convexity-from-osculating-circle

Convexity from the Osculating Circle

Summary: §§311–315. The bare formula $R=(A^2+B^2{)}^{3/2}/(2(A^{2}E-ABD+B^{2}C))$ for the osculating radius involves the radical $\sqrt{A^2+B^2}$ whose sign is ambiguous — the formula does not by itself say whether the curve is concave or convex with respect to the chosen normal direction $N$. Euler resolves the ambiguity by asking whether the next point $m$ on the curve lies on the same side of the tangent $M\mu$ as the perpendicular foot from $N$, or on the opposite side. The decisive sign is that of $(A^{2}E-ABD+B^{2}C)/B$. Two special configurations — $B=0$ (figure 57, ordinate is the tangent) and $A=0$ (figure 58, tangent parallel to the axis) — collapse the test to simpler ratios.

Sources: chapter14 (§§311–315). Figures 55, 57, 58, 59 (in figures55-60).

Last updated: 2026-05-11.

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The sign ambiguity (§311)

The radius of curvature $R=(A^2+B^2)\sqrt{A^2+B^2}/(2(A^{2}E-ABD+B^{2}C))$ contains $\sqrt{A^2+B^2}$, conventionally positive but appearing inside an expression with no fixed sign. The geometric question is whether the curve $Mm$ lies on the same side of the tangent line $M\mu$ as the axis $AN$ (in which case the curve is concave toward $N$, with the center of the osculating circle on segment $MN$), or on the opposite side (in which case the curve is convex with respect to $N$, with the center on the extension of $NM$ beyond $M$).

The test (§312)

Recall that the tangent gives $q\mu =-At/B$ at abscissa $t=Mq$ on the new axis, while the actual ordinate is $qm=u$. Compare:

$qm\lessgtr q\mu ⟺u\lessgtr -\frac{At}{B}.$

Write $u=-At/B-w$ for a tiny correction $w=m\mu$. Substituting into

$0=At+Bu+C{t}^2+Dtu+E{u}^2+\cdots$

and dropping terms in $tw$ and $w^2$ as second-order minute,

$0=-Bw+C{t}^2-\frac{AD{t}^2}{B}+\frac{A^{2}E{t}^2}{B^2}+\cdots ,$

so

$w=\frac{(B^{2}C-ABD+A^{2}E)t^2}{B^3}=\frac{(A^{2}E-ABD+B^{2}C)t^2}{B^3}.$

The sign of $w$ — that is, whether $m$ lies on the $N$-side or the opposite side of the tangent — is the sign of

$\frac{A^{2}E-ABD+B^{2}C}{B}(\text{using that }{B}^2{t}^2>0).$

Rule (§312):

• If $(A^{2}E-ABD+B^{2}C)/B>0$, then $w>0$, and the curve is concave with respect to $N$.
• If $(A^{2}E-ABD+B^{2}C)/B<0$, the curve is convex with respect to $N$.

(One must distinguish $B$ in the denominator from $B^2$ in the radius formula — the former carries a sign; the latter does not.)

Special case I: $B=0$ — ordinate is tangent (§313, figure 57)

When $B=0$, the tangent line at $M$ is the ordinate $PM$ itself. The local equation reduces to $0=At+C{t}^2+Dtu+E{u}^2$ and the radius simplifies to

$R=\frac{A}{2E}.$

To decide concavity, since $Mq=t$ and $qm=u$ with $t$ infinitely smaller than $u$, the terms $t^2$ and $tu$ vanish next to $u^2$, leaving $At+E{u}^2=0$. The sign of $E/A$ tells the story:

• $E/A<0$: curve is concave with respect to $R$;
• $E/A>0$: $t$ would have to be negative for $u$ real, so the curve lies on the other side of the tangent line.

Special case II: tangent inclined (§314, figures 55 / 59)

For the ordinary configuration of figure 55 — tangent $M\mu$ inclined to the axis $AP$ at an acute angle $RM\mu$, with the normal $MN$ meeting the axis beyond $P$ — the abscissa $t>0$ corresponds to a positive ordinate $u>0$. Then $A$ and $B$ have opposite signs and $A/B<0$, equivalently $B/A<0$. The general rule then specializes to:

• Curve concave w.r.t. $N$ iff $(A^{2}E-ABD+B^{2}C)/A<0$, i.e., $(A^{2}E-ABD+B^{2}C)$ and $A$ have opposite signs.

(Figure 59 covers the related case where the tangent intersects the axis on the far side of $P$; the same general formula applies.)

Special case III: $A=0$ — tangent parallel to axis (§315, figure 58)

When $A=0$, the tangent line $MR$ is parallel to the axis $AP$ (and to the curve at $M$). The local equation reduces to $0=Bu+C{t}^2+\cdots$ and, since $u$ is infinitely smaller than $t$, the leading non-vanishing balance is $0=Bu+C{t}^2$. The radius simplifies to

$R=\frac{B}{2C}.$

Concavity rule: if $B$ and $C$ have the same sign ($BC>0$) then $u$ is negative, so the curve sits on the opposite side of the tangent from the positive ordinate direction — that is, the curve is concave with respect to $P$ (which here coincides with $N$). The general rule via the sign of $(A^{2}E-ABD+B^{2}C)/B=B^{2}C/B=BC$ recovers the same conclusion.

Summary table

Configuration	Tangent	Radius	Concave toward $N$ iff
$A=0$	parallel to axis (figure 58)	$B/(2C)$	$BC>0$
$B=0$	ordinate $PM$ (figure 57)	$A/(2E)$	$E/A<0$
general (figure 55, 59)	inclined	$(A^2+B^2{)}^{3/2}/(2(A^{2}E-ABD+B^{2}C))$	sign of $(A^{2}E-ABD+B^{2}C)/B$

Figures

Figures 55–60

Related pages

• chapter-14-on-the-curvature-of-a-curve — chapter summary.
• osculating-circle — the unsigned formula.
• osculating-parabola — derivation that produces the same denominator.
• tangent-by-translation — origin of the special-case configurations $A=0$ and $B=0$.
• inflection-by-vanishing-curvature — case where $A^{2}E-ABD+B^{2}C=0$ and the test is inconclusive at second order.