Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/287

Rh where $$\alpha, \beta, \gamma$$ are the direction angles of the tangent (or curve) at $$P$$. Hence the equations of the tangent line to the curve

at the point (x, y, z) are given by

(69)

and the equation of the normal plane, i.e. the plane passing through (x, y, z) perpendicular to the tangent, is

(70)

X, r, Z being the variable coordinates.

ILLUSTRATIVE EXAMPLE 1. Find the equations of the tangent and the equation of the normal plane to the helix * (0 being the parameter)

fx = s y = Iz =

a sin 0,

(a) at any point ; (b) when 6 = 2 IT.

Solution. = a sin Q = T/, -

dO d9

Substituting in (69) and (70), we get at (x, y, z)

X ~ X = T ~ y = Z ~ Z, tangent line ; y x b

and - y (X x) + x (Y- y) + b (Z - z) = 0, normal plane.

When = 2ir, the point of contact is (a, 0, 26?r),

a 6

or, X=a, bY=aZ 2 afar,

the equations of the tangent line ; and

the equation of the normal plane.

the elements at the same angle.
 * The helix may be defined as a curve traced on a right circular cylinder so as to cut all

Take OZ as the axis of the cylinder, and the point of starting in OX at P Q. Let a = radius of base of cylinder and 0= angle of rotation. By definition,

= -^- = =/fc (const.), or z=ak0.

w~ ,

Let ak = 6 ; then z = b0. Also y = MN= a sin e, x = OM = a cos 0.