Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    Can you guys work these questions out? Please provide workings for both questions. I cant seem to get it right. Thank you.

    Name:  Screenshot_2016-05-23-11-35-12.jpg
Views: 173
Size:  160.2 KB
    Offline

    17
    ReputationRep:
    (Original post by Jyashi)
    Can you guys work these questions out? Please provide workings for both questions. I cant seem to get it right. Thank you.

    Name:  Screenshot_2016-05-23-11-35-12.jpg
Views: 173
Size:  160.2 KB
    Please show us what work you have done so far on both questions.
    Offline

    3
    ReputationRep:
    (Original post by rayquaza17)
    Please show us what work you have done so far on both questions.
    Mind letting me know whether my answer for the 2nd question is right?
    Spoiler:
    Show
    X = 2M/3
    Offline

    9
    ReputationRep:
    (Original post by Dohaeris)
    Mind letting me know whether my answer for the 2nd question is right?
    Spoiler:
    Show
    X = 2M/3
    Post working? I'm not sure if that looks right looking at the question.
    Never mind.
    Offline

    17
    ReputationRep:
    (Original post by Dohaeris)
    Mind letting me know whether my answer for the 2nd question is right?
    Spoiler:
    Show
    X = 2M/3
    That's right.
    Offline

    3
    ReputationRep:
    (Original post by rayquaza17)
    That's right.
    Thanks
    • Thread Starter
    Offline

    2
    ReputationRep:
    Just came back from work so couldnt update post. Here is my working for Q6.

    For Q7 i am getting to the equation of dX÷dt = kM

    Is that a correct expression?


    (Original post by rayquaza17)
    Please show us what work you have done so far on both questions.
    Attached Images
     
    • Thread Starter
    Offline

    2
    ReputationRep:
    Can you tell me how you got to that answer please?

    (Original post by Dohaeris)
    Mind letting me know whether my answer for the 2nd question is right?
    Spoiler:
    Show
    X = 2M/3
    Offline

    3
    ReputationRep:
    (Original post by Jyashi)
    Can you tell me how you got to that answer please?
    I don't mind helping you out. From what I can see, you used partial fractions, but there's no need for that (at least in the way I solved it.) I'll try and take it step-by-step.

    First, they said M is the mass of the fuel that was burned, from this we know that M is a constant.

    Next, we're told that the sum of the mass of waste produced and the mass of unburned fuel remaining is always constant, this'll come into play later on.

    We're given that X is the mass of waste produced.

    Then we're told that the rate of increase of X, dX/dt, is k times that the mass of fuel remaining.

    The mass of fuel remaining will have to be M-X, where M is the original mass of unburned fuel, because we know that the sum of waste produced and unburned fuel remaining is constant. So if you take 'Y' as the unburned fuel remaining, then Y+X=M, so Y=M-X

    From this, we can deduce the equation to be dX/dt = k*(M-X), and you can carry on the to solve the differential equation as you would any other, though you have to remember that M and k are constants.

    Do you get it?
    • Thread Starter
    Offline

    2
    ReputationRep:
    I kind of get it but still dont inderstand it deeply. Can you provide me a working of it in mathmatical form. I find that easier to understand then words.

    (Original post by Dohaeris)
    I don't mind helping you out. From what I can see, you used partial fractions, but there's no need for that (at least in the way I solved it.) I'll try and take it step-by-step.

    First, they said M is the mass of the fuel that was burned, from this we know that M is a constant.

    Next, we're told that the sum of the mass of waste produced and the mass of unburned fuel remaining is always constant, this'll come into play later on.

    We're given that X is the mass of waste produced.

    Then we're told that the rate of increase of X, dX/dt, is k times that the mass of fuel remaining.

    The mass of fuel remaining will have to be M-X, where M is the original mass of unburned fuel, because we know that the sum of waste produced and unburned fuel remaining is constant. So if you take 'Y' as the unburned fuel remaining, then Y+X=M, so Y=M-X

    From this, we can deduce the equation to be dX/dt = k*(M-X), and you can carry on the to solve the differential equation as you would any other, though you have to remember that M and k are constants.

    Do you get it?
    Offline

    1
    ReputationRep:
    Hi,
    Here's my worked solutions to question 7.
    Hope this helps
    Name:  image.jpg
Views: 71
Size:  496.2 KB

    Attachment 537719537721
    Attached Images
     
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 24, 2016

University open days

  • Southampton Solent University
    All faculties Undergraduate
    Sun, 18 Nov '18
  • University of Bradford
    All faculties Undergraduate
    Wed, 21 Nov '18
  • Buckinghamshire New University
    All Faculties Postgraduate
    Wed, 21 Nov '18
Poll
Black Friday: Yay or Nay?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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

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

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