# Is this the correct curvature of the space curve?

So lets say we have a space curve $r=\langle\cos t,\sin t,3t\rangle$ and we need to find the curvature at the point $P(1,0,0)$

$r'(t)=\langle-\sin t,\cos t,3\rangle$

$r''(t)=\langle-\cos t,-\sin t,0\rangle$

$||r'(t)||=\sqrt{(-\sin t)^2+(\cos t)^2+(3)^2} = \sqrt{1+9} = \sqrt{10}$

$||r'(t)||^3=(\sqrt{10})^3=10\sqrt{10}$

$r'(t) \times r''(t)$ = $\langle3\sin t,-3\cos t,1\rangle$

$||r'(t) \times r''(t)|| = \sqrt{(3\sin t)^2+(-3\cos t)^2+1^2}=\sqrt{10}$

$k=\frac{||r(t) \times r''(t)||}{||r'(t)||^3}=\frac{\sqrt{10}}{10\sqrt{10}}=\frac{1}{10}$

Is this correct? I think that's all you need to do for the problem, right?

• Also, sorry, I couldn't figure out how to do the notation for 'cross multiply' so it's a bit messy :( – A Row Jul 27 '17 at 7:27
• Hi A Row, I've just reformatted some of your question. Please feel free to edit it yourself to see the commands I've used, specifically \sin, \cos, \langle, \rangle, and \times. – Theo Bendit Jul 27 '17 at 7:34
• Thanks, I'll check them out :) – A Row Jul 27 '17 at 7:43

Let $r(t) = \langle \cos(t),\sin(t),3t \rangle$. Then the arclength is written: $$s(t) = \int_0^t ||r'(\tau)||_2d\tau=\sqrt{10}t \;\;\;\implies\;\;\;t = \frac{s}{\sqrt{10}}$$ So, using this parametrization, we get: \begin{align} r(s) &= \langle \cos(s/\sqrt{10}),\sin(s/\sqrt{10}),3t/\sqrt{10} \rangle\\ r'(s) &= \langle -\sin(s/\sqrt{10})/\sqrt{10},\cos(s/\sqrt{10})/\sqrt{10},3/\sqrt{10} \rangle\\ r''(s) &= \langle -\cos(s/\sqrt{10})/{10},-\sin(s/\sqrt{10})/{10},0 \rangle \end{align} The curvature is then computable by a simpler formula, i.e: $$\kappa(s) = ||r''(s)||_2=\sqrt{(-\cos(s/\sqrt{10})/{10})^2 + (-\sin(s/\sqrt{10})/{10}})^2 = \frac{1}{10}$$ as you computed as well.