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Parametric curves

  1. May 24, 2009 #1
    1. The problem statement, all variables and given/known data
    a)Consider the parametric curve x = t^2 + t, y = e^t. Find all t such that the tangent line of the curve at (x(t), y(t)) intersects the x-axis at (4,0)

    2. Relevant equations



    3. The attempt at a solution
    I draw out the graph and came out with the points, I was wondering do i eliminate the parameter t to find the equation first or what approach can i take to start this question. Thank you
     
  2. jcsd
  3. May 24, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi mjstyle ! Welcome to PF! :smile:
    The parameter t is the answer, so eliminate it last. :wink:

    Start by writing down the slope of the line from (4,0) to (x(t),y(t)), then get it to equal the slope of the tangent at (x(t),y(t)) …

    what do you get? :smile:
     
  4. May 24, 2009 #3
    Thank you so much for the quick response,

    I was wondering, writing down the slope of the line from (4,0) to (x(t),y(t)),

    m = y(t) - 0 / x(t) - 4

    tangent at (x(t),y(t)):

    dy/dx = dy/dt / dx/dt = e^t / 2t + 1

    then i'm stuck hehe
     
  5. May 24, 2009 #4

    tiny-tim

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    hmm :rolleyes: … that's because (unlike your dy/dx) you haven't yet converted m into a function of just t :wink:
     
  6. May 24, 2009 #5
    omg what am i thinking, that's right

    m = y(t) - 0 / x(t) - 4

    m = e^t / t^2 + t - 4

    tangent at (x(t),y(t)):

    dy/dx = dy/dt / dx/dt = e^t / 2t + 1

    x = t^2 + t, y = e^t

    so right right i fond the slop of the line according to t, how do i find the slop of the tangent line: x = t^2 + t, y = e^t?

    thanks
     
  7. May 24, 2009 #6

    tiny-tim

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    the slope of the tangent line is dy/dx :smile:
     
  8. May 24, 2009 #7
    oh... i always thought that's the tangent line dy/dx then what i do now is

    e^t / 2t + 1 = e^t / t^2 + t - 4 and solve for t?

    which is

    ln t / 2t + 1 = ln t / t^2 + t - 4

    t^2 + t - 4 / 2t + 1 = 1

    t^2 - t - 5 = 0

    using quadratic formula: comes out to be (1 + square root(21)) / 2 and (1 - square root(21)) / 2

    is that correct? thank you so much
     
  9. May 24, 2009 #8

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    Yes, but it would have been a lot quicker just to divide your original equation by et

    (and your "ln t / 2t + 1 = ln t / t^2 + t - 4" is rubbish :rolleyes:)
     
  10. May 24, 2009 #9
    thank you so much for the help!. I actually have one more question,

    Determine the values of t for which the curve x = 2squareroot(1+t), y = intergral from x to t^2 (squareroot(u) - 1)squareroot(1 + squareroot(u)) du, t greater and equal to 0 is concave upward and those for thich is it concave downward.

    Do i find the second derivative of this and then.... hehe
     
  11. May 24, 2009 #10

    tiny-tim

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    (have a square-root: √ and an integral: ∫ :wink:)

    Yes, you need the sign (only) of d2y/dx2

    but since x is monotone increasing (wrt t), that'll be the same as the sign of d/dt (dy/dx) :smile:
     
  12. May 25, 2009 #11
    hey tiny-tim, for the preivous question, i don't think it's right cause i plugged in 1+squareroot(21) / 2 into t in y = e^t and i did not get 0....
     
  13. May 25, 2009 #12

    tiny-tim

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    (what happened to that √ i gave you?)
    sorry … not following you :redface:
     
  14. May 25, 2009 #13
    oh cause, the question asks

    Find all t such that the tangent line of the curve at (x(t), y(t)) intersects the x-axis at (4,0)

    so i'm assuming the t's I fins intersects the x-axis, so let say i plug in the t i find into the y(t) equation, shouldn't i get 0, but plugging in 1+squareroot(21) / 2 into t in y = e^t doesn't give me 0.
     
  15. May 25, 2009 #14

    tiny-tim

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    ?? :confused:

    if you plug in a t you found into the y(t) equation, you get the y coordinate of a point on the curve from which the tangent goes through (4,0)
     
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