## The Calculus Primer (2011)

### Part VIII. Curvature

### Chapter 30. MEANING OF CURVATURE

**8—4.Curvature.** By observing their appearance, it is clear that various curves differ from one another in the degree of curvature, or their rate of turning; furthermore, on a given curve, the rate of turning varies at different points along the curve. For example, if we compare the curves, *y* =*x*^{3} and *y* = 4*x*^{3}, the latter curve is seen to “bend around” or curve more rapidly than the former as *x* increases from zero. Or again, the parabola *y*^{2} = 2*x* is more “sharply” curved in the neighborhood of the vertex than elsewhere, and flattens out, or has less curvature, as the curve recedes from the vertex.

**8—5.Definition of Curvature.** To define precisely the amount of curvature, we proceed as follows. Let *P* be a fixed point on the arc *AB*, and let *Q* be a point in the neighborhood of *P,* at a distance Δs from *P*, measured along *AB.* The tangents at *P* and *Q* make angles of *ω* and *ω* + Δ*ω*, respectively, with the *X*-axis; thus the angle between the two tangents equals Δ*ω*. It will be appreciated that the entire change in direction of the curve in passing from *P* to *Q* is measured by Δ*ω*; furthermore, the *average change of direction per unit arc length,* in passing from *P* to *Q*, is given by the ratio .

Now let *Q* approach *P*; then Δ*s* → 0, and Δ*ω* → 0. Therefore, in general,

where is the value of the derivative at *P*.

The curvature at point *P* is thus defined as the absolute value (numerical value, disregarding sign) of the derivative at point *P*; in symbols,

Curvature at *P* = *K* = ,[1]

where the expression denotes the absolute value of the derivative when evaluated for the particular values of *ω* and *s* that correspond to point *P*. Hence in order to find the value of , the functional relation between *ω* and *s* must be known; but, for curves given in rectangular coordinates, such as *y* = *f*(*x*), this is somewhat awkward. The procedure required will be explained shortly in §8—6.

The notion of curvature as just defined above will be seen to be consistent with our intuitive conception of curvature. In other words, at any point on a curve, the curvature is great or small according as there is a great or small *change in the direction* of the curve as we *pass through that point.*

**8—6.Curvature of a Curve in Rectangular Coordinates.** From the figure, §8—5, we see at once that

Differentiating (1) with respect to *x* by [14], §5—15:

Also, from [1], §8—1, we recall:

Hence, dividing (1) by (2), we obtain:

**8—7.Curvature of the Circle.** Let us consider the circle so placed that its center is on the *Y*-axis, and at a distance from the origin equal to its radius *r.* Let the tangent at *P* make an angle *ω* with the *X*-axis, and let *α* be the central angle of the arc *OP* = *s.* Then, by trigonometry,

*s* = *rα.*

But since the angles *α* and *ω* have their corresponding sides perpendicular, *α* = *ω*; hence

*8* = *rω*,orω = *8*,

and, by differentiating,

In other words, the curvature of a circle at any point *P* on the circle is constant, and equal to the reciprocal of the radius.

**8—8.Curvature of a Curve in Polar Coordinates.** Consider the polar curve, with the radius vector *ρ* to point *P* corresponding to *θ*, and the tangent at *P*, making an angle *ω* with the polar axis. From the figure,

*ω* = *θ* + *ø.*(1)

Differentiating (1) with respect to *θ*:

Further, we may designate the angle between *dρ* and *ds,* in the differential triangle *PRQ,* as *ø*, for when passing to the limit, angle *RQP* equals angle *OPT.* Hence, from triangle *PRQ,* we have

Differentiating (4) with respect to *θ*, by [14], §5—15, and simplifying:

Substituting (5) in (2):

But from §8—3, equation [4], we recall

Dividing (6) by (7), we obtain

**8—9.Finding the Curvature at a Point on a Curve.** This will now be illustrated by several examples.

EXAMPLE 1.Find the curvature of the parabola

*y*^{2} = 12*x* at the point (3,6).

*Solution*.

Substituting these values in [1a] of §8—6, for the point (3,6), we have

or, considering the absolute value, *K* = .

EXAMPLE 2.Find the curvature of the lemniscate

EXAMPLE 3.Find the curvature of the hyperbola *xy* = *k* at any point on the curve. Where does the curve have the greatest curvature?

Since *xy* = *k,* the numerator of this latter fraction in (2) is a constant; therefore *K* will have its greatest value when the denominator is least. But the denominator will have the least value when *x = y*; hence the greatest curvature is at *x* = *y*, or where the 45°-axis cuts the hyperbola.

NOTE. To determine the minimum value of *x*^{2} + *y*^{2}, let *u* = *x*^{2} + *y*^{2}, or *u* = *x*^{2} + , since *xy* = *k*; find , set it equal to zero, and solve for critical values; they are *x* = *y* = ±

**EXERCISE 8—1**

*Find the curvature of each of the following curves at the point indicated:*

**1.** *y*^{2} = 2*px*; (2*p,*2*p*)

**2.** *a*^{2}*x*^{2} + *b*^{2}*y*^{2} = *a*^{2}*b*^{2}; (0,*a*)

**3.** *y* = *x*^{4} − 2*x*^{3} + *x*^{2}; (0,0)

**4.** *y* = log *x*; (*e*,1)

*Find the curvature of each of the following curves at the point where the slope of the curve is zero:*

**5.** *y* = 3*x*^{2} − 6*x* + 5

**6.** *y* = *x*^{3} + 9*x*^{2} − 8

**7.** *y*= (*x* + 2)^{2}

**8.** *y* =

**9.** *y =* cos 2*x*

*Find the curvature at any point:*

**10.** *x*^{2} − *y*^{2} = *a*^{2}

**11.** *ρ* = *e ^{aθ}*

**12.** *y* = *x*^{4}

**13.** *ρ* **=** *aθ*

*Prove:*

**14.** The curvature of a straight line, *y* = *mx* + *b*, is zero at every point on the line.

**15.** The curvature at a point of inflection on any curve equals zero.