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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

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 \(\rho\) and is centred on the origin. This is easy to parametrize:
\begin{align*} &\vr(t)=\rho\cos t\,\hi+ \rho\sin t\,\hj\\ &0\le t\lt 2\pi \end{align*}
Now let's move the circle so that its centre is at some general point \(\vc\text{.}\) To parametrize this new circle, which still has radius \(\rho\) and which is still parallel to the \(xy\)-plane, we just translate by \(\vc\text{:}\)
\begin{align*} &\vr(t)=\vc+\rho\cos t\,\hi+ \rho\sin t\,\hj\\ & 0\le t\lt 2\pi \end{align*}
Finally, let's consider a circle in general position. The secret to parametrizing a general circle is to replace \(\hi\) and \(\hj\) by two new vectors \(\hi'\) and \(\hj'\) which
  1. are unit vectors,
  2. are parallel to the plane of the desired circle and
  3. are mutually perpendicular.
\begin{align*} & \vr(t)=\vc+\rho\cos t\,\hi'+ \rho\sin t\,\hj'\\ & 0\le t\lt 2\pi \end{align*}
To check that this is correct, observe that
  • \(\vr(t)-\vc\) is parallel to the plane of the desired circle because both \(\hi'\) and \(\hj'\) are parallel to the plane of the desired circle and \(\vr(t)-\vc=\rho\cos t\,\hi'+ \rho\sin t\,\hj'\)
  • \(\vr(t)-\vc\) is of length \(\rho\) for all \(t\) because
    \begin{align*} |\vr(t)-\vc\,|^2 &=(\vr(t)-\vc\,)\cdot(\vr(t)-\vc\,)\\ &=(\rho\cos t\,\hi'+ \rho\sin t\,\hj')\cdot (\rho\cos t\,\hi'+ \rho\sin t\,\hj')\\ &=\rho^2\cos^2 t\ \hi'\cdot\hi' + \rho^2\sin^2 t\ \hj'\cdot\hj' +2\rho\cos t\sin t\ \hi'\cdot\hj'\\ &=\rho^2(\cos^2t+\sin^2t)=\rho^2 \end{align*}
    since \(\hi'\cdot\hi'=\hj'\cdot\hj'=1\) (\(\hi'\) and \(\hj'\) are both unit vectors) and \(\hi'\cdot\hj'=0\) (\(\hi'\) and \(\hj'\) are perpendicular).
To find such a parametrization in practice, we need to find the centre \(\vc\) of the circle, the radius \(\rho\) of the circle and two mutually perpendicular unit vectors, \(\hi'\) and \(\hj'\text{,}\) in the plane of the circle. It is often easy to find at least one point \(\vp\) on the circle. Then we can take \(\hi'=\frac{\vp-\vc}{|\vp-\vc|}\text{.}\) It is also often easy to find a unit vector, \(\hk'\text{,}\) that is normal to the plane of the circle. Then we can choose \(\hj'=\hk'\times\hi'\text{.}\) We'll illustrate this now.
