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    Hello! I've attached a picture of the question I'm currently stuck on.

    So I've answered part i and part ii and I've got the partial fraction to be 1/t - t/1+t^2

    then i said 1/m dM/dT = 1/t(1+t^2)

    Then: integral(1/M)dM = ingegral of 1/t(1+t^2) dt

    then i substitututed in my partial fraction so I wrote

    integral(1/M)dM = ingegral of (1/t - t/1+t^2) dt

    Then I integrated this and got

    lnM = lnt- 1/2ln(1+t^2) + c

    Now I'm not quite sure what to write...Do I find C? Did I even need to include C?

    I know I should probably bring up the -1/2 and make it into a power...

    How do I show that M = Kt/sqr(1+t^2) like I'm asked to?

    Where does the K come from? Is it the integration constant C?

    Please help...
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    (Original post by KateAteKarrots)
    Now I'm not quite sure what to write...Do I find C? Did I even need to include C?

    I know I should probably bring up the -1/2 and make it into a power...

    How do I show that M = Kt/sqr(1+t^2) like I'm asked to?

    Where does the K come from? Is it the integration constant C?

    Please help...
    Yes, you should include C - keep working and simplifying and eventually you can just replace whatever function of the arbitrary constant you have with K, another arbitrary constant.
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    (Original post by KateAteKarrots)
    lnM = lnt- 1/2ln(1+t^2) + c
    From here you can already see \displaystyle \ln M = \ln t - \ln \sqrt{1+t^2} + c \Rightarrow \ln M = \ln \frac{t}{\sqrt{1+t^2}} + c

    Now if you take the exponential of both sides: \displaystyle M = \frac{e^c t}{\sqrt{1+t^2}}, now just let K = e^c.
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    (Original post by Zacken)
    From here you can already see \displaystyle \ln M = \ln t - \ln \sqrt{1+t^2} + c \Rightarrow \ln M = \ln \frac{t}{\sqrt{1+t^2}} + c

    Now if you take the exponential of both sides: \displaystyle M = \frac{e^c t}{\sqrt{1+t^2}}, now just let K = e^c.

    Oh thank you for the help! I didn't realise you could take the exponential of both sides?
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    (Original post by KateAteKarrots)
    Oh thank you for the help! I didn't realise you could take the exponential of both sides?
    You can usually do whatever you'd like to both sides of an equation, for example, if you have the equation x^3 = 27, when you're solving this, you're implicitly taking the cube root of both sides to get x = 27^{1/3} = 3.

    If you want another way of looking at it, though, we have:

    \displaystyle \ln M = \ln \frac{t}{\sqrt{1+t^2}} + c  = \ln \frac{t}{\sqrt{1+t^2}} + \ln e^c \Rightarrow \ln M = \ln \frac{e^c t}{\sqrt{1+t^2}}

    Now the logs 'cancel' to get you \displaystyle M = \frac{e^c t}{\sqrt{1+t^2}}, if you're more comfortable working this way.
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    (Original post by Zacken)
    You can usually do whatever you'd like to both sides of an equation, for example, if you have the equation x^3 = 27, when you're solving this, you're implicitly taking the cube root of both sides to get x = 27^{1/3} = 3.

    If you want another way of looking at it, though, we have:

    \displaystyle \ln M = \ln \frac{t}{\sqrt{1+t^2}} + c  = \ln \frac{t}{\sqrt{1+t^2}} + \ln e^c \Rightarrow \ln M = \ln \frac{e^c t}{\sqrt{1+t^2}}

    Now the logs 'cancel' to get you \displaystyle M = \frac{e^c t}{\sqrt{1+t^2}}, if you're more comfortable working this way.
    Yes, that makes complete sense now! Thank you very much for the guidance!
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    (Original post by KateAteKarrots)
    Yes, that makes complete sense now! Thank you very much for the guidance!
    Glad I helped.
 
 
 
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