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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 , n, be its current coordinates, and (x, y, z) be the point of contact through which it passes; then, since it is to be perpen

dicular to the line whose direction cosines are its equation is

(E-x) dx + (n-y) dy + (5—z) dz = 0.

dx

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ds

dy dz ds' ds

(7)

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 does 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

A+ Bn+c=D,

(8)

and let it pass through the three points on the curve (x, y, z), (x + dx, y + dy, z+dz), (x+2dx + d2x, y +2dy + d2y, z + 2 dz + d2≈); whence we have

Ax+BY+ Cz = D,

(9)

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; (13)

=

=

dy d2z-dz d2y dz d2x-dx d2z dx d2y - dy d2x

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

(dy d2z — dz d2y) (§ − x) + (dz d2x — dx d2z) (n—y)

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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

▲ (−x) + B (n− y) +c($—z) = 0;

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

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

whence, by subtraction,

Ad2x + Bd2y+cd2% = 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 binormal, 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 l, m, n be its direction-angles; then, as it is perpendicular to the osculating plane,

=

cos /

=

cos m

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dy d2z - dz d3y dz d2x-dx d2z dx d2y — dy d2x

1

; (15)

{ (dy d2z — dz d2y)2 + (dz d2x — dx d2z)2 + (dx d2y — dy d2x)2 } ✦ The denominator of which last expression may be modified as follows:

(dy d2z-dz d2y)2 + (dz d2x-dx d2z)2 + (dx d2y — dy d2x)2

= (dx2 + dy2+dz2) { (d2x)2 + (d2y)2+(d2z)2 }

- (dx d2x + dy d2y + dz d2z)2; (16)

but since

ds2 = dx2 + dy2+ dz2,

(17)

.. ds d2s dx d2x + dy d2y + dz d2z ;

(18)

and therefore the right-hand member of (16) becomes ds2 {(d2x)2 + (d2y)2 + (d2z)2 — (d2s)2} ;

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dy d2z-dz d2y dz dax-dxd2z

346.] To find the equations to the principal normal.

Let its equations be

ૐ X n-y 5

n-y

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=

(21)

dx day- dy d2x

L

=

=

M

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then, by reason of its being perpendicular to the tangent line, and of its lying in the osculating plane, we have

Ldx + M dy+Ndz = 0,

(23)

L(dyd2z— dzd2y) +м(dzd2x — dxd2z) + N(dxd2y — dyd2x) = 0; (24)

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dx (dx d2x + dyd2y + dzd2z) — d2x (dx2 + dy2 + dz2)

and since

ds2 = dx2 + dy2 + dz2,

=

(26)

ds d2s = dx d2x + dy d2y + dz d2z,

(27)

PRICE, VOL. I.

3 U

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Therefore, if λ, μ, v are the direction-angles of the principal

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347.] Examples on the preceding.

Ex. 1. The curve formed by the intersection of an ellipsoid

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Ex. 2. The helix; see fig. 125.

Let OA = OB = a be the radius of the base-cylinder of the helix, and AON be the angle between the plane of xz and the radius of the cylinder drawn to the point (x, y, z), and whose projection on the plane of xy is on; and let oм = x, MN=y, NP = 2; and let k be the tangent of the angle at which the thread of the helix is inclined to the plane of xy; so that NP = k the arc AN; whereby the equations to the curve are x = a cos 4,

y = a sin 4,

z = kap;

(32)

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the differentiations being performed on the supposition that is equicrescent; therefore the equations to the tangent are

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when

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-a (x) sin + a (n-y) cos + ka (−z) = 0,

nx-y+ka (¿—z) = 0;

=

70,

(35)

= 2; the normal plane therefore cuts the axis of z at a distance from the origin, equal to the z of the helix at which it is drawn.

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therefore if A, μ, v are the direction-angles of the tangent

(36)

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The tangent therefore is always inclined at the same angle to

the axis of z.

or

Hence also the equation to the osculating plane is

ka2 sin(x)-ka2 cos (n-y) + a2 (5—z) = 0,
k (¿y −ŋx) + a ($—z) = 0.

(38)

Also from (37) and (36), taking s to be equicrescent,

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