# parameterization and path integral

• Oct 8th 2010, 01:26 PM
fallofdark
parameterization and path integral
Hey all, I have a question here and I can't seem to get around solving it.

Show that the circle that is the intersection of the plane x + y + z = 0 and the sphere x^2 + y^2 + z^2 = 1 can be expressed as

x(t) = (cos t - sqrt(3)sin t)/sqrt(6)
y(t) = (cos t + sqrt(3)sin t)/sqrt(6)
z(t) = - 2cos t/sqrt(6)

where 0<t<2π

And then, the question continues to...

Calculate the path integral where f(x,y,z)=xy-z and c is a path on the circle in the first part starting from the point (-1/sqrt(2),1/sqrt(2), 0) and finishing at the point (1/sqrt(2),-1/sqrt(2), 0), in the direction of increasing t.

• Oct 8th 2010, 05:46 PM
Prove It
Well, they intersect where they are equal.

So $x+y+z+1 = x^2 + y^2 + z^2$

$1 = x^2 - x + y^2 - y + z^2 - z$

$1 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = x^2 - x + \left(-\frac{1}{2}\right)^2 + y^2 - y + \left(-\frac{1}{2}\right)^2 + z^2 - z + \left(-\frac{1}{2}\right)^2$

$1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 + \left(z - \frac{1}{2}\right)^2$

$\frac{7}{4} = \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 + \left(z - \frac{1}{2}\right)^2$.

Now work with the right hand side and substitute the suggested parameterisations for $x, y, z$. If you get the left hand side after some simplification, then this parameterisation is acceptable.
• Oct 8th 2010, 06:04 PM
Ackbeet
Another approach would be to use rotation matrices. You could assume that, since you have the intersection of a sphere of radius 1 with a plane going through the origin, that that intersection must be described by a circle (locus of points equidistant from a point, and in a plane = circle). To get the desired parametrization, you could start with a parametrization of the circle of radius 1 in the xy plane, and apply to that vector the rotations required to transform the z unit vector into a unit vector in the <1,1,1> direction. I think that would do the trick, more or less. This approach at least has the virtue of being a direct approach.
• Oct 9th 2010, 02:24 AM
fallofdark
Thank you for your inputs! But there are still some stuff I don't understand...

For the first answer by Prove It, I don't quite get why the (-1/2)'s are added to the equation. Also, I don't see how all that will simplify into 7/4 when t does not have an assigned value...
• Oct 9th 2010, 03:03 AM
HallsofIvy
He doesn't add "-1/2", he adds $(-1/2)^2$. He is completing the square to get the equation of the intersection in standard form for a circle.
• Oct 9th 2010, 06:05 PM
fallofdark
Okay, I understand the completing the square thing. But I still don't get how it gets me the parameterization required...
• Oct 9th 2010, 06:21 PM
Prove It
Quote:

Originally Posted by fallofdark
Okay, I understand the completing the square thing. But I still don't get how it gets me the parameterization required...

You're not asked to derive the parameterisation, you're asked to verify it.

So you can substitute the suggested parameterised $x, y, z$ into the RHS.

If you can simplify and get back the LHS, then you can say the parameterisation is appropriate.
• Oct 9th 2010, 06:58 PM
fallofdark
Okay I plugged the suggested x,y,z values into the RHS and got:

7/4 + 2sqrt(3)sin t/sqrt(6)

... which is the LHS with the extra second term. Does it still count as an appropriate parameterisation?
• Oct 9th 2010, 06:59 PM
fallofdark
Okay I plugged the suggested x,y,z values into the RHS and got:

7/4 + 2sqrt(3)sin t/sqrt(6)

... which is the LHS with the extra second term. Does it still count as an appropriate parameterisation?
• Oct 10th 2010, 04:51 AM
HallsofIvy
You plugged the suggested x, y, z values into the RHS of what? None of the equations you give has x, y, or z on the right hand side.

Putting
x(t) = (cos t - sqrt(3)sin t)/sqrt(6)
y(t) = (cos t + sqrt(3)sin t)/sqrt(6)
z(t) = - 2cos t/sqrt(6)

into the LHS of $x^2 + y^2 + z^2 = 1$ gives $\frac{(cos t- sqrt{3}sin t)^2}{6}+ \frac{(cos t+ \sqrt{3}sin t)^2}{6}+ \frac{4cos^2 t}{6}$ $= \frac{cos^2 t- 2\sqrt{3}sin(t)cos(t)+ 3sin^2(t)+ cos^2(t)+ 2\sqrt{3}sin(t)cos(t)+ 3sin^2(t)}{6}+ \frac{4cos^2 t}{6}$ $= \frac{2cos^2 t+ 6 sin^2 t+ 4 cos^2 t}{6}= cos^2 t + sin^2 t= 1$ which is a true statement. The curve described by this parameterization does lie on that sphere.

Putting it into the LHS of x+ y+ z= 0 we have
$\frac{cos t- \sqrt{3}sin t+ cos t+ \sqrt{3}sin t- 2 cos t}{\sqrt{6}}= \frac{2 cos t- 2 cos t}{\sqrt{6}}= 0$
so the curve also lies on that plane.

Another thing that you should check to sure this is "valid" parameterization is that there is no point where all of the derivatives of x, y, and z, with respect to the parameter, t, are simultaneously 0.