Computing curvature of a regular curve from definition

differential-geometrylinear algebra

Let $\gamma:I\to\mathbb{R}^n$ be a regular curve. We define the curvature function $\kappa:I\to[0,\infty)$ to be the function $$\kappa(t)=\frac{|a^{\perp}(t)|}{|v(t)|^2}$$ where $a^{\perp}$ denotes the perpendicular component of the acceleration of $\gamma$ and $v$ is velocity of $\gamma$.

My particular problem is to compute the curvature of the helix $\gamma(t)=(\cos(t),\sin(t),t)$ via the definition provided above. In other words, I cannot reparametrize by arc length. I understand $v(t)=(-\sin(t),\cos(t),1)$ and hence $|v(t)|^2=2$. But what I'm stuck on is finding the perpendicular component of $a(t)=(-\cos(t),-\sin(t),0)$. Do we project $a$ onto $\gamma$ and subtract the resulting vector out of $a$? Or is there a simpler way to compute this?

Best Answer

I will answer your last question in the comments with a little more detail. The standard definition of $\kappa$ comes from an arclength parametrization, so that the velocity vector always has length $1$. A vector function of constant length can change only perpendicular to the vector: If $f(t)\cdot f(t)=\text{constant}$, then the product rule tells us that $2f(t)\cdot f'(t)=0$. So the usual definition of curvature is as the magnitude of the derivative of the unit tangent vector with respect to arclength: $dT/ds = \kappa N$, where here $T$ and $N$ are the unit tangent and principal normal, respectively.

If you have a general regular parametrized curve, then $v(t) = \gamma'(t) = \sigma(t)T(t)$, where $\sigma(t)=|v(t)|=ds/dt$ is the speed and $T$ is the unit tangent. Differentiating, we get $$a(t) = \gamma''(t) = \sigma'(t)T(t) + \sigma(t) T'(t) = \sigma'(t)T(t) + \sigma^2(t)\kappa(t)N(t).$$ Now you see that $\sigma^2(t)\kappa(t)N(t)$ is the component of $a(t)$ orthogonal to $T(t)$, i.e., orthogonal to the velocity vector.

The velocity and the acceleration do not change if we translate our origin by a fixed vector, since differentiating $\gamma(t)+c$ is the same as differentiating $\gamma(t)$. But this shows you that the position vector of the curve is not really relevant to understanding its geometry.

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