2
$\begingroup$

Let $\alpha$ be a unit speed curve with positive curvature $\kappa \gt 0$ and non-zero torsion $\tau \ne 0$, lying on a sphere of radius $r$ centred at $c \in \Bbb{R}^3$.

Show that $\alpha - c = -\frac{1}{\kappa}N - (\frac{1}{\kappa})'\frac{1}{\tau}B$.

Where N is the principle normal vector field and B is the binormal vector field to $\alpha$. I'm not sure how it is possible to show this as all the components on the right hand side of the equation involve derivatives of the curve on the left hand side, any help would be much appreciated.

Then, deduce a formula for the radius of the sphere in terms of $\kappa,\tau$.

$\endgroup$
  • $\begingroup$ This question has been asked and answered numerous times on here. Here is a post where I gave an explicit hint. $\endgroup$ – Ted Shifrin Nov 4 '17 at 22:47
  • $\begingroup$ Thanks for that Ted $\endgroup$ – Helen Peters Nov 4 '17 at 22:58
2
$\begingroup$

Well, we have

$(\alpha - c) \cdot (\alpha - c) = r^2, \tag 1$

since $\alpha$ lies on the sphere of radius $r$ centered at $c$; this is just what equation (1) affirms. If we differentiate (1) with respect to $s$, the arc-length along $\alpha$, we obtain

$2\alpha' \cdot (\alpha - c) = \alpha' \cdot (\alpha - c) + (\alpha - c) \cdot \alpha' = 0, \tag 2$

whence

$\alpha' \cdot (\alpha - c) = 0; \tag 3$

since $\alpha$ is a unit-speed curve, we have

$\alpha' = T, \; \Vert T \Vert = 1, \tag 4$

the unit tangent vector to $\alpha$; thus (3) becomes

$T \cdot (\alpha - c) = 0, \tag 5$

which is nearly self-evident, since $T$ is tangent to the sphere (1), hence normal to the radial vector $\alpha - c$; in any event, we may differentiate (5) once again to obtain

$T' \cdot (\alpha - c) + T \cdot T = T' \cdot (\alpha - c) + T \cdot \alpha' = 0, \tag 6$

or

$T' \cdot (\alpha -c) + \Vert T \Vert^2 = 0; \tag 7$

we now use the Frenet-Serret equation

$T' = \kappa N, \tag 8$

where $N$ is the normal vector to $\alpha$, and (4) to re-write (7) as

$\kappa N \cdot (\alpha - c) + 1 = 0, \tag 9$

which since $\kappa > 0$ implies

$N \cdot (\alpha - c) = -\dfrac{1}{\kappa}; \tag {10}$

we may now differentiate (9) with respect to $s$ to find

$\kappa' N \cdot (\alpha - c) + \kappa N' \cdot (\alpha - c) + kN \cdot \alpha'= 0; \tag{11}$

we have

$N \cdot \alpha' = N \cdot T = 0, \tag{12}$

and also the Frenet-Serret equation

$N' = -\kappa T + \tau B, \tag{13}$

where $\tau$ is the torsion and $B = T \times N$ the binormal vector of $\alpha$; when (10), (12) and (13) are substituted into (11) we obtain

$-\dfrac{\kappa'}{\kappa} + \kappa (-\kappa T + \tau B) \cdot (\alpha - c) = 0, \tag{14}$

or

$-\dfrac{\kappa'}{\kappa} -\kappa^2 T \cdot (\alpha - c) + \kappa \tau B \cdot (\alpha - c) = 0, \tag{15}$

which by virtue of (5) reduces to

$-\dfrac{\kappa'}{\kappa} + \kappa \tau B \cdot (\alpha - c) = 0, \tag{16}$

and since $\kappa > 0 \ne \tau$ we have

$-\dfrac{\kappa'}{\kappa^2 \tau} + B \cdot (\alpha - c) = 0, \tag{17}$

whence

$B \cdot (\alpha - c) = \dfrac{\kappa'}{\kappa^2 \tau} = -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau}; \tag {18}$

(5), (10) and (18) express the components of the radial vector $\alpha - c$ in the orthonormal frame $T$, $N$, $B$, whence

$\alpha - c = -\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B, \tag{19}$

the requisite result.

We may find $r$ in terms of $\kappa$ and $\tau$ by inserting (19) into (1):

$r^2 = (\alpha - c) \cdot (\alpha - c)$ $= \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ) \cdot \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ); \tag{20}$

since

$\Vert N \Vert = \Vert B \Vert = 1, \; N \cdot B = 0, \tag{21}$

(20) reduces to

$r^2 = \dfrac{1}{\kappa^2} + \left ( \left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} \right )^2 = \dfrac{1}{\kappa^2} + \left ( -\dfrac{\kappa'}{\kappa^2} \right )^2 \dfrac{1}{\tau^2} = \dfrac{1}{\kappa^2} + \dfrac{(\kappa')^2}{\kappa^4 \tau^2} = \dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}, \tag{22}$

whence

$r = \sqrt{\dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}} = \dfrac{1}{\kappa^2 \tau} \sqrt{(\kappa')^2 + \kappa^2 \tau^2}, \tag{23}$

as the desired formula for $r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.