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Math Help - Parameterising a certain circle

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    Parameterising a certain circle

    Parameterise the circle that results from intersecting the unit sphere in \mathbb{R}^3 centered at the origin with the plane  [(x,y,z) \in : x = z ]

    I tried x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1
    however the last equation is an ellipse????
    Last edited by FGT12; January 24th 2011 at 11:38 AM.
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    Quote Originally Posted by FGT12 View Post
    Parameterise the circle that results from intersecting the unit sphere in \mathbb{R}^3 centered at the origin with the plane  [(x,y,z) \in : x = z ]

    I tried x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1
    however the last equation is an ellipse????
    the intersection is an ellipse and no circle!
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  3. #3
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    You result is only the projection of the circle to the xy plane, which is really an ellipse.
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    so how would you parameterise the circle in \mathbb{R}^3
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  5. #5
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    It's definitely going to be a circle, because really: how could a plane intersect a sphere and make an ellipse! You're right that it's absurd.

    Have you tried drawing a sketch? Try visualizing the picture with the y axis pointing directly out of the paper, and that may give you some insight. I solved the problem by first coming up with an equation based on the sketch, then proving that it was right. So, like I said. Start with the picture.
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  6. #6
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    ahaok is wrong, a sphere cut by a plane gives a circle, not an ellipse. xxp9 is right- a circle, projected onto a plane at an angle to a line perpendicular to the circle gives an ellipse. That is why you get the equation of an ellipse- you are projecting the circle onto the xy-plane, not parameterizing it.

    Here, your sphere is the unit sphere, x^2+ y^2+ z^2= 1 and the plane is given by x= z. The "standard" parameterization of the unit sphere can be derived from spherical coordinates:
    x= \rho cos(\theta)sin(\phi)
    y= \rho sin(\theta)sin(\phi)
    z= \rho cos(\phi)

    by taking the radius variable, \rho equal to 1:
    x= cos(\theta)sin(\phi)
    y= sin(\theta)sin(\phi)
    z= cos(\phi)

    The fact that z= x means that we must have cos(\theta)sin(\phi)= cos(\phi) so that cos(\theta)= \frac{cos(\phi)}{sin(\phi)}= cot(\phi).

    So one way of getting the parameterization of the circle is to replace \theta by cos^{-1}(cot(\phi)), reducing from two parameters to one:
    x= cos(cos^{-1}(cot(\phi)))sin(\phi)
    y= sin(cos^{-1}(cot(\phi))sin(\phi)
    z= cos(\phi)

    Of course, cos(cos^{-1}(cot(\phi))= cot(\phi). And, we can use sin(\theta)= \sqrt{1- cos^2(\theta)} to say that
    sin(cos^{-1}(cot(\phi))= \sqrt{1- (cos(cos^{-1}(cot(\phi)}= \sqrt{1- cot^2(\phi)} and so have

    x= cot(\phi)sin(\phi)= cos(\phi)
    y= \sqrt{1- cot^2(\phi)}sin(phi)
    z= cos(\phi).
    Last edited by HallsofIvy; January 25th 2011 at 10:28 AM.
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  7. #7
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    Quote Originally Posted by FGT12 View Post
    Parameterise the circle that results from intersecting the unit sphere in \mathbb{R}^3 centered at the origin with the plane  [(x,y,z) \in : x = z ]

    I tried x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1
    however the last equation is an ellipse????
    The vector form for the plane is

    \vec{r}(s,t)=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}t+\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}s



    This gives that x=z=t Using the equation of the sphere gives

    t^2+y^2+t^2=1 \implies y=\pm\sqrt{1-2t^2}

    So a parametric representation of the circle is
    x=t,\quad y=\pm\sqrt{1-2t^2},\quad z=t

    Since the plus or minus is really messy it can be eliminated using trigonometry.

    If we define \sin(\alpha)=\sqrt{1-2t^2} this gives
    \cos(\alpha)=\frac{t\sqrt{2}}{1} \implies t=\frac{1}{\sqrt{2}}\cos(\alpha)

    Using this we get

    \displaystyle \vec{r}(\alpha)=\left( \frac{1}{\sqrt{2}}\cos(\alpha)\right)\vec{i}+\sin(  \alpha)\vec{j}+\left( \frac{1}{\sqrt{2}}\cos(\alpha)\right)\vec{k}

    Parameterising a certain circle-object.jpg


    Edit: too slow
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