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Math Help - differencial equation of flow

  1. #1
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    differencial equation of flow

    ok..so the question is...

    "During a chemical reaction, a compound is being made from two other substances. At time t hours after the start of the reaction, x g of the compound has been produced. Assuming that x = 0 initially, and that

    dx = 2(x-6)(x-3)
    dt

    show that it takes approximately 7 minutes to produce 2 g of the compund."

    We have been started off with...

    the integral of 1 over (x-6)(x-3) dx = the integral of 2 dt

    i don't have any idea where this came from or why it is the first step...can anyone help?
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  2. #2
    Super Member wingless's Avatar
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    You're doing it right. First we need an equation that shows how x is related to time.

    Solve the differential equation. The question states: "Assuming that x = 0 initially".. so x(0)=0. You can find the integration constant from this initial condition. Now we have the equation. Use it to find the answer.

    Remember: In the equation, the unit of t is hours as the question says: "At time t hours". Convert the units if needed.
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  3. #3
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    Hello, Oranges&Lemons!

    Have you never seen Differential Equations or Integration . . . ever?


    During a chemical reaction, a compound is being made from two other substances.
    At time t hours after the start of the reaction, x grams of the compound has been produced.

    Assuming that x = 0 initially, and that: . \frac{dx}{dt} \:=\:2(x-6)(x-3)
    show that it takes approximately 7 minutes to produce 2 g of the compund.

    We have: . \frac{dx}{dt} \;=\;2(x-6)(x-3)

    Separate variables: . \frac{dx}{(x-6)(x-3)} \;=\;2\,dt

    Use Partial Fractions on the left side, and we get: . \frac{1}{3}\left(\frac{1}{x-6} - \frac{1}{x-3}\right)\,dx \;=\;2\,dt

    Integrate: . \frac{1}{3}\bigg[\int\frac{dx}{x-6} - \int\frac{dx}{x-3}\bigg] \;=\;2\int dt

    . . \frac{1}{3}\left[\ln(x-6) - \ln(x-3)\right] \;=\;2t + c \quad\Rightarrow\quad \ln\left(\frac{x-6}{x-3}\right) \;=\;6t + c<br />

    Exponentiate: . \frac{x-6}{x-3} \;=\;e^{6t+c} \;=\;e^{6t}\cdot e^c \quad\Rightarrow\quad \frac{x-6}{x-3}\;=\;Ce^{6t}


    We are told that when t=0,\:x=0

    We have: . \frac{0-6}{0-3} \:=\:Ce^0\quad\Rightarrow\quad C \:=\:2

    . . Hence, the function is: . \frac{x-5}{x-3} \;=\;2e^{6t}



    When x = 2, find t.

    We have: . \frac{2-6}{2-3} \:=\:2e^{6t}\quad\Rightarrow\quad 2e^{6t} \:=\:4\quad\Rightarrow\quad e^{6t} \:=\:2

    . . 6t \:=\:\ln(2)\quad\Rightarrow\quad t \;=\;\frac{1}{6}\ln(2) \;=\;0.1155\text{ hours} \;=\;6.9315\text{ minutes}<br />

    Therefore: . t \;\approx\;7\text{ minutes}

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  4. #4
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    yes, i have seen differentiation and integration before...lots of times..i just couldnt figure out how to do this particular question.

    thanks for your help.
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