Parametrizing Circles These notes discuss a simple strategy for parametrizing circles in th ree dimensions

Parametrizing Circles These notes discuss a simple strategy for parametrizing circles in th ree dimensions - Description

We start with the circle in the xy plane that has radius and is centred on the origin This is easy to parametrize r cos sin Note that we can check that r lies on the desired circle by checking 64257rstly that r lies in the correct plane in this ID: 26299 Download Pdf

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Parametrizing Circles These notes discuss a simple strategy for parametrizing circles in th ree dimensions

We start with the circle in the xy plane that has radius and is centred on the origin This is easy to parametrize r cos sin Note that we can check that r lies on the desired circle by checking 64257rstly that r lies in the correct plane in this

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Parametrizing Circles These notes discuss a simple strategy for parametrizing circles in th ree dimensions




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Parametrizing Circles These notes discuss a simple strategy for parametrizing circles in th ree dimensions. We start with the circle in the xy –plane that has radius and is centred on the origin. This is easy to parametrize: ~r ) = cos sin Note that we can check that ~r ) lies on the desired circle by checking, firstly, that ~r ) lies in the correct plane (in this case, the xy –plane) and, secondly, that the distance from ~r ) to the centre of the circle is ~r cos sin cos +( sin since sin +cos = 1. Now let’s move the circle so that its centre is at some general point ~c

. To parametrize this new circle, which still has radius and which is still parallel to the xy –plane, we just translate by ~c ~c ~r ) = ~c cos sin Finally, let’s consider a circle in general position. The secret to param etrizing a general circle is to replace and by two new vectors and which (a) are unit vectors, (b) are parallel to the plane of the desired circle and (c) are mutually perpendicular. ~c ~r ) = ~c cos sin To check that this is correct, observe that ~r ~c is parallel to the plane of the desired circle because ~r ~c cos sin and both and are parallel to the plane of the desired

circle ~r ~c is of length for all because ~r ~c = ( ~r ~c ~r ~c = ( cos sin cos sin cos sin +2 cos sin (cos +sin ) = since = = 1 ( and are both unit vectors) and = 0 ( and are perpendicular). Joel Feldman. 2003. All rights reserved.
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To find such a parametrization in practice, we need to find the centr ~c of the circle, the radius of the circle and two mutually perpendicular unit vectors, and , in the plane of the circle. It is often easy to find at least one point ~p on the circle. Then we can take ~p ~c ~p ~c . It is also often easy to find a unit

vector, that is normal to the plane of the circle. Then we can choose Example 1 Let be the intersection of the sphere = 4 and the plane . The intersection of any plane with any sphere is a circle. The plane in question passes th rough the centre of the sphere, so has the same centre and same radius as the sphere. So has radius 2 and centre (0 0). The point (2 0) satisfies both = 4 and and so is on . We may choose to be the unit vector in the direction from the centre (0 0) of the circle towards (2 0). Namely = (1 0). Since the plane of the circle is = 0, the vector ) = (0 1) is

perpendicular to the plane of . So we may take (0 1). Then (0 1) (1 0) = (0 1). Subbing in ~c = (0 0), = 2, = (1 0) and (0 1) gives ~r ) = 2cos (1 0)+2sin (0 1) = 2 cos t, sin sin To check this, note that = 2cos 2sin 2sin satisfies both = 4 and Example 2 Let be the circle that passes through the three points (3 0), (0 0) and (0 3). All three points obey = 3. So the circle lies in the plane = 3. We guess, by symmetry, or by looking at the figure below, that the centre of the circle is at the cen tre of mass of the three points, which is [(3 0)+(0 0)+(0 3)] = (1 1). We can check this

by checking that (1 1) is equidistant from the three points: (3 0) (1 1) (2 1) (0 0) (1 1) 1) (0 3) (1 1) 2) This tells us both that (1 1)is indeed the centre and that the radiusof is 6. We may choose to be the unit vector in the direction from the centre (1 1) of the circle towards (3 0). Namely (2 1). Sincetheplaneofthecircleis = 3,thevector ) = (1 1)isperpendiculartotheplaneof So we may take (1 1). Then 18 (1 1) (2 1)= 18 (0 3) = (0 1). Subbing in ~c = (1 1), 6, (2 1) and (0 1) gives ~r ) = (1 1)+ 6cos (2 1)+ 6sin (0 1) 1+2cos t, cos 3sin t, cos 3sin To check this, note that ~r (0) = (3 0),

~r = (0 0) and ~r = (0 3) since cos = cos sin and sin Joel Feldman. 2003. All rights reserved.