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



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

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.

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),
 * $$\frac{dy}{dx} = -\frac{4}{3}, \frac{d^2 y}{dx^2} = \frac{8}{9}$$
 * style="text-align: right;"|&there4; R
 * $$= \frac{ [1 + \frac{16}{9}]^{\frac{3}{2}} }{ \frac{8}{9} } = \frac{125}{24} = 5\frac{5}{24}$$.
 * }
 * $$= \frac{ [1 + \frac{16}{9}]^{\frac{3}{2}} }{ \frac{8}{9} } = \frac{125}{24} = 5\frac{5}{24}$$.
 * }

The circle of curvature crosses the curve at two points.