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Math Help - calc II problem

  1. #1
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    calc II problem

    Can anyone help me with this problem?

    a) The oil consumption rate satisfies the equation C(t) = C0e^rt, where C0 is the consumption rate at t=0 (the number of barrels per year) and r is a constant. If the consumption rate is C0 = 2.5x10^10 barrels per year in 1976 and r= 0.06 how long will it take before 2x10^12 barrels are used up?

    b) As the fuel is almost used up, the prices will probably skyrocket and other sources of energy will be turned to. Let S(t) be the supply left at time t. Assume that S= ds/dt = - (alpha)S, where (alpha) is a constant. Find S(t).
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  2. #2
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    Quote Originally Posted by ewander View Post
    Can anyone help me with this problem?

    a) The oil consumption rate satisfies the equation C(t) = C0e^rt, where C0 is the consumption rate at t=0 (the number of barrels per year) and r is a constant. If the consumption rate is C0 = 2.5x10^10 barrels per year in 1976 and r= 0.06 how long will it take before 2x10^12 barrels are used up?
    C(t) = C_0e^{rt}

    from the problem C_0 = 2.5\times 10^{10} and r= 0.06 so

    C(t) = 2.5\times 10^{10}\times e^{0.06t}

    We need to find t for when C(t) = 2\times 10^{12} so we sub this into what we have and solve for t.

    2\times 10^{12}= 2.5\times 10^{10}\times e^{0.06t}

    \frac{2\times 10^{12}}{2.5\times 10^{10}}= e^{0.06t}

    ln\left(\frac{2\times 10^{12}}{2.5\times 10^{10}}\right) = 0.06t

    Can you finish it from here?
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  3. #3
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    Quote Originally Posted by ewander View Post

    b) As the fuel is almost used up, the prices will probably skyrocket and other sources of energy will be turned to. Let S(t) be the supply left at time t. Assume that S= ds/dt = - (alpha)S, where (alpha) is a constant. Find S(t).
    \frac{ds}{dt}=-\alpha \times s

    First of all separate the variables

    \frac{ds}{s}= -\alpha dt

    Now integrate both sides

    \ln(s)=-\alpha t+c

    You now have no more derivatives, can you get s on its own?
    Last edited by pickslides; July 20th 2009 at 03:05 PM. Reason: typo
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  4. #4
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    a) ln(100) = 0.06t
    t=ln(100)/0.06 = 77
    Thanks!!

    but b) ?????
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  5. #5
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    Quote Originally Posted by ewander View Post

    but b) ?????
    \frac{ds}{dt}=-\alpha \times s

    First of all separate the variables

    \frac{ds}{s}= -\alpha dt

    Now integrate both sides

    \ln(s)=-\alpha t+c

    s(t)=e^{-\alpha t+c}

    any clearer...?
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