Example 7-27. Consider a surface described by the implicit form F(x, y, z) = 0.
Recall that equations of this form define zas a function of xand yand the derivatives
of zwith respect to xand yare given by
∂z
∂x =
−Fx
Fz =
−∂F∂x
∂F
∂z
and
∂z
∂y =
−Fy
Fz =
−∂F∂y
∂F
∂z
Substituting these derivatives into the equation (7.86) and simplifying one finds that
the direction cosines of the unit normal to the surface are given by
nx=
∂F∂x
√
(∂F
∂x
) 2
+
(
∂F
∂y
) 2
+
(∂F
∂z
) 2
, n y=
∂F
√ ∂y
(∂F
∂x
) 2
+
(
∂F
∂y
) 2
+
(∂F
∂z
) 2
, n z=
∂F∂z
√
(∂F
∂x
) 2
+
(
∂F
∂y
) 2
+
(∂F
∂z
) 2
Tangent Plane to Surface
Consider a smooth surface defined by the equation
r =r (u, v ) = x(u, v )ˆe 1 +y(u, v )ˆe 2 +z(u, v )ˆe 3
In order to construct a tangent plane to a regular point r =r (u 0 , v 0 )where the surface
coordinates have the values (u 0 , v 0 ), one must first construct the normal to the surface
at this point. One such normal is
N =∂r
∂u ×
∂r
∂v =N^1 ˆe^1 +N^2 ˆe^2 +N^3 ˆe^3
The point on the surface is
x 0 =x(u 0 , v 0 ), y 0 =y(u 0 , v 0 ), z 0 =z(u 0 , v 0 )
which can be described by the position vector r 0 =r (u 0 , v 0 ). If r represents the
variable point
r =xˆe 1 +yˆe 2 +zˆe 3
which varies over the plane through the point (x 0 , y 0 , z 0 ), then the vector r −r 0 must
lie in the tangent plane and consequently is perpendicular to the normal vector N.
One can then write
(r −r 0 )·N = 0 (7 .87)