Vector calculus problem

**milad83** · April 25th 2012, 10:56 AM

The parametric surface shown in figure 1 hasgrid curves which can be shown to be circles when is constant
x=(2+sin(v))cos(u)
y=(2+sin(v))sin(u)
z=u+cos(v)
find the center and radius of the circle at u=pi
please help me

**earboth** · April 25th 2012, 10:39 PM

Originally Posted by milad83

The parametric surface shown in figure 1 hasgrid curves which can be shown to be circles when is constant
x=(2+sin(v))cos(u)
y=(2+sin(v))sin(u)
z=u+cos(v)
find the center and radius of the circle at u=pi
please help me

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1. Plug in $u = \pi$ . You'll get the vector:

$\langle x,y,z \rangle = \langle -\sin(v) - 2, 0, \cos(v) + \pi \rangle$

2. From

$x= -\sin(v) - 2~\implies~x+2=-\sin(v)$ ...... and

$z = \cos(v)+\pi~\implies~ z-\pi = \cos(v)$

you'll get by squaring:

$(x+2)^2 + (z-\pi)^2 = (-\sin(v))^2+(\cos(v))^2 = 1$

3. This is the equation of a circle in the x-z-plane with $C(-2, 0, \pi)$ and r = 1

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