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Seperate variable and integration

  1. May 1, 2014 #1
    1. The problem statement, all variables and given/known data

    for this Q and t are variables, 10 and surd k are constant, is my working correct?
    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. May 1, 2014 #2
    No.

    ##\frac{\sqrt{K}}{10+(4-\sqrt{K})t}## is NOT the same as ##\frac{\sqrt{K}}{10+(4-\sqrt{K})}\cdot \frac{1}{t}##

    You should think about substitution.
     
  4. May 1, 2014 #3
    i dont understand the circled part, can you expalin?
     

    Attached Files:

  5. May 1, 2014 #4
    Well somebody integrated LHS and RHS of the equation, nothing more.

    LHS of the equation is of course ##\int \frac{dQ}{Q}=ln(Q) + C##, which you already found out. Now you have to integrate the RHS of the equation and you should get exactly the same as it is circled in your last post #3.

    In your original post you already tried to compute the integral on the RHS

    ##-\int_{0}^{t}\frac{\sqrt{k}}{10+(4-\sqrt{k})t}dt##

    but your approach is wrong because you wrote that ##\frac{\sqrt{k}}{10+(4-\sqrt{k})t}=\frac{\sqrt{k}}{10+(4-\sqrt{k})}\cdot \frac{1}{t}## which is NOT true. If anything than ##\frac{\sqrt{k}}{10+(4-\sqrt{k})t}=\frac{\sqrt{k}}{\frac{10}{t}+(4-\sqrt{k})}\cdot \frac{1}{t}## but this does not simplify the integral at all.

    Again: you should consider substitution. Note that ##\int \frac{dx}{x}## is an elementary integral. Try to find such substitution that will bring you to this elementary integral.
     
  6. May 2, 2014 #5
    well, why the following steps contain (10+(4-surd k) t) /10 ... why we should divide 10?
     
  7. May 2, 2014 #6
    Firstly, I never wrote that you should divide (10+(4-surd k) t) by 10.

    Remember that ##\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}##.

    Also take a look at ##\frac{1}{a}\cdot \frac{1}{b}=\frac{1}{a\cdot b}## and ##\frac{1}{a+b}\cdot \frac{1}{c}=\frac{1}{(a+b)\cdot c}=\frac{1}{a\cdot c+b\cdot c}##.

    Now I hope you understand a bit more.
     
  8. May 2, 2014 #7
    please refer fifth post to the post 3 . thanks.
     
  9. May 2, 2014 #8
    Ok, I apologize, I didn't know that we are no longer talking about the circled part. Anyhow, I hope you understand how ton integrate ##-\int_{0}^{t}\frac{\sqrt{k}}{10+(4-\sqrt{k})t}dt##

    Now to answer your last question:

    ##aln(x)=ln(x^a)## also

    ##ln(x)-ln(y)=ln(\frac{x}{y})##

    Note that you RHS integral goes from ##0## to ##t##. Understood?
     
  10. May 2, 2014 #9
    ok! clear now!
     
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