Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    Name:  Photo on 16-03-2016 at 10.27.jpg
Views: 67
Size:  130.2 KB

    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...
    Offline

    22
    ReputationRep:
    (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.
    Offline

    22
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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?
    Offline

    22
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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!
    Offline

    22
    ReputationRep:
    (Original post by KateAteKarrots)
    Yes, that makes complete sense now! Thank you very much for the guidance!
    Glad I helped.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Have you ever participated in a Secret Santa?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.