1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Average Value of a Function - temperature

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data
    If a cup of tea has temperature 98C in a room where the temperature is 20C, then according to Newton's Law of Cooling the temperature of the tea after t minutes is
    [tex]$\displaystyle \Large T(t)=20 + 78 e^{-t/50}.$[/tex]
    What is the average temperature of the tea during the first 38 minutes?


    2. Relevant equations
    [tex]$ fave ={1}/{b-a} \int _a^bf(x)dx$[/tex]


    3. The attempt at a solution
    [tex]$ fave ={1}/{98-20} \int _{20}^{98} T(t)dt$[/tex] [tex]${fave} =\frac{1}{78}\int _{20}^{98} 20 + 78 e^{-t/50}dt?$[/tex]
     
  2. jcsd
  3. Feb 9, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Well, no. It's integral from t=0 to t=38. Why are you putting the bounds for T in instead of t? It's an average temperature over a time interval.
     
  4. Feb 9, 2010 #3
    so [tex]
    ${fave} =\frac{1}{38}\int _{0}^{38} 20 + 78 e^{-t/50}dt?$
    [/tex]

    i dont understand your question.. how should i construct the question?
    in my textbook they wrote like 2 sentences about temperature and i dont get it.. they dont even have an example for this type of questions
     
  5. Feb 9, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's fine. What's your question? You want to average T(t) from t=0 to t=38. That's correct. You just put the T limits in instead of t. That's wrong.
     
  6. Feb 9, 2010 #5
    [tex] ${fave} =\frac{1}{38}\int _{0}^{38} 20 + 78 e^{-t/50}dt$ [/tex]
    my problem is to find the antiderivative of this i think.
    would it be:
    [tex] ${fave} =\frac{1}{38} (\int _{0}^{38} 20dt + \int _{0}^{38}78 e^{-t/50}dt )$ [/tex]
    [tex] ${fave} =\frac{1}{38} (20t+C|^3^8 + \int _{0}^{38}78 e^{-t/50}dt )$ [/tex]
    [tex] ${fave} =\frac{1}{38} (760 + 78 \int _{0}^{38}e^{-t/50}dt )$ [/tex]

    let u = -t/50
    dt = -50du
    [tex] ${fave} =\int _{0}^{38}e^{u}dt $ [/tex]
    [tex] ${fave} =\int _{0}^{38}e^{u}(-50)du $ [/tex]
    [tex] ${fave} =(-50)* \int _{0}^{38}e^{u}du $ [/tex]
    [tex] ${fave} =(-50)* e^{-t/50} + C |^3^8 $ [/tex]

    -50 * e^(-38/50) = -23.38

    [tex] ${fave} \frac{1}{38} (760 + 78 * (-23.38) )$ [/tex] = -27.997

    im sure its wrong.. but where did i f.. up?
     
    Last edited: Feb 9, 2010
  7. Feb 9, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, it's definitely wrong. The mean should be somewhere between 98C and 20C. You are just being sloppy. As to where it happened, I would look at where you have a du integral and you still have t limits. If you change the variable from t to u aren't you supposed to change the limits as well?
     
  8. Feb 9, 2010 #7
    not quite what you did was forget to subtract e^ (-t/50) from e^0.



    [tex]
    ${fave} =\int _{0}^{38}e^{u}(-50)du $
    [/tex]

    is

    evaluates to

    e^ (-38/50) - e ^0
    which is something like -.54
    times your constant which was -78*50/38
    gives you a positive number, somewhere around 55.
    add that to

    [tex]
    ${fave} =\int _{0}^{38}du $
    [/tex]

    The correct exact answer is 74.63, which makes perfect sense.
     
    Last edited: Feb 9, 2010
  9. Feb 9, 2010 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are forgetting to change the t limits to u limits as well.
     
  10. Feb 9, 2010 #9
    cant believe how stupid my mistake was :bugeye: this small e^0 changed everything...
    well.. at least i get how to do this kind of question now! its actually not that hard when i think about it now :rolleyes:

    thanks once more guys :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Average Value of a Function - temperature
Loading...