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Convert f(x) to parametric holding something constant

  1. Oct 17, 2006 #1
    Hi all, if I have a problem like:

    The path of a particle is described by [tex]y=4x^2[/tex], and it has a constant velocity of 5 m/s.

    How do I make a parametric equation out of this?

    I tried doing:
    [tex]r = xi + f(x)j[/tex]
    [tex]r = ti + 4t^2j[/tex],

    but then [tex]v = \frac{dr}{dt} = i + 8tj[/tex], so [tex]|v|=\sqrt{1+64t^2}[/tex], and at, for example, time [tex]t = 0[/tex], [tex]|v| = \sqrt{1+0} = 1[/tex], not the constant 5 m/s! Anyone know how I can make a parametric equation that will work?
     
    Last edited: Oct 17, 2006
  2. jcsd
  3. Oct 17, 2006 #2

    Zurtex

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    So you have y = 4x2 and |v(t)| = 5 (and v(t) > 0)

    Letting x = (x(t),y(t))

    You know that dv/dt = x

    So you know that:

    | integral(x,dt) | = 5

    Think about it for a bit, sure it will come to you :smile:
     
  4. Oct 17, 2006 #3
    I don't get it!

    I tried:

    [tex]|(\int x\,dt)| = 5[/tex]

    [tex]\sqrt{(xt)^2+C^2}= 5[/tex]

    [tex]x^2t^2+C^2=25[/tex]

    It doesn't seem to work! When x = 0, C=5, so
    [tex]|v| = x^2t^2 +5[/tex]?
     
    Last edited: Oct 17, 2006
  5. Oct 17, 2006 #4

    HallsofIvy

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    Then length is NOT [itex]| \int x dx|[/itex]! Also, x is itself a function of t so your integration is wrong.

    If x= f(t) and y= g(t) then since y= 4x2, g(t)= 4(f(t)2 so
    [tex]\frac{dg}{dt}= 8f(t)\frac{df}{dt}[/tex]
    The length f(t)i+ g(t)j is
    [tex]\sqrt{f^2(t)+ 64f^2(t)\left(\frac{df}{dt}\right)^2}= 5[/tex]
    Solve that differential equation for f(x) and then find g(x).
     
    Last edited: Oct 18, 2006
  6. Oct 17, 2006 #5
    Oh, thanks. Wait, how do I solve for f(x)? All I see is f(t)!
     
  7. Oct 18, 2006 #6
    Ah, nevermind. Thanks for the help!
     
  8. Oct 16, 2010 #7

    nvn

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    Re: Convert cartesian equation f(x) to parametric equations

    There appear to be typographic mistakes in post 4. For example,

    [tex]\sqrt{f^2(t)+64f^2(t)\left(\frac{df}{dt}\right)^2}\;=\:5\quad\ \ \mbox{should instead be}[/tex]

    [tex]\sqrt{\left(\frac{df}{dt}\right)^2+64f^2(t)\left(\frac{df}{dt}\right)^2}=\:5.[/tex]​

    I would instead solve the given problem in post 1 as follows.

    Given y = 4*x2, ds/dt = 5 m/s. At x = 0, t = 0.

    [tex]\mathbf{\vec s}=x\,\mathbf{i}+y\,\mathbf{j}[/tex]

    [tex]d\mathbf{\vec s}=dx\,\mathbf{i}+dy\,\mathbf{j}[/tex]

    [tex]ds=[(dx)^2+(dy)^2]^{0.5}[/tex]​

    [tex]\mbox{But since }y=4x^2,\ \ \therefore \ dy=8x\,dx.\ \mbox{Therefore,}[/tex]

    [tex]\begin{equation*}\begin{split}ds&=[(dx)^2+64x^2(dx)^2]^{0.5}\\
    &=(1+64x^2)^{0.5}\,dx\end{equation*}\end{split}[/tex]

    [tex]\frac{ds}{dt}=(1+64x^2)^{0.5}\,\frac{dx}{dt}[/tex]​

    [tex]\mbox{But }\frac{ds}{dt}=5\ \mbox{m/s}.[/tex]

    [tex]5=(1+64x^2)^{0.5}\,\frac{dx}{dt}[/tex]

    [tex]\begin{equation*}\begin{split}dt&=\frac{1}{5}(1+64x^2)^{0.5}\,dx\\
    &=\frac{8}{5}(\frac{1}{64}+x^2)^{0.5}\,dx\\
    &=\frac{8}{5}[\left(\frac{1}{8}\right)^2+x^2]^{0.5}\,dx\end{equation*}\end{split}[/tex]

    [tex]\int dt=\frac{8}{5}\int[\left(\frac{1}{8}\right)^2+x^2]^{0.5}\,dx[/tex]

    [tex]t=\frac{8}{5}\int[\left(\frac{1}{8}\right)^2+x^2]^{0.5}\,dx[/tex]​

    Therefore, from any integral table,

    [tex]t\:=\:\frac{4}{5}x\,(\frac{1}{64}+x^2)^{0.5}\ +\ \frac{1}{128}\ln|x+(\frac{1}{64}+x^2)^{0.5}|\ +\ c[/tex]​

    [tex]\mbox{At}\ x=0,\ t=0;\ \ \therefore \ c=-\frac{1}{128}\ln(\frac{1}{8}).[/tex]

    [tex]\mbox{Now, let }\:x=\frac{1}{8}\sinh z.\ \ \mbox{Therefore,}[/tex]

    [tex]\begin{equation*}\begin{split}t\:&=\:\frac{1}{10}(\sinh z)(\frac{1}{64}+\frac{1}{64}\sinh^2z)^{0.5}\ \\
    &+\ \frac{1}{128}\ln\left[\frac{1}{8}\sinh z+(\frac{1}{64}+\frac{1}{64}\sinh^2z)^{0.5}\right]\ -\ \frac{1}{128}\ln(\frac{1}{8})\\
    &=\:\frac{1}{80}(\sinh z)(1+\sinh^2z)^{0.5}\ \\
    &+\ \frac{1}{128}\ln\left(\frac{1}{8}[\sinh z+(1+\sinh^2z)^{0.5}]\right)\ -\ \frac{1}{128}\ln(\frac{1}{8})\\
    &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\ln\left[\frac{1}{8}(\sinh z+\cosh z)\right]\ -\ \frac{1}{128}\ln(\frac{1}{8})\\
    &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\ln(\frac{1}{8}\,e^z)\ -\ \frac{1}{128}\ln(\frac{1}{8})\\
    &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\ln(\frac{1}{8})\ +\ \frac{1}{128}\ln(e^z)\ -\ \frac{1}{128}\ln(\frac{1}{8})\\
    &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\,z\\
    &=\:\frac{1}{160}\sinh(2z)\;+\;\frac{1}{128}\,z\end{split}\end{equation*}[/tex]​


    Unfortunately, this equation cannot be solved analytically for z (to my knowledge). But you could solve it for z numerically, for any given value of t. And then you can compute the corresponding value of x and y, because x = 0.125*sinh(z), and y = 4*x2.
     
  9. Oct 16, 2010 #8

    HallsofIvy

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    Thanks, but this thread was 4 years old!
     
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