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Quick limit question

  1. Jan 28, 2007 #1
    I already found this equation:
    T= 15/1.36(.6*sin(t)-cos(t))+100-23.971/e^(.6t)

    Now I'm supposed to find the range of temperatures (T) as t approaches infinity.

    I tried 100+(15/1.36*.6*sin(90))
    and 100-(15/1.36*cos(0))

    but that's not right. What am I missing?

    Thanks.
     
  2. jcsd
  3. Jan 29, 2007 #2
    If I got it right your equation looks like this:
    [tex]\frac{15}{1.36}[0.6sin(t)-cos(t)]+100-\frac{23.971}{e^{0.6t}}[/tex]

    Since [tex]\frac{23.971}{e^{0.6t}}[/tex] goes to [tex]0[/tex] when t goes to [tex]\infty[/tex] you should calculate the maximum and the minimum of the expression [tex]\frac{15}{1.36}[0.6sin(t)-cos(t)]+100[/tex] and this will be your range of temperatures.
     
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