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CLP-4 Vector Calculus

Section 1.12 Optional — Parametrizing Circles

We now discuss a simple strategy for parametrizing circles in three dimensions, starting with the circle in the xy-plane that has radius ρ and is centred on the origin. This is easy to parametrize:
r(t)=ρcostıı^+ρsintȷȷ^0t<2π
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 ρ and which is still parallel to the xy-plane, we just translate by c:
r(t)=c+ρcostıı^+ρsintȷȷ^0t<2π
Finally, let’s consider a circle in general position. The secret to parametrizing a general circle is to replace ıı^ and ȷȷ^ by two new vectors ıı^ and ȷȷ^ which
  1. are unit vectors,
  2. are parallel to the plane of the desired circle and
  3. are mutually perpendicular.
r(t)=c+ρcostıı^+ρsintȷȷ^0t<2π
To check that this is correct, observe that
  • r(t)c is parallel to the plane of the desired circle because both ıı^ and ȷȷ^ are parallel to the plane of the desired circle and r(t)c=ρcostıı^+ρsintȷȷ^
  • r(t)c is of length ρ for all t because
    |r(t)c|2=(r(t)c)(r(t)c)=(ρcostıı^+ρsintȷȷ^)(ρcostıı^+ρsintȷȷ^)=ρ2cos2t ıı^ıı^+ρ2sin2t ȷȷ^ȷȷ^+2ρcostsint ıı^ȷȷ^=ρ2(cos2t+sin2t)=ρ2
    since ıı^ıı^=ȷȷ^ȷȷ^=1 (ıı^ and ȷȷ^ are both unit vectors) and ıı^ȷȷ^=0 (ıı^ and ȷȷ^ are perpendicular).
To find such a parametrization in practice, we need to find the centre c of the circle, the radius ρ of the circle and two mutually perpendicular unit vectors, ıı^ and ȷȷ^, in the plane of the circle. It is often easy to find at least one point p on the circle. Then we can take ıı^=pc|pc|. It is also often easy to find a unit vector, k^, that is normal to the plane of the circle. Then we can choose ȷȷ^=k^×ıı^. We’ll illustrate this now.

Example 1.12.1.

Let C be the intersection of the sphere x2+y2+z2=4 and the plane z=y.
  • The intersection of any plane with any sphere is a circle. The plane in question passes through the centre of the sphere, so C has the same centre and same radius as the sphere. So C has radius 2 and centre (0,0,0).
  • Notice that the point (2,0,0) satisfies both x2+y2+z2=4 and z=y and so is on C. We may choose ıı^ to be the unit vector in the direction from the centre (0,0,0) of the circle towards (2,0,0). Namely ıı^=(1,0,0).
  • Since the plane of the circle is zy=0, the vector (zy)=(0,1,1) is perpendicular to the plane of C. So we may take k^=12(0,1,1).
  • Then ȷȷ^=k^×ıı^=12(0,1,1)×(1,0,0)=12(0,1,1).
Substituting in c=(0,0,0), ρ=2, ıı^=(1,0,0) and ȷȷ^=12(0,1,1) gives
r(t)=2cost(1,0,0)+2sint12(0,1,1)=2(cost,sint2,sint2)0t<2π
To check this, note that x=2cost, y=2sint, z=2sint satisfies both x2+y2+z2=4 and z=y.

Example 1.12.2.

Let C be the circle that passes through the three points (3,0,0), (0,3,0) and (0,0,3).
  • All three points obey x+y+z=3. So the circle lies in the plane x+y+z=3. We guess, by symmetry, or by looking at the figure below, that the centre of the circle is at the centre of mass of the three points, which is 13[(3,0,0)+(0,3,0)+(0,0,3)]=(1,1,1). We must check this and can do so by checking that (1,1,1) is equidistant from the three points:
    |(3,0,0)(1,1,1)|=|(2,1,1)|=6|(0,3,0)(1,1,1)|=|(1,2,1)|=6|(0,0,3)(1,1,1)|=|(1,1,2)|=6
    This tells us both that (1,1,1) is indeed the centre (as only the centre is equidistant from any three distinct points on a circle) and that the radius of C is 6.
  • We may choose ıı^ to be the unit vector in the direction from the centre (1,1,1) of the circle towards (3,0,0). Namely ıı^=16(2,1,1).
  • Since the plane of the circle is x+y+z=3, the vector (x+y+z)=(1,1,1) is perpendicular to the plane of C. So we may take k^=13(1,1,1).
  • Then
    ȷȷ^=k^×ıı^=118(1,1,1)×(2,1,1)=118(0,3,3)=12(0,1,1)
Substituting in c=(1,1,1), ρ=6, ıı^=16(2,1,1) and ȷȷ^=12(0,1,1) gives
r(t)=(1,1,1)+6cost16(2,1,1)+6sint12(0,1,1)=(1+2cost,1cost+3sint,1cost3sint)
To check this, note that r(0)=(3,0,0), r(2π3)=(0,3,0) and r(4π3)=(0,0,3) since cos2π3=cos4π3=12, sin2π3=32 and sin4π3=32.