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If we let $\alpha(s) = (x(s), y(s))$ be a unit speed curve in the plane, how can we define, $(t, n, k)$ where $k$ is the signed curvature?

ii) Also, I need to prove that $t' = kn$, and $n = -kt$.

I note that frenet serret says exactly this but then this leads into circular reasoning by just saying what I need to prove.

iii) and how can I show that $k = \theta'$, where $\theta$ = angle between $t$ and the positive $x$ axis?

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up vote 2 down vote accepted

i) The unit tangent is defined by $t(s) = \alpha'(s)$ since the curve is parameterized by arc-length. In general one defines $k$ and $n$ via $$ \alpha''(s) = k(s) n(s) $$ where $ n(s) = \dfrac{\alpha''(s) }{|\alpha''(s)|}$ and $ k(s) = |\alpha''(s)|.$ However in the plane it is often more useful to have a notion of signed curvature, which allows the curvature to be positive or negative, depending on whether the curve is locally going clockwise or anti-clockwise. At a point $\alpha(s)$ there are only two possible unit normals and we let $n_0(s)$ be the unit normal such that the basis $(t,n_0)$ of $\mathbb{R}^2$ has the same orientation as the standard basis. You can think of it as picking the normal the you obtain by turning the tangent vector $\pi/2$ anti-clockwise. Now that $n_0(s)$ is defined, we define the signed curvature to be such that $\alpha''(s) = k_0(s) n_0(s).$ From now on I will drop the subscript zeros and $k$ will always denote signed curvature.

ii) Since $t(s) = \alpha'(s)$, $t'(s) = \alpha''(s)$ so $t'(s) = k(s) n(s)$ by how $k$ and $n$ are defined. For the next one, I believe you made a typo, forgetting the derivative and actually meant $n'(s) = -k(s) t(s).$ To see why that should be true, note that since $t$ and $n$ form a basis of $\mathbb{R}^2$ we can write $n' = At + B n$ for some scalar functions $A$ and $B.$ Since $n$ is a unit normal and thus has constant length $1$, $|n|^2 = n \cdot n =1.$ By the product rule $(u\cdot v)' = u'\cdot v + u \cdot v'$ we must have $n\cdot n'=0.$ So taking dot products with $n$ gives $ 0 = n\cdot n' = A t\cdot n + B n\cdot n= B$ since $t$ is orthogonal to $n.$ Similarly, note that differentiating $t\cdot n=0$ gives $t'\cdot n + t\cdot n'=0$ so $t\cdot n' = -t'\cdot n = -kn \cdot n = -k$ and thus, taking dot products with $n$ gives $ -k = t\cdot n' = A t\cdot t = A$ and thus, $n' = -kt.$

iii) Let $\theta(s)$ be a continuous function that describes the angle the tangent $t(s)$ makes with the horizontal line through $\alpha(s).$ Thus $\alpha'(s) = t(s) = (\cos \theta(s), \sin \theta(s)).$ The normal is chosen to be an angle of $\pi/2$ anti-clockwise from the tangent so $n(s) = (\cos (\theta(s) + \pi/2), \sin( \theta(s) + \pi/2 ) )= (-\sin \theta(s), \cos \theta(s) ).$ Now, $$kn = \alpha''(s) = \frac{d\alpha'(s)}{d\theta} \frac{d\theta}{ds} = \frac{d\theta}{ds} (-\sin \theta(s), \cos \theta(s) ) $$ so $k=\dfrac{d\theta}{ds}.$

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