Second method for finding center of curvature

Here we shall make use of the definition of circle of curvature given in §12.6. Draw a figure showing the tangent line, circle of curvature, radius of curvature, and center of curvature $ C=(\alpha, \beta)$ corresponding to the point $ P=(x,y)$ on the curve. For example, in Figure 14.1, replace $ P_2$ by $ P$, replace $ (\alpha',\beta')$ by $ (\alpha,\beta)$, and imagine the tangent line to the curve drawn at $ P$. Call the origin in the plane $ O$, the projection of $ P$ to the $ x$-axis $ D$, the projection of $ C$ to the $ x$-axis $ A$, and call $ B$ the projection of $ P$ onto the segment $ CA$. This is depicted in Figure 14.2.

Figure 14.2: Circle of curvature.
\includegraphics[height=5cm,width=6cm]{circle-curvature4.eps}

Then

$\displaystyle \alpha = OA = OD - AD = OD - BP = x - BP,
$

$\displaystyle \beta = AC = AB + BC = DP + BC = y + BC.
$

But $ BP = R\sin \tau$, $ BC = R\cos \tau$. Hence

$\displaystyle \alpha = x -R\sin \tau,\ \ \ \ \beta = y + R\cos \tau.$ (14.8)

From (9.8) [§9.4], and (12.5) [12.5],

$\displaystyle \sin \tau
= \frac{ \frac{dy}{dx} }{ \left[ 1
+ \left( \frac{dy}{dx} \right)^2 \right]^{\frac{1}{2}} },
$

$\displaystyle \cos \tau
= \frac{1}{\left[ 1 +
\left( \frac{dy}{dx} \right)^2 \right]^{\frac{1}{2}}},
$

$\displaystyle R = \frac{ \left[ 1
+ \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} }.
$

Substituting these back in (14.8), we get

$\displaystyle \alpha = x - \frac{ \frac{dy}{dx}\left[ 1 + \left( \frac{dy}{dx} ...
... \ \ \beta = y + \frac{1 + \left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{dx^2}}.$ (14.9)

From Lemma 8.8.18.8], we know that at a point of inflection (as Q in Figure 14.3)

$\displaystyle \frac{d^2 y}{dx^2} = 0.
$

Figure 14.3: Geometric visualization of a change in direction.
\includegraphics[height=4.5cm,width=4cm]{circle-of-curvature2.eps}

Therefore, by (12.3) [§12.4], the curvature $ K = 0$; and from (12.5) [§12.5], and (14.9) [§14.2], we see that in general $ \alpha$, $ \beta$, $ R$ increase without limit as the second derivative approaches zero. That is, if we suppose $ P$ with its tangent to move along the curve to $ P'$, at the point of inflection $ Q$ the curvature is zero, the rotation of the tangent is momentarily arrested, and as the direction of rotation changes, the center of curvature moves out indefinitely and the radius of curvature becomes infinite.

Example 14.2.1   Find the coordinates of the center of curvature of the parabola $ y^2 = 4px$ corresponding (a) to any point on the curve; (b) to the vertex.

Solution. $ \frac{dy}{dx} =\ \frac{2p}{y}$; $ \frac{d^2 y}{dx^2} = -\frac{4p^2}{y^3}$.

(a) Substituting in (14.7) [§14.1],

$\displaystyle \alpha = x + \frac{y^2 + 4 p^2}{y^2} \cdot \frac{2p}{y}
\cdot \frac{y^3}{4 p^2} = 3x + 2p.
$

$\displaystyle \beta = y - \frac{y^2 + 4 p^2}{y^2} \cdot \frac{y^3}{4 p^2}
= -\frac{y^3}{4 p ^2}.
$

Therefore $ \left( 3x + 2p, -\frac{y^3}{4 p^2} \right)$ is the center of curvature corresponding to any point on the curve.

(b) $ (2p,0)$ is the center of curvature corresponding to the vertex $ (0, 0)$.

david joyner 2008-11-22