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Topos Uranos
English

Curvature and radius of curvature

Summary

This lesson measures how much the graph of a function bends at a point. The tangent is the line that has, at the point, the same value and the same derivative as the function; the osculating circle is the one that has, in addition, the same second derivative. Differentiating the equation of a circle twice shows that, at each of its points, a combination of the first and the second derivative equals exactly its radius; with this, the lesson proves that, if the second derivative of the function does not vanish, the osculating circle exists and is unique, and gives its center and its radius, the radius of curvature. Its reciprocal is the curvature, which vanishes where the second derivative vanishes. It then proves that the center of curvature is the limit of the points where nearby normals meet, as Huygens saw, and that the curvature does not change when the axes are swapped, which makes it possible to measure it where the tangent is vertical. The problems decide whether the osculating circle crosses the graph, find the point of maximum curvature of the exponential, the points of maximum and minimum curvature of an ellipse and the radius of curvature of the catenary, and study the jump in curvature between a line and an arc.

Learning objectives

  1. Define the osculating circle to the graph of a function at a point, prove that it exists and is unique when the second derivative does not vanish, and compute its center and its radius.
  2. Compute the curvature and the radius of curvature of a graph at a point, interpret zero curvature and find the points of maximum or minimum curvature of a graph.
  3. Obtain the center of curvature as the limit of the intersections of the normals, and measure the curvature where the tangent is vertical by swapping the roles of the axes.
  4. Decide, with the higher-order derivative test, whether the osculating circle crosses the graph or stays on one side of it.

The osculating circle

The tangent to the graph of ff at (a,f(a))(a, f(a)) is the line that has at aa the same value and the same derivative as ff. The lesson on concavity, convexity and inflection points showed that, if f′′(a)≠0f''(a) \neq 0, the graph lies near aa on one side only of its tangent: above it if f′′(a)>0f''(a) > 0 and below it if f′′(a)<0f''(a) < 0. But it did not say how far the graph moves away from it, that is, how much it bends. To measure this, it is convenient to compare it with the curves that always bend by the same amount, the circles: a small one bends a lot, and a large one, little. A circle is not the graph of a function, because each vertical line through its interior meets it twice; but each of its two halves, without their endpoints, is.

LemmaThe derivatives of a semicircle

Let (h,k)(h, k) be a point of the plane, r>0r > 0, σ\sigma equal to 11 or to −1-1, and u(x)=k+σr2−(x−h)2u(x) = k + \sigma \sqrt{r^2 - (x - h)^2} for xx in (h−r,h+r)\ioo{h - r}{h + r}, whose graph is the upper semicircle if σ=1\sigma = 1, or the lower one if σ=−1\sigma = -1, with center (h,k)(h, k) and radius rr, without its endpoints. Then uu, u′u' and u′′u'' are differentiable on (h−r,h+r)\ioo{h - r}{h + r}, u(x)≠ku(x) \neq k, and at each point xx of the interval:

(x−h)+(u(x)−k)u′(x)=0,1+(u′(x))2+(u(x)−k)u′′(x)=0(x - h) + \left(u(x) - k\right) u'(x) = 0, \qquad 1 + \left(u'(x)\right)^2 + \left(u(x) - k\right) u''(x) = 0

Consequently, u′′(x)u''(x) does not vanish and has the sign of −σ-\sigma, so the lower semicircle is strictly convex and the upper one strictly concave, and at all its points:

(1+(u′(x))2)32∣u′′(x)∣=r\frac{\left(1 + \left(u'(x)\right)^2\right)^{\frac{3}{2}}}{\abs{u''(x)}} = r

Proof

  1. (x−h)2+(u(x)−k)2=r2,x∈(h−r,h+r)(x - h)^2 + \left(u(x) - k\right)^2 = \dato{circ}{r^2}, \qquad x \in \ioo{h - r}{h + r}

