Turn on thread page Beta

p3 parametrics watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    part 1:

    a curve is given by:

    x=1+cos t + sin t
    y= 2+ cos t-sin t
    t is between 0 and 2pi.

    find the gradient of the curve with parameter t (done) and show that the equation of the normal is:

    (sin t + cos t)y + (sin t-cos t)x= 3 sin t +cos t

    part 2:

    by 1st writing in
    cos t + sin t=x-1;
    cos t-sin t=y-2;

    solve the eq. for cos t and sint in terms of x and y.
    Offline

    0
    ReputationRep:
    (Original post by innitman_uk)
    part 1:

    a curve is given by:

    x=1+cos t + sin t
    y= 2+ cos t-sin t
    t is between 0 and 2pi.

    find the gradient of the curve with parameter t (done) and show that the equation of the normal is:

    (sin t + cos t)y + (sin t-cos t)x= 3 sin t +cos t

    part 2:

    by 1st writing in
    cos t + sin t=x-1;
    cos t-sin t=y-2;

    solve the eq. for cos t and sint in terms of x and y.
    Well for the normal, divide -1 by your gradient to get its gradient. Put that into y = mx + c, use the first 2 equations to find c and it will probably rearrange to give what they want.
    Offline

    2
    ReputationRep:
    (Original post by innitman_uk)
    part 1:

    a curve is given by:

    x=1+cos t + sin t
    y= 2+ cos t-sin t
    t is between 0 and 2pi.

    find the gradient of the curve with parameter t (done) and show that the equation of the normal is:

    (sin t + cos t)y + (sin t-cos t)x= 3 sin t +cos t
    dx/dt = cos t - sin t
    dy/dt = -(sin t + cos t)

    So dy/dx = dy/dt / dx/dt = (sin t + cos t) / (sin t - cos t).

    Gradient of normal
    = -1/Gradient of tangent
    = -1/(dy/dx)
    = (cos t - sin t) / (sin t + cos t).

    Equation of normal:

    y
    = [Gradient of normal]*x + Constant
    = [(cos t - sin t) / (sin t + cos t)]*x + Constant
    = [(cos t - sin t) / (sin t + cos t)]*(x - (1 + cos t + sin t)) + 2 + cos t - sin t
    = [(cos t - sin t) / (sin t + cos t)]*x + 2 + cos t - sin t - [(cos t - sin t) / (sin t + cos t)] - (cos t - sin t)
    = [(cos t - sin t) / (sin t + cos t)]*x + 2 - [(cos t - sin t) / (sin t + cos t)]

    Multiplying through by (sin t + cos t),

    (sin t + cos t)*y
    = (cos t - sin t)*x + 2sin t + 2cos t - cos t + sin t
    = (cos t - sin t)*x + 3sin t + cos t.

    Rearranging,

    (sin t - cos t)*x + (sin t + cos t)*y = 3sin t + cos t.

    (Original post by innitman_uk)
    part 2:

    by 1st writing in
    cos t + sin t=x-1;
    cos t-sin t=y-2;

    solve the eq. for cos t and sint in terms of x and y.
    x - 1 = cos t + sin t
    y - 2 = cos t - sin t

    Adding,

    x + y - 3 = 2cos t
    => cos t = x/2 + y/2 - 3/2.

    Subtracting,

    x - y + 1 = 2sin t
    => sin t = x/2 - y/2 + 1/2.
    • Thread Starter
    Offline

    0
    ReputationRep:
    cheers
 
 
 

University open days

  1. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  2. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
  3. Birmingham City University
    General Open Day Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?

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.