Example 1. Calculate the curvature of the helical line

$$
\mathbf{r}=a \cos t \mathbf{i}+a \sin t \mathbf{j}+h t \mathbf{k}
$$
Solution. Since

$$
\begin{aligned}
\frac{d \mathbf{r}}{d t} & =-a \sin t \mathbf{i} + a \cos t \mathbf{j} + h \mathbf{k} \\
\frac{d^{2} \mathbf{r}}{d t^{2}} & =-a \cos t \mathbf{i} - a \sin t \mathbf{j}
\end{aligned}
$$

then the vector product

$$
\left[\frac{d \mathbf{r}}{d t}, \frac{d^{2} \mathbf{r}}{d t^{2}}\right]=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-a \sin t & a \cos t & h \\
-a \cos t & -a \sin t & 0
\end{array}\right|=a h \sin t \mathbf{i} - a h \cos t \mathbf{j} + a^{2} \mathbf{k}
$$

Therefore,

$$
\left|\left[\frac{d \mathbf{r}}{d t}, \frac{d^{2} \mathbf{r}}{d t^{2}}\right]\right|=a \sqrt{a^{2}+h^{2}}, \quad\left|\frac{d \mathbf{r}}{d t}\right|=\sqrt{a^{2}+h^{2}}
$$

By formula (3),

$$
\begin{gathered}
K=\frac{1}{R}=\frac{a}{a^{2}+h^{2}} \\
R=\frac{a^{2}+h^{2}}{a}=\text { const. }
\end{gathered}
$$

or

Thus, the helical line has a constant radius of curvature.

## Problems for Independent Solution

Find the radius of curvature of the given lines:

36. $\mathbf{r}=\ln \cos t \mathbf{i} + \ln \sin t \mathbf{j} + \sqrt{2} t \mathbf{k}$
37. $\mathbf{r}=t^{2} \mathbf{i} + 2 t^{3} \mathbf{j}$.

38. $\mathbf{r}=3 t^{2} \mathbf{i} + (3 t - t^{3}) \mathbf{j} + 2 \mathbf{k}$ at $t=1$.

39. $\mathbf{r}=a(\cos t + t \sin t) \mathbf{i} + a(\sin t - t \cos t) \mathbf{j}$ at $t=\frac{\pi}{2}$.
40. $\mathbf{r}=a \operatorname{ch} t \mathbf{i} + a \operatorname{sh} t \mathbf{j} + a t \mathbf{k}$ at any point $t$.

## §6. Osculating Plane. Binormal. Torsion. Frenet Formulas

A plane passing through the tangent line and the principal normal to a given curve $L$ at a point $M$ is called the osculating plane at point $M$.

For a plane curve, the osculating plane coincides with the plane of the curve.

If the vector $\mathbf{r}=\mathbf{r}(t)$ has a continuous derivative $\frac{d \mathbf{r}}{d t}$ in the neighborhood of the point $t_{0}$ and, in addition, a second derivative $\frac{d^{2} \mathbf{r}\left(t_{0}\right)}{d t^{2}}$ such that

$$
\left[\frac{d \mathbf{r}\left(t_{0}\right)}{d t}, \frac{d^{2} \mathbf{r}\left(t_{0}\right)}{d t^{2}}\right] \neq 0
$$

then at the point $t=t_{0}$ there exists an osculating plane to the curve $\mathbf{r}=\mathbf{r}(t)$, the vector equation of which is

$$
\left(\boldsymbol{p}-\mathbf{r}\left(t_{0}\right),\left[\frac{d \mathbf{r}\left(t_{0}\right)}{d t}, \frac{d^{2} \mathbf{r}\left(t_{0}\right)}{d t^{2}}\right]\right)=0
$$

where $\rho=\rho(t)$ is the radius-vector of the current point of the plane.

The normal to the curve at point $M$, perpendicular to the osculating plane of the curve at this point, is called the binormal of the curve at the given point $M$.

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-10.jpg?height=225&width=252&top_left_y=608&top_left_x=106)

Fig. 10
Let $b^{0}$ denote the unit vector of the binormal, oriented such that the vectors $\tau^{0}, \mathbf{n}^{0}, \mathbf{b}^{0}$ form a right-handed triplet (Fig. 10). Then

$$
\mathbf{b}^{0 \cdot 2}=1, \quad \mathbf{b}^{0}=\left[\tau^{0}, \mathbf{n}^{0}\right]
$$

For the derivative $\frac{d b^{0}}{d s}$ we get

$$
\frac{d b^{0}}{d s}=\left[\tau^{0}, \frac{d n^{0}}{d s}\right]
$$

The vector $\frac{d b^{0}}{d s}$ is perpendicular to both the vector $\tau^{0}$ and the vector $\mathbf{b}^{0}$, i.e., it is collinear with the vector $\mathbf{n}^{0}$. Let

$$
\left|\frac{d b^{0}}{d s}\right|=\frac{1}{T}
$$

then we will have

$$
\frac{d \mathbf{b}^{0}}{d s}=\frac{1}{T} \mathbf{n}^{0}
$$

The quantity $\frac{1}{T}$ is called the torsion of the curve, and the quantity $T$ is called the radius of torsion of the curve.

The torsion of the curve is given by the formula

$$
\frac{1}{T}=R^{2}\left(\frac{d \mathbf{r}}{d s}, \frac{d^{2} \mathbf{r}}{d s^{2}}, \frac{d^{3} \mathbf{r}}{d s^{3}}\right)
$$

where the symbol $(a, b, \mathbf{c})$ denotes the mixed product of the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, i.e., $(a, b, c)=(a,[b, c])$.

In the case when the curve is given by a vector-parametric equation $\mathbf{r}=\mathbf{r}(t)$, we have

$$
\frac{1}{T}=\frac{\left(\frac{d \mathbf{r}}{d t}, \frac{d^{2} \mathbf{r}}{d t^{2}}, \frac{d^{3} \mathbf{r}}{d t^{3}}\right)}{\left|\left[\frac{d \mathbf{r}}{d t}, \frac{d^{2} \mathbf{r}}{d t^{2}}\right]\right|^{2}}
$$