What is radius of curvature? — Mathematics

বইগুলো এখনো বাংলা ভাষায় পাওয়া যায় না — ইংরেজি সংস্করণ দেখানো হচ্ছে।

Definition 18.14 University Mathematics — Year 2 · অধ্যায় 18 — Curves

When $\kappa(s) \neq 0$ , the radius of curvature is $R(s) = 1/\abs{\kappa(s)}$ and the center of curvature is $\tilde\gamma(s) + \frac{1}{\kappa(s)} N(s)$ ; the circle with that center and radius $R(s)$ is the osculating circle, the best circular approximation of the curve at $\tilde\gamma(s)$ .

The parabola y = x2/2, its moving Frenet frame (T, N), and the osculating circle at the vertex (radius 1, since (0) = 1). The frame turns as the point moves; curvature is the rate of that turning per unit of arc length. — **Figure 18.1.** The parabola $y = x^2/2$ , its moving Frenet frame $(T, N)$ , and the osculating circle at the vertex (radius $1$ , since $\kappa(0) = 1$ ). The frame turns as the point moves; curvature is the rate of that turning per unit of arc length.

উদাহরণ

Example 18.15 (The osculating circle of the exponential)

For $y = \eu^x$ at the point $(0, 1)$ : $f'(0) = f''(0) = 1$ , so by the graph formula below,

\kappa(0) = \frac{1}{(1 + 1)^{3/2}} = \frac1{2\sqrt2}, \qquad R = 2\sqrt2 .

The unit tangent is $T = \frac{(1, 1)}{\sqrt2}$ , the direct normal $N = \frac{(-1, 1)}{\sqrt2}$ , and the center of curvature is

(0, 1) + 2\sqrt2\cdot\frac{(-1, 1)}{\sqrt2} = (-2,\ 3) :

the osculating circle has equation $(x + 2)^2 + (y - 3)^2 = 8$ . As a check of the “best circular approximation” claim: solving the circle’s equation for $y$ near $(0,1)$ and expanding gives $y = 1 + x + \frac{x^2}2 + O(x^3)$ — exactly the second-order Taylor expansion of $\eu^x$ . The osculating circle matches value, slope and second derivative; an ordinary tangent circle would match only the first two.

Example 18.16 (The evolute of a circle is its center)

For the circle of radius $R$ traversed counterclockwise, $\kappa = 1/R$ and $N$ points toward the center, so the center of curvature $\tilde\gamma + \frac1\kappa N$ is the center of the circle, for every $s$ : the osculating circle of a circle is the circle itself, and the locus of centers of curvature collapses to a point. This degenerate case calibrates Exercise 18.6: there the evolute’s velocity is $-\frac{\kappa'}{\kappa^2}N$ , which vanishes identically precisely when $\kappa$ is constant.

Example 18.18 (Circle, line, parabola)

A line has $\kappa = 0$ (and conversely: $T' = 0$ means $T$ constant, so $\tilde\gamma(s) = \tilde\gamma(0) + sT$ , a line). The circle of radius $R$ traversed counterclockwise has $\kappa = 1/R$ : with $\gamma(t) = (R\cos t, R\sin t)$ , the formula gives $\kappa = R^2/R^3$ . For the parabola $y = x^2/2$ : $\kappa(x) = 1/(1 + x^2)^{3/2}$ , maximal at the vertex — the parabola is most sharply bent where it turns around.

প্রসঙ্গসহ পড়ো →