Difference between revisions of "Osculating circle"
From Encyclopedia of Mathematics
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| − | ''at a given point | + | {{TEX|done}} |
| + | ''at a given point $M$ of a curve $l$'' | ||
| − | The circle that has contact of order | + | The circle that has contact of order $n\geq2$ with $l$ at $M$ (see [[Osculation]]). If the curvature of $l$ at $M$ is zero, then the osculating circle degenerates into a straight line. The radius of the osculating circle is called the radius of curvature of $l$ at $M$, and its centre the centre of curvature (see Fig.). If $l$ is the plane curve given by an equation $y=f(x)$, then the radius of the osculating circle is given by |
| − | + | $$\rho=\left|\frac{(1+y'^2)^{3/2}}{y''}\right|.$$ | |
<img style="border:1px solid;" src="https://www.encyclopediaofmath.org/legacyimages/common_img/o070540a.gif" /> | <img style="border:1px solid;" src="https://www.encyclopediaofmath.org/legacyimages/common_img/o070540a.gif" /> | ||
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Figure: o070540a | Figure: o070540a | ||
| − | If | + | If $l$ is the spatial curve given by equations |
| − | + | $$x=x(u),\quad y=y(u),\quad z=z(u),$$ | |
then the radius of the osculating circle is given by | then the radius of the osculating circle is given by | ||
| − | + | $$\rho=\frac{(x'^2+y'^2+z'^2)^{3/2}}{\sqrt{(y'z''-z'y'')^2+(z'x''-x'z'')^2+(x'y''-y'z'')^2}}$$ | |
| − | (where the primes denote differentiation with respect to | + | (where the primes denote differentiation with respect to $u$). |
| + | ====References==== | ||
| + | <table> | ||
| + | <TR><TD valign="top">[a1]</TD> <TD valign="top"> R.S. Millman, G.D. Parker, "Elements of differential geometry" , Prentice-Hall (1977) pp. 39</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top"> D.J. Struik, "Lectures on classical differential geometry" , Dover, reprint (1988) pp. 14</TD></TR> | ||
| + | </table> | ||
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Latest revision as of 09:43, 15 August 2023
at a given point $M$ of a curve $l$
The circle that has contact of order $n\geq2$ with $l$ at $M$ (see Osculation). If the curvature of $l$ at $M$ is zero, then the osculating circle degenerates into a straight line. The radius of the osculating circle is called the radius of curvature of $l$ at $M$, and its centre the centre of curvature (see Fig.). If $l$ is the plane curve given by an equation $y=f(x)$, then the radius of the osculating circle is given by
$$\rho=\left|\frac{(1+y'^2)^{3/2}}{y''}\right|.$$
Figure: o070540a
If $l$ is the spatial curve given by equations
$$x=x(u),\quad y=y(u),\quad z=z(u),$$
then the radius of the osculating circle is given by
$$\rho=\frac{(x'^2+y'^2+z'^2)^{3/2}}{\sqrt{(y'z''-z'y'')^2+(z'x''-x'z'')^2+(x'y''-y'z'')^2}}$$
(where the primes denote differentiation with respect to $u$).
References
| [a1] | R.S. Millman, G.D. Parker, "Elements of differential geometry" , Prentice-Hall (1977) pp. 39 |
| [a2] | D.J. Struik, "Lectures on classical differential geometry" , Dover, reprint (1988) pp. 14 |
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How to Cite This Entry:
Osculating circle. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Osculating_circle&oldid=14753
Osculating circle. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Osculating_circle&oldid=14753
This article was adapted from an original article by BSE-3 (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article