Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

C4 parametric equations help

Announcements Posted on
    • Thread Starter
    Offline

    ReputationRep:
    A curve has parametric equations: x = 2cos2t y = 6sint

    a) find the gradient of the curve at the point where t = pi/3 (done this)

    b) find a cartesian equation of the curve in the form:
    y = f(x), -k < x < k


    No idea what to do for b =S
    Offline

    ReputationRep:
    Is there any way you can eliminate t from the equations, maybe using a trigonometric identity?
    • Thread Starter
    Offline

    ReputationRep:
    Tried using cos2t as 1 - 2sin^2t, but it didn't seem to work out.
    Offline

    ReputationRep:
    As far as I'm aware that should work out...

    What did you get for 2sin^2(t) then?
    • Thread Starter
    Offline

    ReputationRep:
    Erm... y = 6 - root(18x)
    Can't find a mark scheme to check, but no idea where the K comes into it x_X
    Offline

    ReputationRep:
    it should be y=3root(2-x)
    x=2(1-2sin^2t) y=6sint
    y^2=36sin^2t
    (y^2)/36=sin^2t

    x=2-4sin^2t
    sub y
    x=2-((Y^2)/9)

    multiply through by 9

    9x=18-(Y^2)
    Y=root(18-9x)

    thats what i get atleast

    so -2<x<2
    Offline

    ReputationRep:
    (Original post by nitinkf)
    it should be y=3root(2-x)
    x=2(1-2sin^2t) y=6sint
    y^2=36sin^2t
    (y^2)/36=sin^2t

    x=2-4sin^2t
    sub y
    x=2-((Y^2)/9)

    multiply through by 9

    9x=18-(Y^2)
    Y=root(18-9x)

    thats what i get atleast

    so -2<x<2
    I concur.
    Offline

    ReputationRep:
    Put x=2cos2t into the form x=2(2(cost)^2-1) or x=4(cost)^2-2

    then y=6sint into Y^2=36(sint)^2

    Then into identity (Sint)^2 + (cost)^2 = 1 in order to eliminate paramiter

    The constant bit (K) is to show that you undersatnd the maximum and minimum x values of the cartesian equation.
    • Thread Starter
    Offline

    ReputationRep:
    Got it now, thanks all
    Offline

    ReputationRep:
    For this one I got


    EDIT: Just realised I forgot the 6! Dammit...
    Offline

    ReputationRep:
    I've got the same question but can anyone help me with part a? I used the chain rule on it and and the whole dx/dt=1/(dt/dx) but I want to check my answer? :3
    Offline

    (Original post by revilowaldow)
    I've got the same question but can anyone help me with part a? I used the chain rule on it and and the whole dx/dt=1/(dt/dx) but I want to check my answer? :3

    I make it -\dfrac{\sqrt{3}}{2}

    If you get anything else and want someone to check it, then post your working.

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: July 19, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
How do you sleep?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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.