Results 1 to 2 of 2

Math Help - Logistic function story problem

  1. #1
    Member wiseguy's Avatar
    Joined
    Jul 2010
    Posts
    101

    Logistic function story problem

    "Suppose that the water in a hot water heater is 70 Celsius above room temperature. When you first turn on the faucet, the water coming out is only 2 above room temperature. Ten seconds later it has warmed to 15 above room temperature.

    The general equation for a logistic function is:

    T(x)=c/(1+ae^-bx)

    Use the three given values to find algebraically the particular equation of this function.



    I spent some time and worked it out:


    t(x)=70/(1+ae^(-bx))

    10=70/(1+ae^(-b*15))
    10+(1+ae^(-b*15))=70
    10+10ae^(-b*15)=70
    10ae^(-b*15)=60
    a=60/(10e^(-b*15))

    a(10e^(-b*15))=60
    10ae^(-b*15)=60
    ae^(-15b)=6
    log(ae^(-15b))=log(6)
    log(a)-15b*log(e)=log(6)
    -15b*log(e)=-log(a)+log(6)
    b=(log(a)-log(6))/(15log(e))
    b=(log(60/(10e^(-b*15)))-log(6))/(15log(e))
    b=(log(60/(10e^(-b*15)))-0.7781512504)/(6.514417229)
    b=(log(6(1e^15b))-0.7781512504)/(6.514417229)
    b=(log(16.30969097^15b)-0.7781512504)/(6.514417229)
    6.514417229b=log(16.30969097^15b)-0.7781512504
    7.292568479=log(16.30969097^15b)
    10^7.292568479=16.30969097^15b
    19614104.26=16.30969097^15b
    log(base 16.30969097)19614104.26=15b
    6.01475867=15b
    b=0.4009839113

    a=60/(10e^(-b*15))
    a=60/(10e^(-0.4009839113*15))
    a=60/10e^-6.01475867
    a=2456.562119

    t(x)=70/(1+2456.562119e^(-0.4009839113x))




    Unfortunately, when I tested a point, it was wrong.

    What did I do incorrectly?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,579
    Thanks
    1418
    Quote Originally Posted by wiseguy View Post
    "Suppose that the water in a hot water heater is 70 Celsius above room temperature. When you first turn on the faucet, the water coming out is only 2 above room temperature. Ten seconds later it has warmed to 15 above room temperature.

    The general equation for a logistic function is:

    T(x)=c/(1+ae^-bx)

    Use the three given values to find algebraically the particular equation of this function.



    I spent some time and worked it out:


    t(x)=70/(1+ae^(-bx))
    Okay. Any time you cite a formula you should specify what all the variables represent but I think I am with you up to here. I think that "T(x)" is the temperature (above room temperature) after the tap has been open x seconds.

    10=70/(1+ae^(-b*15))
    Now you have lost me! Are you saying that "T" is the time and "x" is the temperature? I recommend you verify that. If so, you are going to have the temperature going to infinity as the time goes to some constant rather than the other way around! I think you have "x" and "T" reversed.

    10+10ae^(-b*15)=70
    10ae^(-b*15)=60
    a=60/(10e^(-b*15))

    a(10e^(-b*15))=60
    10ae^(-b*15)=60
    ae^(-15b)=6
    log(ae^(-15b))=log(6)
    log(a)-15b*log(e)=log(6)
    -15b*log(e)=-log(a)+log(6)
    b=(log(a)-log(6))/(15log(e))
    b=(log(60/(10e^(-b*15)))-log(6))/(15log(e))
    b=(log(60/(10e^(-b*15)))-0.7781512504)/(6.514417229)
    b=(log(6(1e^15b))-0.7781512504)/(6.514417229)
    b=(log(16.30969097^15b)-0.7781512504)/(6.514417229)
    6.514417229b=log(16.30969097^15b)-0.7781512504
    7.292568479=log(16.30969097^15b)
    10^7.292568479=16.30969097^15b
    19614104.26=16.30969097^15b
    log(base 16.30969097)19614104.26=15b
    6.01475867=15b
    b=0.4009839113

    a=60/(10e^(-b*15))
    a=60/(10e^(-0.4009839113*15))
    a=60/10e^-6.01475867
    a=2456.562119

    t(x)=70/(1+2456.562119e^(-0.4009839113x))




    Unfortunately, when I tested a point, it was wrong.

    What did I do incorrectly?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 3 differential equations, including the logistic function
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: December 20th 2010, 02:40 AM
  2. Logistic Discriminant Function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 2nd 2010, 03:48 AM
  3. [SOLVED] Power function story problem!
    Posted in the Pre-Calculus Forum
    Replies: 23
    Last Post: July 25th 2010, 09:05 PM
  4. Logistic Problem - Help Please!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 7th 2009, 09:23 PM
  5. Logistic function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 9th 2009, 04:54 AM

/mathhelpforum @mathhelpforum