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

    0
    ReputationRep:
    Hello there,
    Have they got an error on here? Question No. 7 part a of the question . Was that meant to be -4? thanks!

    http://www.london-oratory.org/maths/...s/E4Qint_P.pdf
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    (Original post by laurawoods)
    Hello there,
    Have they got an error on here? Question No. 7 part a of the question . Was that meant to be -4? thanks!

    http://www.london-oratory.org/maths/...s/E4Qint_P.pdf
    No, 4 is correct, You're trying to find an area which is positive.

    If you look at the integral they've given, everything in it is positive, so the answer will be positive. If you had a -4 you would be getting a negative answer.

    check the limits on your integral and make sure you have them the right way round
    Offline

    14
    ReputationRep:
    (Original post by laurawoods)
    Hello there,
    Have they got an error on here? Question No. 7 part a of the question . Was that meant to be -4? thanks!

    http://www.london-oratory.org/maths/...s/E4Qint_P.pdf
    hahah you stress me out with your posts about Alevels all the time
    Offline

    18
    ReputationRep:
    (Original post by laurawoods)
    Hello there,
    Have they got an error on here? Question No. 7 part a of the question . Was that meant to be -4? thanks!

    http://www.london-oratory.org/maths/...s/E4Qint_P.pdf
    No, 4 is correct.

    When you have a negative integral you can switch the limits and make it positive.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by davros)
    No, 4 is correct, You're trying to find an area which is positive.

    If you look at the integral they've given, everything in it is positive, so the answer will be positive. If you had a -4 you would be getting a negative answer.

    check the limits on your integral and make sure you have them the right way round
    sorry i don't think I am understanding this! When we initially write down we get (y) * (dx/dt) = -4sin^2t

    Isn't it so?

    And also where did they get pi/4 and 0 from ? are we meant to work that out ourselves or simply take that from the answer they have given? thanks for your help!
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    (Original post by laurawoods)
    sorry i don't think I am understanding this! When we initially write down we get (y) * (dx/dt) = -4sin^2t

    Isn't it so?

    And also where did they get pi/4 and 0 from ? are we meant to work that out ourselves or simply take that from the answer they have given? thanks for your help!
    Your derivative is correct

    Remember when you integrate to work out an area you're going from the leftmost x-value to the rightmost x-value.

    In your graph, the leftmost x-value is 0 at the point O, so your bottom limit of integration is the value of t that makes x=0.

    At the rightmost point (call it P) you know that y=0. So what value of t makes y=0? This gives you the top limit of integration.

    Then note that when you have a minus sign in the function you're integrating you can convert this to a positive sign by swapping the limits of the integral round. This should get you to something that looks like the answer!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by davros)
    Your derivative is correct

    Remember when you integrate to work out an area you're going from the leftmost x-value to the rightmost x-value.

    In your graph, the leftmost x-value is 0 at the point O, so your bottom limit of integration is the value of t that makes x=0.

    At the rightmost point (call it P) you know that y=0. So what value of t makes y=0? This gives you the top limit of integration.

    Then note that when you have a minus sign in the function you're integrating you can convert this to a positive sign by swapping the limits of the integral round. This should get you to something that looks like the answer!
    Hello, I think I get it now! thanks for the answer! these limits are driving me mad :>
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    (Original post by laurawoods)
    Hello, I think I get it now! thanks for the answer! these limits are driving me mad :>
    no problem!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by davros)
    no problem!
    Hello , pls can I ask a quick question :
    When we have an exponential equation and they ask us for the long term speed etc is it the value that comes like "+C" for example :
    V = -5e^-1/5t + 60


    so would the long term value be the 60 ? is this always the case?
    Offline

    4
    ReputationRep:
    Yay Maths
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    )
    (Original post by laurawoods)
    Hello , pls can I ask a quick question :
    When we have an exponential equation and they ask us for the long term speed etc is it the value that comes like "+C" for example :
    V = -5e^-1/5t + 60


    so would the long term value be the 60 ? is this always the case?
    Yes if you've got a negative exponential like that - as t gets bigger and bigger, e^{-kt} gets smaller and smaller if k>0, so the total value gets closer and closer to the constant (in this case 60).
 
 
 
  • 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
    Would you like to hibernate through the winter months?
    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.