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Parametric 2'nt deratives

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data
    find the slope and concavity of the funtion at the given point.
    x=t^2
    Y=t^2+t+1
    (0,0)

    2. Relevant equations

    t=x^(1/2)

    3. The attempt at a solution
    t=0 when x=0
    x'=2t
    y'=2t+1
    M=2t/(2t+1)=0
    for the second deritive would you take the deritive of 2t/(2t+1) devided by the deritive of 2t+1
    once i find the second deritive i would plug in o for t and if it was + than it would be up -would be down. right?
     
  2. jcsd
  3. Jan 6, 2009 #2

    Defennder

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    Homework Helper

    I assume that by M you mean [tex]\frac{d^2y}{dx^2}[/tex]. It should be y'/x' instead here.
    Why should it be the derivative of y'=2t+1? How would you use the chain rule to determine a correct expression for d^2y/dx^2 ?
    Not quite. You'll want to find the value of t for which x=y=0. For x=0,t=0 so that's correct. For y, setting t=0 gives y=1. Setting y=0 and solving for t gives you the correct t-values.
     
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