Results 1 to 9 of 9

Math Help - average temperature

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    148

    average temperature

    Hi everyone,

    Could you please tell me if this is correct?

    If a cup of coffe has temperature 95 C in a room where the temperature is 20 C, then according to Newton's Law of Cooling, the temperature of the coffee after t minutes is T(t)=20+75e^-t/50. What is the average temperature of cofee during the first half hour?

    u=-t/50
    du=-1/50dt
    -50du=-1/50dt

    int. 20+75e^-t/50 from 0 to 30
    20t-3750e^u from 0 to 30
    -1458.044

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by chocolatelover View Post
    int. 20+75e^-t/50 from 0 to 30
    20t-3750e^u from 0 to 30
    -1458.044

    Thank you very much
    Two comments:
    1) The coefficient of the exponential is -\frac{75}{50} not -75 \cdot 50.

    2) You forgot to divide by the time range.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2007
    Posts
    148
    Hi,

    1) The coefficient of the exponential is not .
    Could you please show me how you got -75/50?

    I would then have 1/30(20t-75/50e^-t/50) right?

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2007
    Posts
    148
    Hi,

    1) The coefficient of the exponential is not .
    Could you please show me how you got that?

    I got 19.973C. Does that look right?

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by chocolatelover View Post
    Hi,



    Could you please show me how you got -75/50?

    I would then have 1/30(20t-75/50e^-t/50) right?

    Thank you very much
    Sorry. I screwed up the coefficient.

    T_{ave} = \frac{1}{30 - 0} \int_0^{30}(20 + 75e^{-t/50})~dt

    Let u = -\frac{t}{50} \implies dt = -50 du <-- Typically we keep the negative sign in the exponent, but this is fine too.
    T_{ave} = \frac{1}{30} \int_0^{-30/50}(20 + 75e^{u})~(-50du)

    T_{ave} = -\frac{50}{30} \int_0^{-3/5}(20 + 75e^{u})~du <-- Your expression here was slightly off: you didn't multiply the 20 by -50.

    T_{ave} = -\frac{5}{3} \int_0^{-3/5}(20 + 75e^{u})~du

    T_{ave} = \frac{5}{3} \int_{-3/5}^0(20 + 75e^{u})~du

    T_{ave} = \frac{5}{3} (20(0)+ 75e^0) - \frac{5}{3} \left (20 \cdot \frac{-3}{5} + 75e^{-3/5} \right )

    T_{ave} = \frac{5}{3} \cdot 75 - \frac{5}{3} (-12 + 75e^{-3/5})

    T_{ave} =76.3985

    -Dan

    PS By the way, you should be able to tell that your answer is incorrect. The temperature can't go below 75 C, so the average temperature can't be that low either.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    148
    Thank you very much

    Can you explain why you changed the limits? Wouldn't it stay from 0 to 30?

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by chocolatelover View Post
    Thank you very much

    Can you explain why you changed the limits? Wouldn't it stay from 0 to 30?

    Thank you very much
    No because you have changed your variable by a factor of 50. So your limits should change as well to reflect that.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Aug 2007
    Posts
    148
    No because you have changed your variable by a factor of 50. So your limits should change as well to reflect that
    One more question, does this only happen when you are finding the average or do you also have to change the limits when you have a numberdu when solving an integral? How come for the upper limit you divided by -50 instead of multiplying?

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by chocolatelover View Post
    One more question, does this only happen when you are finding the average or do you also have to change the limits when you have a numberdu when solving an integral? How come for the upper limit you divided by -50 instead of multiplying?

    Thank you very much
    Any time that you use a substitution in a definite integral you need to alter the limits. In the general case we have:
    \int_a^b f(kx)~dx

    We obviously want to use a substitution of u = kx \implies du = k~dx. Now, we see that any value of x corresponds to a value of u such that x = \frac{u}{k}. The limits, belonging to a span of the x axis, are "x" values. So we need to change them to "u" values. Hence:
    \int_a^b f(kx)~dx = \int_{a/k}^{b/k} f(u)~\frac{du}{k}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: May 26th 2011, 06:11 AM
  2. Replies: 2
    Last Post: March 13th 2010, 05:49 AM
  3. Temperature
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 3rd 2009, 03:11 AM
  4. temperature PDE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 9th 2009, 09:32 PM
  5. Temperature
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 3rd 2009, 10:37 AM

Search Tags


/mathhelpforum @mathhelpforum