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Thread: Rate of Change

  1. #1
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    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
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  2. #2
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    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 $\displaystyle t_1$ minutes and 100,000 bacteria at $\displaystyle (t_1+10)$ minutes.

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

    Let $\displaystyle P$ = population of bacteria.

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

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

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

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


    $\displaystyle \begin{array}{ccccc}\text{When }t = t_1,\;P = 20,\!000\!: & 10,\!000\,e^{kt_1} &=&20,\!000 & [1] \\

    \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]: .$\displaystyle \frac{10,\!000\,e^{kt_1}\cdot e^{10k}}{10,\!000\,e^{kt_1}} \;=\;\frac{100,\!000}{20,\!000} $

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


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

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



    B) How many bacteria were there after 20 minutes?

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




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

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

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

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

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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    thank you very much for your explanation but i just have one question. I followed you perfectly up until where you found $\displaystyle k$.

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

    thank you.
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