Results 1 to 4 of 4

Math Help - Parametric Curves

  1. #1
    Newbie
    Joined
    Jan 2007
    Posts
    16

    Parametric Curves

    "Given the Parametric equations x=tsint and y=tcost find dy/dx at the point ((pi/2),0)

    I remember that one must differentiate y'/x', but I don't remember what to do with the given point.

    Y' = cost + -tsint
    X' = sint + tcost

    Do you plug in pi/2 into X', and 0 into Y', or?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Xavier20 View Post
    "Given the Parametric equations x=tsint and y=tcost find dy/dx at the point ((pi/2),0)

    I remember that one must differentiate y'/x', but I don't remember what to do with the given point.

    Y' = cost + -tsint
    X' = sint + tcost

    Do you plug in pi/2 into X', and 0 into Y', or?
    dy/dx = dy/dt dt/dx

    Now differentiate x=t sin(t) wrt x to get:

    1 = dt/dx sin(t) + t cos(t) dt/dx

    so:

    dt/dx = 1/[sin(t) +y]

    Now differentiate y= t cos(t) wrt t to get:

    dy/dt = cos(t) - t sin(t) = cos(t) - x

    so dy/dx = [cos(t) - x]/[sin(t) +y]

    Now the point ((pi/2),0) corresponds to t=p/2 and plugging this into the
    above gives us dy/dx at this point is:

    [-pi/2]/[1] =-pi/2

    RonL
    Last edited by CaptainBlack; April 25th 2007 at 01:02 PM. Reason: correct typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, Xavier!

    You have the right idea . . . but be careful.


    Given the parametric equations: .x = t新in(t) and y = t搾os(t)
    find dy/dx at the point (π/2 ,0)
    . . . . . . . . . . . . . .dy . . . dy/dt
    The derivative is: . --- .= .-------
    . . . . . . . . . . . . . .dx . . . dx/dt

    Your derivatives were correct: .dy/dt .= .cos(t) - t新in(t)
    . . . . . . . . . . . . . . . . . . . . . .dx/dt .= .sin(t) + t搾os(t)

    . . . .dy . . . cos(t) - t新in(t)
    So: .--- .= .-------------------
    . . . .dx . . . sin(t) + t搾os(t)


    The point (π/2, 0) means: .x = π/2, y = 0

    . . Hence: .t新in(t) = π/2 . . t = π/2


    . . . . . . . . .dy . - . cos(π/2) - (π/2)新in(π/2) . - . .0 - (π/2)(1) . . . . .π
    Therefore: . --- .= .------------------------------ .= .--------------- .= .- --
    . . . . . . . . .dx . . . sin(π/2) + (π/2)搾os(π/2) . . . 1 + (π/2)(0) . - . . 2

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2007
    Posts
    16
    Thank you! =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 14th 2010, 01:56 PM
  2. Parametric Curves
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 20th 2009, 06:14 PM
  3. Parametric Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 6th 2009, 10:34 PM
  4. parametric curves
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 6th 2008, 10:35 PM
  5. Parametric curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 27th 2008, 10:11 PM

Search Tags


/mathhelpforum @mathhelpforum