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Using parametric differentiation to evaluate the slope of a curve - attempted

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data

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

    y(t) = (2t) / (t^2 +1)

    at the point t=1

    2. Relevant equations

    Line equation = y-y1 = m(x-x1)

    chan rule = (dy/dt) / (dx/dt) = dy/dx

    3. The attempt at a solution

    I find the y1 and x1 values by subing in t=1 to the x(t) and y(t) equations
    I get the point (0,1) when t=1

    I have used the Quotient rule to find dx/dt & dy/dx (Is this right?)
    Doing the above I get:

    dy/dt = (-2(t^2 -1)) / (t^2+1)^2

    dx/dt = (4t) / (t^2 +1)^2

    So dy/dx = (dy/dt) / (dx/dt) however when I sub in t=1 to (dy/dt) I get 0 as the numerator

    I know that the m of the line equation is equal to dy/dx

    does anyone know what I'm doing wrong? I'll try uploading a pic of my work
    cheers
     
  2. jcsd
  3. Feb 7, 2013 #2
    pic of my working
     

    Attached Files:

  4. Feb 7, 2013 #3
    What makes you think something is wrong? have you plotted the x v y graph for t having a range of values? say t=-1, 0, 1, 2, 3
     
  5. Feb 7, 2013 #4
    if t=1 does this mean that m= 0?
     
  6. Feb 7, 2013 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Using parametric differentiation to evaluate the slope of a curve

    The graph of this function is an ellipse. t= 1 is the point (0, 1) where the tangent line is, in fact, horizontal so the dy/dx= 0 there.
     

    Attached Files:

  7. Feb 7, 2013 #6
    Re: Using parametric differentiation to evaluate the slope of a curve

    cheers
     
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