Let \(C\) be the intersection of the sphere \(x^2+y^2+z^2=4\) and the plane \(z=y\text{.}\)
  • 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 \((0,0,0)\text{.}\)
  • Notice that the point \((2,0,0)\) satisfies both \(x^2+y^2+z^2=4\) and \(z=y\) and so is on \(C\text{.}\) We may choose \(\hi'\) to be the unit vector in the direction from the centre \((0,0,0)\) of the circle towards \((2,0,0)\text{.}\) Namely \(\hi'=(1,0,0)\text{.}\)
  • Since the plane of the circle is \(z-y=0\text{,}\) the vector \(\vnabla(z-y)=(0,-1,1)\) is perpendicular to the plane of \(C\text{.}\) So we may take \(\hk'=\frac{1}{\sqrt{2}}(0,-1,1)\text{.}\)
  • Then \(\hj'=\hk'\times\hi' =\frac{1}{\sqrt{2}}(0,-1,1)\times(1,0,0)=\frac{1}{\sqrt{2}}(0,1,1)\text{.}\)
Substituting in \(\vc=(0,0,0)\text{,}\) \(\rho=2\text{,}\) \(\hi'=(1,0,0)\) and \(\hj'=\frac{1}{\sqrt{2}}(0,1,1)\) gives
\begin{align*} \vr(t)&=2\cos t\,(1,0,0)+ 2\sin t\,\frac{1}{\sqrt{2}}(0,1,1)\\ &=2\big(\cos t, \frac{\sin t}{\sqrt{2}},\frac{\sin t}{\sqrt{2}}\big)\\ & 0\le t\lt 2\pi \end{align*}
To check this, note that \(x=2\cos t\text{,}\) \(y=\sqrt{2}\sin t\text{,}\) \(z=\sqrt{2}\sin t\) satisfies both \(x^2+y^2+z^2=4\) and \(z=y\text{.}\)
Let \(C\) be the circle that passes through the three points \((3,0,0)\text{,}\) \((0,3,0)\) and \((0,0,3)\text{.}\)
  • All three points obey \(x+y+z=3\text{.}\) So the circle lies in the plane \(x+y+z=3\text{.}\) 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 \(\frac{1}{3}[(3,0,0)+(0,3,0)+(0,0,3)]=(1,1,1)\text{.}\) We must check this and can do so by checking that \((1,1,1)\) is equidistant from the three points:
    \begin{alignat*}{2} \big|(3,0,0)-(1,1,1)\big|&=\big|(2,-1,-1)\big|&&=\sqrt{6}\\ \big|(0,3,0)-(1,1,1)\big|&=\big|(-1,2,-1)\big|&&=\sqrt{6}\\ \big|(0,0,3)-(1,1,1)\big|&=\big|(-1,-1,2)\big|&&=\sqrt{6} \end{alignat*}
    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 \(\sqrt{6}\text{.}\)
  • We may choose \(\hi'\) to be the unit vector in the direction from the centre \((1,1,1)\) of the circle towards \((3,0,0)\text{.}\) Namely \(\hi'=\frac{1}{\sqrt{6}}(2,-1,-1)\text{.}\)
  • Since the plane of the circle is \(x+y+z=3\text{,}\) the vector \(\vnabla(x+y+z)=(1,1,1)\) is perpendicular to the plane of \(C\text{.}\) So we may take \(\hk'=\frac{1}{\sqrt{3}}(1,1,1)\text{.}\)
  • Then
    \begin{align*} \hj'&=\hk'\times\hi' =\frac{1}{\sqrt{18}}(1,1,1)\times(2,-1,-1)=\frac{1}{\sqrt{18}}(0,3,-3)\\ & =\frac{1}{\sqrt{2}}(0,1,-1) \end{align*}
Substituting in \(\vc=(1,1,1)\text{,}\) \(\rho=\sqrt{6}\text{,}\) \(\hi'=\frac{1}{\sqrt{6}}(2,-1,-1)\) and \(\hj'=\frac{1}{\sqrt{2}}(0,1,-1)\) gives
\begin{align*} \vr(t)&=(1,1,1)+\sqrt{6}\cos t\,\frac{1}{\sqrt{6}}(2,-1,-1) + \sqrt{6}\sin t\,\frac{1}{\sqrt{2}}(0,1,-1)\cr &=\big(1+2\cos t, 1-\cos t+\sqrt{3}\sin t,1-\cos t-\sqrt{3}\sin t\big) \end{align*}
To check this, note that \(\vr(0)=(3,0,0)\text{,}\) \(\vr\big(\frac{2\pi}{3}\big)=(0,3,0)\) and \(\vr\big(\frac{4\pi}{3}\big)=(0,0,3)\) since \(\cos\frac{2\pi}{3}=\cos\frac{4\pi}{3}=-\frac{1}{2}\text{,}\) \(\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}\) and \(\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}\text{.}\)
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