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

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

$x=\phi (t),y=\psi (t),z=\chi (t)$

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

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

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 helix may be defined as a curve traced on a right circular cylinder so as to cut all

the elements at the same angle.

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.