Multivariable Calculus 13.4 Textbook Notes

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13.4 Curvature and Normal Vectors of a Curve

This section is omitted in the department syllabus, but it’s worthwhile for our honors sections to do a brief discussion of at least part of this section. Then central concept here is the curvature of a curve and the principal normal vector of a curve (under some condition). Intuitively, the curvature of a curve at a point measures how fast its “direction angle” is turning per unit length of movement along the curve. Technically, if $T(t)$ is the unit tangent vector of the curve $r(t)$ (defined when $r\prime (t)\ne 0$, namely, for a smooth curve): $T(t)=\frac{r^{\prime }(t)}{||r^{\prime }(t)||}$, then the curvature should be $||\frac{dT}{ds}||=||v(t{)}^{-1}\frac{dT}{dt}||=v(t{)}^{-1}||\frac{d}{dt}\left(\frac{r^{\prime }(t)}{||r^{\prime }(t)||}\right)||$.

What remains is to develop an effective way of computing this quantity. Example 2 demonstrates how to do this kind of computation for a circle of radius $a$. It’s instructive to compute the curvature of the helix curve (see Example 5). Exercises 5, 6 gives formulae for computing the curvature of a plane curve given as a graph or in parametric form.

Reading Questions/Quizzes

1. Compute the curvature of the curve given by $r(t)=〈1-t^2,3+3t^2,t^2〉$ wherever it is defined. Another observation about $T$ is that $T\cdot T=1$ for all $t$ (and $s$), so we get $0=\frac{d}{ds}(T\cdot T)=2\frac{dT}{ds}\cdot T$, which means that the vector $\frac{dT}{ds}$ is always orthogonal to the tangent vector $T$, therefore is a vector normal to the curve. If this vector $\frac{dT}{ds}$ is not zero, then it has a unit vector pointing in its direction: $||\frac{dT}{ds}|{|}^{-1}\frac{dT}{ds}$, which is called the principal normal vector to $r(t)$ at this point. Denote it by $N$, then we have $N=||\frac{dT}{ds}|{|}^{-1}\frac{dT}{ds}$ , from which we deduce $\frac{dT}{ds}=||\frac{dT}{ds}||N=\kappa N$, where $\kappa =\Vert \frac{dT}{ds}\Vert$ is the curvature.
   1. From my 13.3 textbook notes, $T(t)=\frac{<-1,3,1>}{\sqrt{11}}$ when $t>0$ and $T(t)=\frac{<1,-3,-1>}{\sqrt{11}}$ when $t<0$. Since $\frac{dT}{ds}$ can be represented as $\frac{dT}{dt}\cdot \frac{dt}{ds}$ and $\frac{dT}{dt}=0$, the curvature $\kappa$ is 0.