Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    Please I need help completing these questions I know my part c isnwrong which will affect my part d hence why I haven't attempted all of part d. I seriously don't know what I've done wrong with part b that has lead to the mistake in c) any help please

    Name:  ImageUploadedByStudent Room1420750783.567092.jpg
Views: 107
Size:  170.0 KB

    Above attachment are the questions and part c)

    Name:  ImageUploadedByStudent Room1420750829.834378.jpg
Views: 77
Size:  173.8 KB

    These are the answers, all help is appreciated thanks very much 😬


    Posted from TSR Mobile
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    Please I need help completing these questions I know my part c isnwrong which will affect my part d hence why I haven't attempted all of part d. I seriously don't know what I've done wrong with part b that has lead to the mistake in c) any help please

    Name:  ImageUploadedByStudent Room1420750783.567092.jpg
Views: 107
Size:  170.0 KB

    Above attachment are the questions and part c)

    Name:  ImageUploadedByStudent Room1420750829.834378.jpg
Views: 77
Size:  173.8 KB

    These are the answers, all help is appreciated thanks very much 😬


    Posted from TSR Mobile
    wow
    I did this question with one of my students this morning (June 2001 paper)
    expand sin2t

    then either recognition or substitution u = sint
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TeeEm)
    wow
    I did this question with one of my students this morning (June 2001 paper)
    expand sin2t

    then either recognition or substitution u = sint
    Is this for part b) or c)

    For part b) I clearly have it in the form they mean, meaning its correct but then that isn't the answer apparently.
    If it is the answer, I used the formula in the c3 booklet where -2Sin(A+B)/2 x Sin (A-B)/2

    but just realised I did it wrong still whilst typing this as. its -27 as the coefficient so if I want it to work I would have to divide my integral by 27/2 meaning my cosA-cosb will have a coefficient of 27/2.


    But still I'm stuck, can you explain part b) possibly and we'll take it from there? thanks a lot
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    Is this for part b) or c)

    For part b) I clearly have it in the form they mean, meaning its correct but then that isn't the answer apparently.
    If it is the answer, I used the formula in the c3 booklet where -2Sin(A+B)/2 x Sin (A-B)/2

    but just realised I did it wrong still whilst typing this as. its -27 as the coefficient so if I want it to work I would have to divide my integral by 27/2 meaning my cosA-cosb will have a coefficient of 27/2.


    But still I'm stuck, can you explain part b) possibly and we'll take it from there? thanks a lot
    what is the problem?
    I understood that it was part (c)
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TeeEm)
    what is the problem?
    I understood that it was part (c)
    Sorry if I'm not being clear. To begin with part b) is the problem as that seems to be apparently wrong so what do I do? or is it right, because my friend told me its wrong and he may be wrong
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    Sorry if I'm not being clear. To begin with part b) is the problem as that seems to be apparently wrong so what do I do? or is it right, because my friend told me its wrong and he may be wrong
    I will check (b)
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    Sorry if I'm not being clear. To begin with part b) is the problem as that seems to be apparently wrong so what do I do? or is it right, because my friend told me its wrong and he may be wrong
    limits are the wrong way round in part b

    that the only mistake
    A = +27
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TeeEm)
    limits are the wrong way round in part b

    that the only mistake
    A = +27
    +27? oh I got -27 from doing integral of y dx/dt

    y = 9sin 2t x = 3cos t

    dx/dt = -3sin t as cos differentiates to -sin


    therefore y dx/dt = 9sin 2t * -3sin t = -27sin2tsint? dt
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    +27? oh I got -27 from doing integral of y dx/dt

    y = 9sin 2t x = 3cos t

    dx/dt = -3sin t as cos differentiates to -sin


    therefore y dx/dt = 9sin 2t * -3sin t = -27sin2tsint? dt
    READ THE LAST POST

    the limits are the wrong way round
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TeeEm)
    READ THE LAST POST

    the limits are the wrong way round
    Ohhhhh, I see the limits for A=-27 should be the other way round.


    So is that a rule because I haven't learnt that if you change the limites so it was meant to be 0 and pi/2 for my A=-27 with 0 being at the top
    but instead its pi/2 0 with pi/2 being at the top. When you change the limits like this is it a rule that you change sign?


    I dont get why a change in sign will be needed please, thanks
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    Ohhhhh, I see the limits for A=-27 should be the other way round.


    So is that a rule because I haven't learnt that if you change the limites so it was meant to be 0 and pi/2 for my A=-27 with 0 being at the top
    but instead its pi/2 0 with pi/2 being at the top. When you change the limits like this is it a rule that you change sign?


    I dont get why a change in sign will be needed please, thanks
    there is no magic

    you integrate in cartesian left to right
    at x=0 use the x=... equation and you get pi/2
    when y=0 or (x=3, can you see why?) you get t=0
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TeeEm)
    there is no magic

    you integrate in cartesian left to right
    at x=0 use the x=... equation and you get pi/2
    when y=0 or (x=3, can you see why?) you get t=0

    Hahahah there is no magic, sorry I get that but then my integral at the top in the pic below can be written in the form of the integral at the bottom in the pic below.
    I'm asking is tht a general rule, for you to change signs if you want to change limits?

    Name:  ImageUploadedByStudent Room1420755536.517732.jpg
Views: 48
Size:  152.3 KB


    Posted from TSR Mobile
    Offline

    19
    ReputationRep:
    (Original post by hudl)
    hahahah there is no magic, sorry i get that but then my integral at the top in the pic below can be written in the form of the integral at the bottom in the pic below.
    I'm asking is tht a general rule, for you to change signs if you want to change limits?

    Name:  ImageUploadedByStudent Room1420755536.517732.jpg
Views: 48
Size:  152.3 KB


    posted from tsr mobile
    absolutely
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TeeEm)
    absolutely
    Beautiful, thanks a lot, appreciate it wish I could rep you but seems you've helped me to many times before!
    Offline

    19
    ReputationRep:
    (Original post by Hudl)
    Beautiful, thanks a lot, appreciate it wish I could rep you but seems you've helped me to many times before!
    not too worry.

    it was my pleasure
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 8, 2015
Poll
Could you cope without Wifi?
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

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.