Results 1 to 3 of 3

Math Help - Rate of Change

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    19

    Rate of Change

    The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experiment, a culture had 10,000 bacteria initially, 20,000 a time t1 minutes and 100,000 bacteria at (t1+10) minutes.
    A)in terms of t only, find the number of bacteria in the culture a any time t minutes, t≥0.
    B)How many bacteria were there after 20 minutes.
    C)How many minutes had elasped when the 20,000 bacteria were observed
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,807
    Thanks
    697
    Hello, calc_help123!

    The rate of change in the number of bacteria in a culture is proportional to the number present.
    In a certain laboratory experiment, a culture had 10,000 bacteria initially,
    20,000 a time t_1 minutes and 100,000 bacteria at (t_1+10) minutes.

    A) In terms of t only, find the number of bacteria in the culture a any time t \geq 0.

    Let P = population of bacteria.

    We have: . \frac{dP}{dt} \:=\:kP \quad\Rightarrow\quad \frac{dP}{P} \:=\:k\,dt \quad\Rightarrow\quad \ln P \:=\: kt + c \quad\Rightarrow\quad P \:=\:e^{kt+c}

    . . . . P \:=\:e^{kt}\cdot e^c \:=\:e^{kt}\cdot C \quad\Rightarrow\quad P \:=\:Ce^{kt}

    When t = 0,\:P = 10,\!000\!:\quad 10,\!000 \:=\:Ce^0 \quad\Rightarrow\quad C \:=\:10,\!000

    The function (so far) is: . P \;=\;10,\!000\,e^{kt}


    \begin{array}{ccccc}\text{When }t = t_1,\;P = 20,\!000\!: & 10,\!000\,e^{kt_1} &=&20,\!000 & [1] \\<br /> <br />
\text{When }t = t_1\!+\!10,\;P = 100,\!000\!: & 10,\!000\,e^{k(t_1+10)} &=& 100,\!000 & [2] \end{array}

    Divide [2] by [1]: . \frac{10,\!000\,e^{kt_1}\cdot e^{10k}}{10,\!000\,e^{kt_1}} \;=\;\frac{100,\!000}{20,\!000}

    . . e^{10k} \:=\:5 \quad\Rightarrow\quad 10k \:=\:\ln 5 \quad\Rightarrow\quad k \:=\:\frac{\ln5}{10}


    The function is: . P \;=\;10,\!000\,e^{\frac{\ln5}{10}t} \;=\;10,\!000\,\left(e^{\ln 5}\right)^{\frac{t}{10}}

    . . Hence: . P \;=\;10,\!000\cdot5^{\frac{1}{10}t}



    B) How many bacteria were there after 20 minutes?

    When t=20\!:\;\;P \;=\;10,\!000\cdot5^{(\frac{1}{10})(20)} \;=\;10,\!000\cdot5^2 \;=\;250,\!000




    C) How many minutes had elasped when the 20,000 bacteria were observed?

    We have: . 10,000\cdot5^{\frac{1}{10}t_1} \:=\:20,000 \quad\Rightarrow\quad 5^{\frac{1}{10}t_1} \:=\:2

    Take logs: . \ln\left(5^{\frac{1}{10}t_1}\right) \:=\:\ln2 \quad\Rightarrow\quad \tfrac{1}{10}\,t_1\cdot \ln(5) \:=\:\ln 2

    Therefore: . t_1 \;=\;\frac{10\ln 2}{\ln 5} \;\approx\; 4.3\text{ minutes}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
    Joined
    Sep 2006
    Posts
    157
    thank you very much for your explanation but i just have one question. I followed you perfectly up until where you found k.

    i do not get how P \;=\;10,\!000\,\left(e^{\ln 5}\right)^{\frac{t}{10}} is true. and how did you get P \;=\;10,\!000\cdot5^{\frac{1}{10}t}?

    thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 12th 2011, 09:51 AM
  2. rate change
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 11th 2008, 03:11 AM
  3. rate of change
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 20th 2008, 07:53 AM
  4. rate of change
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 3rd 2007, 09:39 AM
  5. Rate of change
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 18th 2007, 07:27 AM

Search Tags


/mathhelpforum @mathhelpforum