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Math Help - Coordinate Geometry - intrinsic coordinates problems

  1. #1
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    Coordinate Geometry - intrinsic coordinates problems

    Hi

    I'm having some trouble working out how to convert from intrinsic to cartesian coordinates, and from cartesian to intrinsic, when the function is not given explicitly in terms of y. Let me show you a couple of problems which I have been unable to solve to illustrate this,

    (1) The intrinsic equation of the curve is s = \psi^2. The curve passes through the point (2,0). Show that x = 2(\cos\psi + \psi\sin\psi) and find y in terms of \psi

    (2) For the curve with equation \sin y = e^x, where s is measured from the point (0,\frac{\pi}{2}), show that e^s = \tan \frac{\psi}{2}

    For the first I tried to let \sqrt{s} = \psi and then take the cosine of both sides since \frac{dx}{ds} = \cos\psi however I ended up with a bit of a mess and didn't really get anywhere.

    For the second I tried to find an expression for \frac{dy}{dx} and set it equal to \tan\psi and also find the length of s, but I ended up with a horrible integral and again didn't really get anywhere.

    I'm really at a loose end on how to solve these now, I think there must be something simple I'm missing, but I just can't see it.

    Any help would be much appreciated, thanks

    Stonehambey
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  2. #2
    Super Member Rebesques's Avatar
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    Just to get you going. For 1), you know that \frac{{\rm d}x}{{\rm d}s}={\rm cos}\psi, so (somewhat heuristically) we get

    \left(\frac{{\rm d}s}{{\rm d}x}\right)^{-1}={\rm cos}\psi

    and remembering s=\psi^2,

    \left(\frac{{\rm d}(\psi^2)}{{\rm d}x}\right)^{-1}={\rm cos}\psi\Rightarrow\left(2\psi\frac{{\rm d}\psi}{{\rm d}x}\right)^{-1}={\rm cos}\psi\Rightarrow\frac{{\rm d}x}{{\rm d}\psi}=2\psi{\rm cos}\psi.

    Integrate for \psi to get x = 2(\cos\psi + \psi\sin\psi)+c for some c\in \mathbb{R}, and use the given initial condition x=2 when \psi=0 to get c=0.
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  3. #3
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    A little less heuristically:

    (And first some clarification. \psi here is the angle the line from (0,0) to a given point on the curve makes with the positive x-axis (so tan(\psi) is the slope of the tangent line- the derivative of y with respect to x) and s is the arc length to that point from some fixed point on the graph.)

    s= \int\sqrt{1+ \left(\frac{dy}{dx}\right)^2}dx so
    \frac{ds}{d\psi}= \sqrt{1+ \left(\frac{dy}{dx}\right)^2}\frac{dx}}d\psi}= 2\psi since s= \psi^2. But \frac{dy}{dx}= tan(\psi) so we have the differential equation
    \sqrt{1+ tan^2\psi}\frac{dx}{d\psi}= sec(\psi)\frac{dx}{d\psi}= 2\psi
    \frac{dx}{d\psi}= \frac{2\psi}{sec(\psi)}= 2\psi cos(\psi)
    so dx= 2\psi cos(\psi)d\psi)

    Integrating the right side by parts, using u= 2\psi, dv= cos(\psi)d\psi, we have [tex]du= 2d\psi[tex], [tex]v= sin(\psi) so
    x= 2\psi sin(\psi)- 2\int sin(\psi)d\psi= 2\psi sin(\psi)+ 2 cos(\psi)+ C.

    Using the fact that x= 2 when \psi= 0 (the point (2, 0) is on the x-axis, of course, so the line from (0,0) to (2,0) makes angle \psi= 0 with the x-axis) C= 0 giving x= 2\psi sin(\psi)+ 2 cos(\psi) as required.

    By the chain rule, \frac{dy}{dx}= \frac{dy}{d\psi}\frac{d\psi}{dx} and we already know that \frac{dx}{d\psi}= 2\psi cos(\psi) so \frac{d\psi}{dx}= \frac{1}{2\psi cos(\psi)} and \frac{dy}{dx}= tan(\psi) so we have
    \frac{dy}{d\psi}= \psi tan(\psi)cos(\psi)= \psi sin(\psi) and you can again use integration by parts to find y as a function of \psi.
    Last edited by HallsofIvy; July 18th 2009 at 04:29 PM.
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