sure because the question is asked in the chapter on integration but it doesn't look like I need to use it.
     
  2. jcsd
  3. Aug 18, 2008 #2

    CompuChip

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    I think you have to find the average of the function by integration. Adding and dividing by two works if the function is linear. For example, suppose that between t = 2 and t = 4.9 the temperature is 40 and between t = 4.9 and t = 5 it is -10. Then surely you would disagree if I said that the average temperature between t = 2 and t = 5 was (40 - 10)/2 = 15 degrees?
     
    Last edited: Aug 18, 2008
  4. Aug 18, 2008 #3
    You have to use integration. The reason for this is because your function [tex]t: [a,b] \rightarrow R[/tex] is continuous. By the Mean Value Theorem for Integrals then there is a point [tex]x_{0}[/tex] in [a,b] such that

    [tex]\frac{1}{b-a} \int_a^{b} t = t(x_{0})[/tex] where [a,b] is [2,5].
     
  5. Aug 18, 2008 #4
    I tried using the Mean Value Theorem and came up with this.
    [tex]
    \frac{1}{5-2}\int_2^{5} 61+18\cos\frac{\pi t}{6} dt = 61+18\cos\frac{\pi t_{0}}{6}[/tex]

    [tex]
    \frac{1}{3} 61t+\frac{108}{\pi}sin\frac{\pi t}{6} \mid_2^{5} = \frac{-\sqrt3+1}{\pi}[/tex]

    [tex]
    61+18cos\frac{\pi t_{0}}{6} = \frac{-\sqrt3+1}{\pi}[/tex]

    [tex]t_{0}\approx 3.449[/tex]

    When I plug this into [tex]61+18cos\frac{\pi t}{6}[/tex] I get approximately 56.8056. Is this right?
     
    Last edited: Aug 18, 2008
  6. Aug 18, 2008 #5
    assuming that you got initial equations correct, I also got same answer ..

    >> int ('1/3*(61+18*cos(pi*t/6))','t',2,5)

    ans =

    -(-61*pi+18*3^(1/2)-18)/pi


    >> eval(ans)

    ans =

    56.8057

    >> (ans-61)/18

    ans =

    -0.2330

    >> acos(ans)

    ans =

    1.8060

    >> ans*6/pi

    ans =

    3.4492

    >>

    EDIT: ONLY THING I WOULD BE WORRIED ABOUT IS USING ACOS VALUE OF CALCULATOR, IT IS WRONG SOMETIMES ... (I will see into this) but 3.4492 looks reasonable
     
    Last edited: Aug 18, 2008
  7. Aug 19, 2008 #6

    CompuChip

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    The answer is correct. No need for the mean value theorem:
    [tex]\operatorname{avg}\limits_{[a,b]} f(x) = \frac{1}{b - a} \int_a^b f(x) \, \mathrm dx[/tex]
    so in this case the average temperature is
    [tex]61 + \frac{1}{3} \int 18 \cos(\pi t / 6) \, \mathrm dt = 61 + 6 \left[ \frac{6}{\pi} \sin(\pi t / 6) \right]_{t = 2}^{t = 5} = 61 + \frac{18}{\pi} \left( 1 - \sqrt{3} \right) [/tex]
    which evaluates numerically to approximately 56.8057.
     
  8. Aug 19, 2008 #7

    HallsofIvy

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    I must say that I am not at all crazy about the way you write things!
     
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