    If xx is in the interval, r2−(x−h)2>0r^2 - (x - h)^2 > 0, so u(x)−k=σr2−(x−h)2u(x) - k = \sigma \sqrt{r^2 - (x - h)^2} does not vanish, and uu is differentiable by the corollary on derivatives of standard compositions, from the lesson on the chain rule. Squaring, the graph of uu lies on the circle:

  2. 2(x−h)+2(u(x)−k)u′(x)=0,1+(u′(x))2+(u(x)−k)u′′(x)=02(x - h) + 2\left(u(x) - k\right) u'(x) = 0, \qquad 1 + \left(u'(x)\right)^2 + \left(u(x) - k\right) u''(x) = \dato{dos}{0}

    The two sides are the same function of xx on the interval, and their derivatives coincide: this is the implicit differentiation of the lesson on the derivative of the inverse function and implicit differentiation, applied to a function we already know to be differentiable. Solving, u′(x)=−x−hu(x)−ku'(x) = -\frac{x - h}{u(x) - k} is a quotient of differentiable functions with nonzero denominator, differentiable by the quotient rule, from the lesson on the differentiation rules; differentiating again gives the second equality, from which u′′(x)=−1+(u′(x))2u(x)−ku''(x) = -\frac{1 + \left(u'(x)\right)^2}{u(x) - k} is also a quotient of differentiable functions.

  3. 1+(u′(x))2=r2(u(x)−k)2,∣u′′(x)∣=r2∣u(x)−k∣3→(1+(u′(x))2)32∣u′′(x)∣=r1 + \left(u'(x)\right)^2 = \frac{r^2}{\left(u(x) - k\right)^2}, \qquad \abs{u''(x)} = \frac{r^2}{\abs{u(x) - k}^3} \rightarrow \resaltar{\frac{\left(1 + \left(u'(x)\right)^2\right)^{\frac{3}{2}}}{\abs{u''(x)}} = r}

    Since 1+(u′(x))2>01 + \left(u'(x)\right)^2 > 0, the second derivative does not vanish and has the sign opposite to that of u(x)−ku(x) - k, which is that of σ\sigma; the corollary on convexity and second derivative gives strict convexity or strict concavity. By the first equality, 1+(u′(x))21 + \left(u'(x)\right)^2 is the sum of (x−h)2(x - h)^2 and (u(x)−k)2\left(u(x) - k\right)^2, which equals r2r^2, divided by (u(x)−k)2\left(u(x) - k\right)^2.

On a circle, the slope of the tangent and the second derivative change from one point to another, but the combination in the lemma does not change: it always equals the radius. This suggests using it to measure how much any graph bends. First, it is convenient to find the circle that best fits the graph at a point, the one that agrees with it up to the second derivative.

DefinitionOsculating circle

Let ff be a function differentiable on an open interval II and twice differentiable at a point aa of II. A circle with center (h,k)(h, k) and radius r>0r > 0 is osculating to the graph of ff at (a,f(a))(a, f(a)) if it passes through that point without the point being its leftmost or its rightmost one, that is, if (a−h)2+(f(a)−k)2=r2(a - h)^2 + \left(f(a) - k\right)^2 = r^2 and f(a)≠kf(a) \neq k, and if the semicircle that contains the point, the graph of the function uu of the lemma with σ\sigma equal to the sign of f(a)−kf(a) - k, has at aa the same value, the same derivative and the same second derivative as ff:

u(a)=f(a),u′(a)=f′(a),u′′(a)=f′′(a)u(a) = f(a), \qquad u'(a) = f'(a), \qquad u''(a) = f''(a)

Its center is called the center of curvature of the graph at that point.

The tangent line satisfies the first two equalities; the osculating circle, the third as well. The name comes from the Latin osculari, to kiss: the circle does not merely touch the graph, like the other tangent circles, but bends as the graph does. The figure shows, with f(x)=x32−x+1f(x) = \frac{x^3}{2} - x + 1 and a=1.2a = 1.2, three circles tangent to the graph and, among them, the osculating one.

