By D. Somasundaram

Differential Geometry: a primary path is an creation to the classical concept of area curves and surfaces provided on the Graduate and submit- Graduate classes in arithmetic. in line with Serret-Frenet formulae, the idea of area curves is constructed and concluded with an in depth dialogue on basic lifestyles theorem. the idea of surfaces comprises the 1st primary shape with neighborhood intrinsic houses, geodesics on surfaces, the second one primary shape with neighborhood non-intrinsic homes and the basic equations of the outside idea with numerous purposes.

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Differential geometry arguably deals the smoothest transition from the traditional college arithmetic series of the 1st 4 semesters in calculus, linear algebra, and differential equations to the better degrees of abstraction and evidence encountered on the top department by way of arithmetic majors. this present day it truly is attainable to explain differential geometry as "the research of constructions at the tangent space," and this article develops this standpoint.

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**Example text**

1) Differentiating (1) with respect to s we get dvx dsx , , ... (2) The equation (2) shows that the tangent to the involute C is parallel to the principal normal at A on the curve. Let us choose the positive direction along the involute such that tx = n. Using this in (2), we have /7c. (5) Taking dot product of (5) with itself on both sides, K2(c - sfK2 = T2 + K2 which T2+K2 2 gives KX = —= r- K2(C-J)2 To find torsion, we must have an expression for b which we shall find as follows. _£l. (8) 9 ' /C'T ~ K(C-S)(K2 x— + r2) If we take /c= —, and r= —, then K: = —y- and r = p a p 5-.

Thus the osculating sphere at every point of the curve is the same sphere so that the curve lies on a sphere. Example 4. Show that a necessary and sufficient condition that a curve lies on a sphere is O ds at every point of the curve. To prove the necessity of the condition, let us assume that the curve lies on a sphere. Then the sphere is the osculating sphere at every point of the curve so that the radius of the osculating sphere is constant. (2) ds Since p' * 0, we get p + a — {op') = 0 proving the necessity of the condition.

Let the curve be the intersection of the two surfaces/(jc, y> z) = 0 and g(x, ytz) = 0 Let h = V/x Vg. If A is the operator Ar = h, then 32 Differential Geometry—A First Course M -Ah-AH \H\2 |h| 3 Proof. (2) are the outward drawn normals to the surfaces (1). Since the unit tangent at P to the curve of intersection of the two surfaces is perpendicular to both the vectors in (2), it is parallel to V/x Vg. Hence V/x Vg is a scalar multiple of t. (3) Let us assume h = (h{i h2, h3). Further |h| = A.