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Math Help - Differentiation of parametric equation

  1. #1
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    Exclamation Differentiation of parametric equation

    2day i doing this question...den i stuck.

    x=t^2 , y = 1/(1+t)
    1-2t

    My solution,
    dx/dt =2t(1-t)/(1-2t)^2 , dy/dt= -1/(1+t)^2


    Then,
    dy/dx = dy/dt * dt/dx

    =-1/(1+t)^2 * (1-2t)^2/2t(1-t)
    =?????
    at there i stuck.
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  2. #2
    Senior Member vincisonfire's Avatar
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    Reply

    You've got an answer why do you say you're stuck.
    Maybe you could simplify it a little but
    dx/dt = {2t*(1-2t)+2t*t^2}/(1-2t)^2
    dy/dt = -1/(1+t)^2 as you said
    You did the good calculation dy/dx = (dy/dt)/(dx/dt)
    Simplify as much as you can and that is your answer : (1/2)*{(2t-1)^2}/{t*(t^2-1)*(t+1)}
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  3. #3
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    Quote Originally Posted by vincisonfire View Post
    You've got an answer why do you say you're stuck.
    Maybe you could simplify it a little but
    dx/dt = {2t*(1-2t)+2t*t^2}/(1-2t)^2
    dy/dt = -1/(1+t)^2 as you said
    You did the good calculation dy/dx = (dy/dt)/(dx/dt)
    Simplify as much as you can and that is your answer : (1/2)*{(2t-1)^2}/{t*(t^2-1)*(t+1)}
    Ok, oic...like tat ar, becouse i expand all and wan to simplify but canot.
    Haha...didnt see tat. Thx a alot...my problem solved. I tot the answer should be very simple.
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