Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/185

104. Circle of curvature. Consider any point P on the curve C. The tangent drawn to the curve at P has the same slope as the curve itself at P (§ 64). In an analogous manner we may construct for each point of the curve a circle whose curvature is the same as the curvature of the curve itself at that point. To do this, proceed as follows. Draw the normal to the curve at P on the concave side of the curve. Lay off on this normal the distance PC = radius of curvature (= R) at P. With C as a center draw the circle passing through P. The curvature of this circle is then

$K={\frac {1}{R}},$

which also equals the curvature of the curve itself at P. The circle so constructed is called the circle of curvature for the point P on the curve.

In general, the circle of curvature of a curve at a point will cross the curve at that point. This is illustrated in the above figure.

Just as the tangent at P shows the direction of the curve at P, so the circle of curvature at P aids us very materially in forming a geometric concept of the curvature of the curve at P, the rate of change of direction of the curve and of the circle being the same at P.

In a subsequent section (§ 116) the circle of curvature will be defined as the limiting position of a secant circle, a definition analogous to that of the tangent given in § 32.

Illustrative Example 1. Find the radius of curvature at the point (3, 4) on the equilateral hyperbola $xy=12$, and draw the corresponding circle of curvature.

Solution.	${\frac {dy}{dx}}=-{\frac {y}{x}},{\frac {d^{2}y}{dx^{2}}}={\frac {2y}{x^{2}}},$
For (3, 4),	${\frac {dy}{dx}}=-{\frac {4}{3}},{\frac {d^{2}y}{dx^{2}}}={\frac {8}{9}}$
∴ R	$={\frac {[1+{\frac {16}{9}}]^{\frac {3}{2}}}{\frac {8}{9}}}={\frac {125}{24}}=5{\frac {5}{24}}$.

The circle of curvature crosses the curve at two points.