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we shall have the equations to the tangent in terms of the partial differential-coefficients of the intersecting surfaces. Similarly may the direction cosines in (3) be determined.

342.] To find the equation to the normal plane to a curve in space.

The plane perpendicular to the tangent line, and passing through the point of contact, is called the normal plane. Let (, 7j, ( be its current coordinates, and y, z) be the point of contact through which it passes; then, since it is to be perpendicular to the line whose direction cosines are —r,

as as as

its equation is

343.] To find the equation to the osculating plane to a curve in space.

In curves such as we have discussed in previous Chapters, all the points lie in one plane; and therefore the curves are called plane curves. This property however docs not hold good for all curves in space; although every three consecutive points must be in one plane, yet the fourth may be out of it; or in other words, every two consecutive tangents are in the same plane, but the next consecutive tangent is in general in a different one; our object is to determine the equation to the plane which contains two consecutive tangents, and which is called the osculating plane, and is defined as follows:

The osculating plane is the plane containing three consecutive points on a curve.

Let the equation to the plane be

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A B c (13)

dy d2z—dzd2y dzd2x dx d2z d-x dly dy d'x'

whence, dividing (12) by the several terms of equality (13), we have

(dyd2z dzd2y) (£—x) + (dz d2x dx d'2z) (y—y)

+ (dx d2y - dy d2x) (f-z) = 0; (14) which is the equation to the osculating plane.

344.] The method by which we have deduced this equation is the same as if we had defined the osculating plane to be that in which two consecutive tangents lie, as will be apparent from what follows.

Let the equation to the plane passing through {x, y, z) be A(f-j7)+B(r,-y)+c(C-z) =0;

and since it is to be that in which two consecutive tangents lie, whose direction cosines are respectively

dx dy dz dx + d2x dy + d2y dz + d2z
ds' ds' ds' ds + d2s' ds + d2s' ds + d'2s'

we have the conditions

\dx + Bdy + cdz = 0,

A(dx + d2x) + B(dy + d2y) + c(dz + d2z) = 0;

whence, by subtraction,

xd2x + Bd2y + cd2z = 0;

which two relations between A, B, C are the same as those above marked (10) and (11), whence equality (13) follows, and therefore the equation to the osculating plane is the same.

345] It is manifest from (7) that all straight lines passing through a point of contact, and perpendicular to the tangent line, lie in the normal plane; two of these normal lines have peculiar properties in relation to the osculating plane, viz. that which is perpendicular to it, and that which lies in it, and is therefore the line of intersection of it by the normal plane. The latter is called the principal normal, and the former has the distinctive name of binortnal, being, as it is, perpendicular to two consecutive elements of the curve, while all other normals are perpendicular to only one.

To find the equations to the binormal.

Let /, m, n be its direction-angles; then, as it is perpendicular to the osculating plane,

cos / cos m cos n

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and therefore the equations to the principal normal are

£—* _ v-y (—*

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Therefore, if A, ft, v are the direction-angles of the principal normal, we have

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therefore the equations to the tangent line are

f—g q—y f— g

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