# What is the curvature at a certain point?

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?

• I added the "differential-geometry" tag to your post. Cheers! – Robert Lewis Sep 27 '18 at 3:49
• 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. – Christian Blatter Sep 27 '18 at 7:26

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 $$s$$.

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