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C4 Integration Question Watch

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    I already have answers for these, I'm just doubtful about the way I've worded them and it seems kinda wrong to me :/

    Could someone work through these and show me how they'd do them? Thx in advance for any help

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    (Original post by Louisb19)
    Rate of concentration is given by  \dfrac{\mathrm{d}C}{\mathrm{d}t} , the concentration is given by  C .

    Since  \dfrac{\mathrm{d}C}{\mathrm{d}t} goes up as  C goes down (negatively proportional) the relationship can be modeled by the differential equation

     \dfrac{\mathrm{d}C}{\mathrm{d}t} = k C , where k is a negative constant.

    So the relationship can therefore me modelled by  \dfrac{\mathrm{d}C}{\mathrm{d}t} = -k C where k is positive constant.

    ii)  t = \ln{\dfrac{1}{kC}} + \ln{A} = \ln{\dfrac{A}{kC}}

      C = \dfrac{A}{k} e^{-t}
    Yeah, thanks for that. That's what I did woo

    iii) Not sure exactly how you are supposed to handle the A and k constants together (if you are supposed to merge them or what) so I'll leave this for someone else.
    Damn, this was the bit I needed the help with... I'm having the same problem in that you can't really define k due to that extra constant :/
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    Can you post your workings because for the general solution I got C=Ae^(-kt)
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    But anyway if at t=0 C=C_0
    This means that the constant you have in front of the exponential term will be C_0.
    At t=4, C= 0.1C_0
    Then just solve for k using logarithms.
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    When I I integrate (cos (x))^4.
    Why cant i use inverse chain rule and why cant i use inverse chain rule if i want to integrate,

    (Cos (4x))^4
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    (Original post by justthatboy)
    When I I integrate (cos (x))^4.
    Why cant i use inverse chain rule and why cant i use inverse chain rule if i want to integrate,

    (Cos (4x))^4
    By using reverse chain rule here you will get a function that includes the product of cosines and sine. But if you differentiate a function that is a product of dimes and cosines you is the product rule - not the chain rule. So you would have to use the reverse product rule (integration by parts or use trig identities or substitution).
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    (Original post by Louisb19)
    ii)  t = \ln{\dfrac{1}{kC}} + \ln{A} = \ln{\dfrac{A}{kC}}
    (Original post by Alexion)
    Yeah, thanks for that. That's what I did woo
    That's not correct, unfortunately, you get \displaystyle \int \frac{\mathrm{d}C}{C} = -kt + c, so C = \exp\left(-kt + c\right) = Ae^{-kt}.

    You know that t = 0 \Rightarrow C = C_0, so C = C_0 e^{-kt}.

    Now, plug in the info for t=4 and you should be able to solve for k.
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    (Original post by Zacken)
    That's not correct, unfortunately, you get \displaystyle \int \frac{\mathrm{d}C}{C} = -kt + c, so C = \exp\left(-kt + c\right) = Ae^{-kt}.

    You know that t = 0 \Rightarrow C = C_0, so C = C_0 e^{-kt}.

    Now, plug in the info for t=4 and you should be able to solve for k.
    Ah. Oops. Just a coincidence that Louis and I made the same mistake then haha

    Thanks for the help
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    (Original post by Alexion)
    Ah. Oops. Just a coincidence that Louis and I made the same mistake then haha

    Thanks for the help
    Louis Van Gaal?
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    (Original post by TeeEm)
    Louis Van Gaal?
    @Louisb19 ya silly poop

    But who's to say that Van Gaal didn't make the same mistake in a maths A-level?
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    (Original post by Alexion)
    @Louisb19 ya silly poop

    But who's to say that Van Gaal didn't make the same mistake in a maths A-level?
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