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C3 help - differentiating lnx and e^x

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    The question is:

    9) In a biological compound, bacteria are being grown in a culture. The mass of the bacteria at time t hours is P milligrams. At time t = 0, p = 3 and dP/dt = 6.

    (i) A standard model for this situation is given by P = Ae^kt, where A and K are constants.

    a) Write down dp/dt in terms of A, k and t. Find the values of A and k.

    I don't understand what it wants me to do really, because it isn't simply differentiating P is it?
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    (Original post by Gilsenan)

    I don't understand what it wants me to do really, because it isn't simply differentiating P is it?

    Yes it is

    You have both P and P' at t=0 so can find all of the constants
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    You can differentiate P to give Ake^kt
    Then you can use that with t=0 to find values for A and K
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    You can't differentiate e^x? Lawl.
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    (Original post by Gilsenan)
    The question is:

    9) In a biological compound, bacteria are being grown in a culture. The mass of the bacteria at time t hours is P milligrams. At time t = 0, p = 3 and dP/dt = 6.

    (i) A standard model for this situation is given by P = Ae^kt, where A and K are constants.

    a) Write down dp/dt in terms of A, k and t. Find the values of A and k.

    I don't understand what it wants me to do really, because it isn't simply differentiating P is it?
    I don't think you have copied the question correctly.

    Do you mean at t=0, p = 3

    or do you mean that when p = 3 , dp/dt = 6
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    (Original post by steve2005)
    I don't think you have copied the question correctly.

    Do you mean at t=0, p = 3

    or do you mean that when p = 3 , dp/dt = 6
    The question is correct.

    At t=0, p=3 and dp/dt=6.
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    (Original post by Gilsenan)
    The question is:

    9) In a biological compound, bacteria are being grown in a culture. The mass of the bacteria at time t hours is P milligrams. At time t = 0, p = 3 and dP/dt = 6.

    (i) A standard model for this situation is given by P = Ae^kt, where A and K are constants.

    a) Write down dp/dt in terms of A, k and t. Find the values of A and k.

    I don't understand what it wants me to do really, because it isn't simply differentiating P is it?
     \displaystyle P=Ae^{kt}

    Sub in P=3 and t=0, in the above equation to find the value of A.

    Now differentiate,  \displaystyle P=Ae^{kt} , with respect to 't' and then sub in t=0 and dP/dt=6 to find the value of 'k'.
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    (Original post by raheem94)
     \displaystyle P=Ae^{kt}

    Sub in P=3 and t=0, in the above equation to find the value of A.

    Now differentiate,  \displaystyle P=Ae^{kt} , with respect to 't' and then sub in t=0 and dP/dt=6 to find the value of 'k'.
    Why are you so sure the question is correct. The problem is that when t = 0 the k is impossible to evaluate. ( At least that's what I think ..)
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    Thanks guys, got it now The question is correct, I've double checked.
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    (Original post by steve2005)
    Why are you so sure the question is correct. The problem is that when t = 0 the k is impossible to evaluate. ( At least that's what I think ..)
    'k' can be evaluated when you find dP/dt.

    See the spoiler
    Spoiler:
    Show

     \displaystyle P=Ae^{kt}
    When we sub in t=0 and P=3, we can find A, A=3.

     \displaystyle \frac{dP}{dt}=kAe^{kt}=3ke^{kt}
    When t=0, dP/dt=6,
     \displaystyle \frac{dP}{dt}=3ke^{kt} \implies 6=3ke^0 \implies \boxed{k=2}

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    (Original post by Gilsenan)
    Thanks guys, got it now The question is correct, I've double checked.

    Is this the solution?




    Uploaded with ImageShack.us
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    (Original post by steve2005)
    Is this the solution?




    Uploaded with ImageShack.us
    Yes, its correct. I have a similar solution in the spoiler in my post.

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