THE EVOLUTE - Curvature - The Calculus Primer

The Calculus Primer (2011)

Part VIII. Curvature

Chapter 32. THE EVOLUTE

8—14.Definition and Properties of the Evolute. It is clear that every point of a given curve has its center of curvature. The locus of these centers of curvature is called the evolute of the curve. In other words, as a point Pmoves along a given curve, the center of curvature Q moves along another curve; this path is the evolute of the original curve. The original curve is the involute of its evolute.

If the given curve is given by y = f(x), then the right-hand members of [1] and [2] of §8—13 may be stated in terms of x, whence they become

X = ø(x), Y = ψ(x),[3]

where X and Y are the coordinates of any point on the curve y = f(x), instead of the coordinates (α,β) of some particular point.

The evolute of a curve presents two interesting properties:

(1) The normals to a curve are tangent to the evolute, each at the center of curvature that lies upon it.



(2) The difference in length between any two radii of curvature is equal to the length of the arc of the evolute included by them; i.e., length of arc Q1Q2 = r2r1.

Property (2), sometimes referred to as the “string property” of the evolute, may be stated informally as follows. Let an inextensible but flexible string be attached to a fixed point on the evolute and wrapped tightly around a given arc AQ1 of the evolute, and let the remaining portion of the string be held taut. Now as the free end (P) of the string is allowed to move so that the string remains tangent to the evolute, the free end P will trace the original curve or involute, and the point of tangency will be the corresponding center of curvature Q. From the diagram, it will be seen that the curve P1 · · · P5 is not the only possible involute of the given evolute; for, as the string is unwrapped, similar curves M1 · · · M5, N1 · · · N5, etc, are traced. Thus, while a curve has but one evolute, it has an infinite number of involutes.



Find the coordinates of the center of curvature of the following curves at the point indicated:

1. y = ex; point (0,1).

2. y = log x; at the point where x = 2.

3. x2 + y2 = k2; point (0,k).

4. y2 = 4x; point (1,2).

5. y = sin x; point images.



1. Find images:

(a) y3 − 3y + x = 0.

(b) x3 + xy2 + y3 = 4.

2. Find images:

(a) x2y2 = 1.

(b) y2 − 2xy + a2 = 0.

3. Find images:

(a) y = arc sin images.

(b) y = log images.

4. Find maximum or minimum values:

(a) y = x1/x.

(b) y = sin x (1 + cos x).

(c) y = x3 − 3x2 + 6x.

5. Find the radius of curvature at the given point:

(a) y2 = 4x, at (0,0).

(b) y images, at (0,1).

6. Find the equation of the tangent to the given curve at the point indicated:

(a) ay2 = x3; (a,a).

(b) xy = 4; (4,1).

(c) y2 == 4x; (1,2).

7. If the sides of a rectangle are a and b, find the rectangle of maximum area that can be drawn so as to have its sides pass through the vertices of the given rectangle.

8. Find the equation of the line tangent to the curve x2 + y2 = x3 at the point (2,2).

9. If Z = images, and xy = a2, prove that images

10. Assuming that a raindrop is a perfect sphere, and that it increases in volume at a rate proportional to its surface area, prove that the radius increases at a constant rate.