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So I am asked to do the following:

Find the curvature of $r(t) = \langle 3t,2\sin t, 2\cos t\rangle$ at the point $\left( \frac{5\pi}{2},1,-\sqrt{3} \right)$.

I find $r′(t)=\langle 3,−2cos(t),-2sin(t)\rangle$ and $||r′(t)||=\sqrt{13}$. I then find $T'(t)$ to be $-2/\sqrt{13} \langle 0,-sin(t),-cos(t)\rangle $.

I then find the expression for curvature to be $-(2/13) \langle 0,-sin(t),-cos(t)\rangle $. However, I don't know how to find a numerical answer for curvature at the given point since I am given the point and not a value of $t$.

Can anyone help me? What am I missing here?

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  • $\begingroup$ I added the "differential-geometry" tag to your post. Cheers! $\endgroup$ – Robert Lewis Sep 27 '18 at 3:49
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    $\begingroup$ Two things: (i) You can find the $t$-value belonging to the given point by inspection of the $x$-coordinate. (ii) The curve is a helix, hence the curvature is constant. $\endgroup$ – Christian Blatter Sep 27 '18 at 7:26
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One generally relatively easy, pragmatic method to calculate the curvature of any curve such as $r(t)$ is to exploit the first Frenet-Serret equation

$T'(s) = \kappa(s) N(s), \tag 1$

where $T(s)$ is the unit tangent vector, $N(s)$ the unit normal vector, $\kappa(s)$ the curvature, and $s$ the arc-length or distance along the curve $r(t)$. Here $\kappa(s) > 0$ by definition, so since $\Vert N(s) \Vert = 1$, we see that

$\Vert T'(s) \Vert = \Vert \kappa(s) N(s) \Vert = \kappa(s) \Vert N(s) \Vert = \kappa(s); \; \tag 2$

we may indeed regard (1)-(2) as giving the definition of $\kappa(s)$.

So . . . in order to apply (2) to the problem at hand, we must first re-parametrize $r(t)$ by the arc-length $s$; we may express $s$ in terms of $t$ by means of the definitive formula

$\dfrac{ds}{dt} = \Vert r'(t) \Vert; \tag 3$

which identifies $ds/dt$ with the magnitude of the velocity vector $r'(t)$; in the present case we have

$r(t) = (3t, 2\sin t, 2\cos t), \tag 4$

whence

$r'(t) = (3, 2\cos t, -2\sin t), \tag 5$

and we see that

$\Vert r'(t) \Vert^2 = 3^2 + 4\cos^2 t + 4\sin^2 t = 9 + 4 = 13, \tag 6$

or

$\Vert r'(t) \Vert = \sqrt{13}; \tag 7$

thus

$\dfrac{ds}{dt} = \sqrt{13}; \tag 8$

it then follows that

$s = \sqrt{13} t + s_0 \tag 9$

for some constant $s_0$, and taking $t = 0$ in (9) we find

$s_0 = s(0); \tag{10}$

we may arbitrarily choose $s_0$, the value of $s$ at $t = 0$;

$s_0 = 0 \tag{11}$

is obviously a convenient choice; then

$s = \sqrt{13} t, \tag{12}$

or

$t = \dfrac{s}{\sqrt{13}}; \tag{13}$

thus we may re-write $r(t)$ in terms of arc-length as

$r(s) = \left ( \dfrac{3s}{\sqrt{13}}, 2 \sin \dfrac{s}{\sqrt{13}}, 2\cos \dfrac{s}{\sqrt{13}} \right ); \tag{14}$

then

$T(s) = r'(s) = \left ( \dfrac{3}{\sqrt{13}}, \dfrac{2}{\sqrt{13}} \cos \dfrac{s}{\sqrt{13}}, -\dfrac{2}{\sqrt{13}} \sin \dfrac{s}{\sqrt{13}} \right ), \tag{15}$

and

$\kappa(s) N(s) = T'(s) = \left ( 0, -\dfrac{2}{13} \sin \dfrac{s}{\sqrt{13}}, -\dfrac{2}{13} \cos \dfrac{s}{\sqrt{13}} \right ). \tag{16}$

Finally, using formula (2) we find

$\kappa(s) = \Vert T'(s) \Vert = \dfrac{2}{13} \tag{17}$

at all points on $r(s)$, and in particular at the given point $( 5 \pi / 2, 1, -\sqrt 3)$.

Lastly, I note that my answer differs from that of my colleague Joaquin Liniado by a factor of $1 / 13$; this arises from the two factors of $1 / \sqrt {13}$ which would be introduced by using two factors $dt / ds$ via the chain rule to compensate for taking derivatives with respect to $t$ instead of the arc-length $s$.

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    $\begingroup$ Makes sense! Thank you! $\endgroup$ – darklord0530 Sep 27 '18 at 17:34
  • $\begingroup$ My pleasure, my friend. And thanks for the "acceptance"! $\endgroup$ – Robert Lewis Sep 27 '18 at 17:36

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