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Parametric Equations describing curves

  1. Dec 4, 2014 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    ScreenShot2014-12-04at62850AM_zpscc55710d.png

    2. Relevant equations


    3. The attempt at a solution

    For part A) my answer was:

    [itex]\int_a^b \sqrt{(dx/dt)^2+(dy/dt)^2}dt[/itex]

    The work I used for part A was based off this sites explanation: http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

    For part B)
    I simple took the derivatives with respect to t of the giving x and y functions then squared them and solved the integral from 0 to 1.
    The answer I recieved was 3/2.
    I have strong verification that this is correct.

    Part C)
    Part C is where I am stuck. When they say y=F(x) I usually connect a capital F with the antiderivative. Does this hold true?
    How do we describe the curve using y=F(x)?

    Part D seems rather straightforward once part C is understood.

    Thank you for any help.
     
  2. jcsd
  3. Dec 4, 2014 #2

    HallsofIvy

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    You are given [tex]x= (4/3)t^{3/2}[/tex] and [tex]y= t- (1/2)t^2[/tex] and you want to write y as a function of x.

    Solve the first equation for t as a function of x and replace the t in the second equation with that.
     
  4. Dec 4, 2014 #3

    RJLiberator

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    [(3/4)*x]^(2/3)=t

    y = [(3/4)*x]^(2/3)-(1/2)([(3/4)*x]^(2/3))^2

    This is writing y as a function of x, correct?

    So now with this information I can plug in x = 0 and x = 1 to get the range of x?

    And then I can use this equation to solve for part D? I assume.

    Kind regards.
     
  5. Dec 4, 2014 #4

    LCKurtz

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    Correct? Hard to say. If you would write it in Latex it would be readable.



    No. Those are the end point ##t## values. What are the corresponding ##x## values?
     
  6. Dec 4, 2014 #5

    RJLiberator

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    Perhaps I need to use

    (3x/4)^(2/3) = t and let 1 be 1 and 0 to find the range.

    Thus the range would be from 0 to 4/3
     
  7. Dec 5, 2014 #6

    HallsofIvy

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    Yes, the range is from x= 0 to x= 4/3. But I see no reason to solve for t to find that. You were given that [tex]x= (4/3)t^{3/2}[/tex] just set t equal to 0 and 1 in that.
     
  8. Dec 5, 2014 #7

    RJLiberator

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    Excellent. Yes, thank you for confirming the result and showing the way.

    :)
     
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