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This integral of k dt needed to get this mark-M1?

  1. Mar 2, 2007 #1

    inv

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    [Solved]This integral of k dt needed to get this mark-M1?

    1. The problem statement, all variables and given/known data
    Hi.The material,the question and the marking scheme respectively for my pure maths paper 3 question x1 only is shown below,it's an A2-level Oct 2002 paper if u know what that is.

    [​IMG]

    [​IMG]

    Is the integral circled in red needed to integrate?I find I don't need it.Am I wrong,if so why?


    2. Relevant equations
    Simple integration rules only.


    3. The attempt at a solution
    I did attempt by doing :
    Integral of dt/da = integral of (.1 a^-1 + .1(10-a)^-1)
    = .1 ln(a/(10-a))

    Ps- there, I did it without the integral circled in red.
     
    Last edited: Mar 2, 2007
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  3. Mar 2, 2007 #2

    Dick

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    You did the integral, but what's the solution to the problem? I.e. What is a as a function of t? I don't see any t dependence in your answer.
     
  4. Mar 2, 2007 #3

    inv

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    The question is asking for t to be expressed in t=ca+d ,where a is a variable and the rest is constant.

    Solution:
    dt/da=250 {a(10-a)}^-1

    t=integral of dt/da
    =what I integrated

    There's the t dependence in my answer too.
     
    Last edited: Mar 2, 2007
  5. Mar 2, 2007 #4

    Dick

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    Think clearly about this. If you believe you solved it, you ought to be able to tell me the value of a at t=1. How do you do that?
     
  6. Mar 2, 2007 #5

    inv

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    Previous post I said "what I integrated" is on my hard copy. This is what it is-
    t=25ln{a/(10-a)} .


    I'm afraid if t=1 , I don't know how to find a.
     
    Last edited: Mar 2, 2007
  7. Mar 2, 2007 #6

    Dick

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    Ok. If you think about it, you did integrate k*dt. That's where the 250 came from in your expression. k=1/250. The integration is pretty trivial, but in solving equations like this it can easily come out to be not so. That's why it's written this way. The LHS could have been f(t)*dt for some complicated function f.
     
  8. Mar 2, 2007 #7

    inv

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    What do you mean
    ?
     
  9. Mar 2, 2007 #8

    Dick

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    I'm saying that if you do other differential equations by separation of variables, the 't' integration can be just as hard as the 'a' integration after you've separated the equation into f(t)*dt=g(a)*da. In this case the t dependence was so simple you didn't notice that you had integrated it.
     
  10. Mar 2, 2007 #9

    inv

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    U're implying that for every integration question like this on my paper 3 pure maths I need to specify the integral of $#%$ da = integral of k*dt like this case?

    Edit :"$#%$" there wasn't a bad word ,I meant whatever the equation may be type "$#%$" da Dick. U're implying that?
     
    Last edited: Mar 2, 2007
  11. Mar 2, 2007 #10
    Not just for the maths papers. Deciding on the integrals is the next step. You are less likely to make a mistake if you write down each step.

    Dick gave good advice.

    Getting into the habit of doing this even for a simple case will help you when the going gets more difficult.
     
    Last edited: Mar 2, 2007
  12. Mar 2, 2007 #11

    inv

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    Hello Dick,u're implying it ?
     
  13. Mar 2, 2007 #12

    cristo

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    You want to solve the equation [tex]\frac{da}{dt}=ka(10-a)[/tex]. This is a 1st order separable differential equation, and so to solve, we separate the variables so that we obtain [tex]\int \frac{da}{a(10-a)}=\int kdt[/tex]. The fact that we can separate this equation means that we can solve it by direct integration. Now, in this case, the t integral is trivial; namely the RHS=kt, however there are cases where this will not be so.

    For example, suppose that the equation was [tex]\frac{da}{dt}=ka(10-a)t[/tex]. This is still separable, but now the t integral is non-trivial, and so on separation we obtain [tex]\int \frac{da}{a(10-a)}=\int kt dt[/tex]. This can still be solved, but now the RHS = k/2 t2.

    The point that Dick was trying to make, and that the solution makes, is that you should get into the habit of separating the equation properly before integrating since, in general, the t integral is non-trivial.

    Does this make any more sense?

    edit: and yes, for the sake of the exam, I would say you need to include the separation step, since there will probably be a mark for noting that the equation is separable. (In fact, the "M1" in the margin indicates that there is a method mark for separating the equation)
     
    Last edited: Mar 2, 2007
  14. Mar 2, 2007 #13

    HallsofIvy

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    you need to integrate kdt to have an equation! If you only integrate the left side, you would have (1/10)ln(a/(10-a)) NOT equal to anything! Where would you go from there? As it is, you have (1/10)ln(a/(10-a))= kt+ C or ln(a/(10-a))= 10kt+ C which can be rewritten a/(10-a)= Ce^(10kt).
     
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