As the point moves on the curve, the centres of osculating circle and osculating sphere trace out curves namely the locus of centre of curvature and the locus of the centre of spherical curvature. The two loci are the results of point property of the curves. These tangents will generate a surface and we consider curves on this surface. These notions lead to the definitions of involutes and consequently evolutes of a given curve. To start with, it is worthwhile to point out the basic differences between the evolutes of a plane curve and those of a space curve.

However the concept of involute of a plane curve has natural generalisation to the space curves. Once we obtain the involute C of a curve C, we define C to be the evolute of C. The surface generated by the tangent lines to the given curve C is called the tangent surface to C. Using this definition, let us find the position vector of a point P on the tangent surface.

## tetamarfe.ml

Let A be any point on the curve at an arcual distance s from a fixed point O on C. Since P is a point on the tangent surface, AP is tangent to C. Since R is a function of two parameters u and s, we denote the position vector by R s, u so that we can ;write R. A curve which lies on the tangent surface of C and intersects the generators of the tangent surface orthogonally is called the involute of C denoted by C.

From the definition it follows that the tangents of C are normal to C. This means that the tangent to C at a point P is orthogonal to the tangent at the corresponding point of C.

The following theorem gives the equation of an involute. We use the suffix 1 for the quantities pertaining to C. Since the involute lies on the tangent surface, the position vector rx of a point Pj on the involute Fig. That is Using this and taking dot product with t on both sides of 2 we get. For different choices of c, we get different involutes of the system. Corollary 1. The distance between corresponding points of two involutes is constant.

Let P be a fixed point on C. Thus the length between two such corresponding points is constant. Corollary 2. Since t is the same for different involutes, the tangents at the corresponding points of the involutes are parallel. C-S 2 - Till! Show that the involutes of a circular helix are plane curves. Hence the involute of a circular helix must be a plane curve. Next we shall define the evolute of a space curve and derive its equation. If C is an involute of a given curve C, then C is defined to be evolute of C. Given the involute C of C, we find the equation of C in the following theorem.

Let P be a point on C corresponding to the point Q on C. PQ is a tangent at P orthogonal to C. Hence PQ is perpendicular to the tangent at Q to C. We use this fact repeatedly in the proof. Using this position vector of any point rx on C Fig. Differentiating 2 with respect to st we get dr. The above equation represents an infinite system of evolutes of the given curve, one evolute arising from each choice of c. The tangents to two different evolutes corresponding to two constants cx and c2 drawn from the same point of the given curve are inclined to each other at a constant angle cx - c2. Comparing 1 and 2 , an evolute cannot be obtained as the locus of centres of curvature.

Hence an evolute cannot be obtained as the locus of centres of spherical curvature. However we have the following fact as mentioned earlier.

The locus of the centres of spherical curvature is an evolute if and only if the curve is a plane curve. Now we shall consider two curves having the same principal normals called Betrand curves. A pair of curves y, yx which have the same principal normals are called Betrand curves. We use the suffix 1 for the entities pertaining toy,.

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From the definition, we note the following properties of Betrand curves. We prove that X is constant.

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Let it be a. Hence the angle between the binormals at the corresponding points is constant. If y and yx are Betrand curves, then i there exists a linear relation between the curvature and torsion of either curve. Proof, i We shall find the linear relation first. From the definition of Betrand curves, the principal normals coincide.

By property iii , bx makes a constant angle a with the positive direction of b and consequently it makes an angle 90 - a with t.

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## A First Course In Differential Geometry

The point r is at a distance -A along the normal at rx and t is inclined at - a to t,. A2 Let t, make a constant angle a with t along the positive direction. Thus for yx we have the following relation. If P, Pj are the corresponding points of the Betrand curves and C, Cx are their centres of curvatures, then the cross-ratio PC Px Cx is constant and equal to sec2 a where a is the angle between the tangents at P and Px to the two curves 7and yx.

Prove the following.

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Hence every curve has infinity of Betrand mates. Hence the curve yis of constant curvature. Thus each of the curves 7and yx is the locus of the centres of curvature of the other curve. When fcand T are constants, there are an infinte number of Betrand associates. Find the Betrand associates of circular helices. Let us find the Betrand associates of Helices. Since the constant A is arbitrary, we have an infinite number of Betrand associates. Instead of having the same normal, we can construct curves whose position vectors are either t, n or b.

Since t, n, b are unit vectors, all these curves lie on a unit sphere and they are called spherical indicatrices which we shall define one by one as follows. The locus of a point whose position vector is the unit tangent t to the curve yis called the spherical indicatrix of the tangent to y From the definition, we note the following properties, i Let O be the centre of the unit sphere.

## Differential geometry: a first course in curves and surfaces

Let us draw through O the unit tangent vectors of different points of y in the positive direction of the tangent. Then the curve traced on the unit sphere by the extremities of these unit tangents through O is the spherical indicatrix of the tangent to the curve y ii The tangent to the indicatrix of the tangent to y is parallel to the principal normal at the corresponding points of y Proof.

To find torsion, we must get b! The locus of a point whose position vector is the unit binormal b of a curve y is called the spherical indicatrix of the binormal. The curvature KX and torsion xx of the spherical indicatrix of b to 7 are. From the above two theorems, we get the following corollary.

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Let P, and P2 be the corresponding points with the position vector rt and r 2 on the spherical indicatrix of the tangent and binormal to y. Next we define the spherical indicatrix of the principal normal to y. The locus of the point whose position vector is equal to the unit principal normal at every point of the given curve is called the spherical indicatrix of the principal normal. We note the following properties from the definition. Differentiating the above with respect to s, we have dvx dsx dn.