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# C4 calculus help? watch

1. 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?

2. (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?

Yes, you should include C - keep working and simplifying and eventually you can just replace whatever function of the arbitrary constant you have with , another arbitrary constant.
3. (Original post by KateAteKarrots)
lnM = lnt- 1/2ln(1+t^2) + c
From here you can already see

Now if you take the exponential of both sides: , now just let .
4. (Original post by Zacken)
From here you can already see

Now if you take the exponential of both sides: , now just let .

Oh thank you for the help! I didn't realise you could take the exponential of both sides?
5. (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 , when you're solving this, you're implicitly taking the cube root of both sides to get .

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

Now the logs 'cancel' to get you , if you're more comfortable working this way.
6. (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 , when you're solving this, you're implicitly taking the cube root of both sides to get .

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

Now the logs 'cancel' to get you , if you're more comfortable working this way.
Yes, that makes complete sense now! Thank you very much for the guidance!
7. (Original post by KateAteKarrots)
Yes, that makes complete sense now! Thank you very much for the guidance!

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