TheoremExistence and uniqueness of the osculating circle

Let ff be a function differentiable on an open interval II and twice differentiable at a point aa of II, and let m=f′(a)m = f'(a) and q=f′′(a)q = f''(a). If q≠0q \neq 0, the graph of ff has at (a,f(a))(a, f(a)) exactly one osculating circle, with center (h,k)(h, k) and radius ρ\rho:

h=a−m(1+m2)q,k=f(a)+1+m2q,ρ=(1+m2)32∣q∣h = a - \frac{m\left(1 + m^2\right)}{q}, \qquad k = f(a) + \frac{1 + m^2}{q}, \qquad \rho = \frac{\left(1 + m^2\right)^{\frac{3}{2}}}{\abs{q}}

Its center lies on the normal line to the graph at the point, above the point if q>0q > 0 and below it if q<0q < 0. If q=0q = 0, the graph has no osculating circle at (a,f(a))(a, f(a)).

Proof

  1. (a−h)+(f(a)−k)m=0(a - h) + \left(f(a) - k\right) m = \dato{normal}{0}

    Suppose that a circle with center (h,k)(h, k) and radius rr is osculating, with its semicircle uu. The first equality of the lemma, at x=ax = a, with u(a)=f(a)u(a) = f(a) and u′(a)=mu'(a) = m, says that the center satisfies the equation of the normal line to the graph at the point, according to the definition of tangent line and normal line in the lesson on the derivative as a limit, both when m≠0m \neq 0 and when the normal is vertical.

  2. 1+m2+(f(a)−k)q=0→f(a)−k=−1+m2q,a−h=m(1+m2)q,r2=(1+m2)3q21 + m^2 + \left(f(a) - k\right) q = 0 \rightarrow \resaltar{f(a) - k = -\frac{1 + m^2}{q}}, \qquad a - h = \frac{m\left(1 + m^2\right)}{q}, \qquad r^2 = \frac{\left(1 + m^2\right)^3}{q^2}

    The second equality of the lemma, at x=ax = a, with u′′(a)=qu''(a) = q, is impossible if q=0q = 0, because 1+m2>01 + m^2 > 0: there is no osculating circle. If q≠0q \neq 0, it determines f(a)−kf(a) - k; the first equality then gives a−ha - h, and the equation of the circle, the radius. Therefore, there is at most one osculating circle, and its center and its radius are those of the statement.

  3. u′(a)=−a−hf(a)−k=m,u′′(a)=−1+m2f(a)−k=qu'(a) = -\frac{a - h}{f(a) - k} = \resaltar{m}, \qquad u''(a) = -\frac{1 + m^2}{f(a) - k} = \resaltar{q}

    Conversely, let hh, kk and ρ\rho be those of the statement. The circle with center (h,k)(h, k) and radius ρ\rho passes through (a,f(a))(a, f(a)), by the calculation above, and f(a)−kf(a) - k does not vanish and has the sign of −q-q, so σ=−1\sigma = -1 if q>0q > 0, and the center lies above the point, and σ=1\sigma = 1 if q<0q < 0. The function uu of the lemma satisfies u(a)=k+σ∣f(a)−k∣=f(a)u(a) = k + \sigma\abs{f(a) - k} = f(a), and the two equalities of the lemma, at x=ax = a, give its derivatives: the circle is osculating.

If q>0q > 0, the osculating circle rests on the graph from above along its lower semicircle, which is convex; if q<0q < 0, along the upper one, which is concave. In both cases, the proposition on the graph and its tangent, from the lesson on concavity and convexity, with n=2n = 2, applied to ff and to uu, which have the same tangent and second derivatives of the same sign, places the graph and the circle, near aa, on the same side of that tangent. If f′′(a)=0f''(a) = 0, no circle will do, because the second derivative of a semicircle never vanishes; the tangent line, whose second derivative is zero, satisfies the three equalities, and plays the role of a circle of infinite radius.