The gradient has fundamental geometric significance illustrated by the following
picture.
PICT
In this picture, the surface is a piece of a level surface of a function of three variables
f
(x,y,z)
. Thus the surface is defined by f
(x,y,z)
= c or more completely
as
{(x,y,z) : f (x,y,z) = c}
. For example, if f
(x,y,z)
= x^{2} + y^{2} + z^{2}, this
would be a piece of a sphere. There are two smooth curves in this picture which
lie in the surface having parameterizations, x_{1}
(t)
=
(x1 (t),y1(t),z1(t))
and
x_{2}
(s)
=
(x2(s),y2 (s),z2(s))
which intersect at the point
(x0,y0,z0)
on this
surface.^{2}
This intersection occurs when t = t_{0} and s = s_{0}. Since the points x_{1}
(t)
for t in an
interval lie in the level surface, it follows
f (x (t),y (t),z (t)) = c
1 1 1
for all t in some interval. Therefore, taking the derivative of both sides and using the
chain rule on the left,
is perpendicular to both the direction vectors of the two
indicated curves shown. Surely if things are as they should be, these two direction vectors
would determine a plane which deserves to be called the tangent plane to the level
surface of f at the point
(x0,y0,z0)
and that ∇f
(x0,y0,z0)
is perpendicular to this
tangent plane at the point
(x0,y0,z0)
.
Example 21.10.1Find the equation of the tangent plane to the level surface
f (x,y,z) = 6
of the function f
(x,y,z)
= x^{2} + 2y^{2} + 3z^{2}at the point
(1,1,1)
.
First note that
(1,1,1)
is a point on this level surface. To find the desired plane it
suffices to find the normal vector to the proposed plane. But ∇f
(x,y,z)
=
(2x,4y,6z)
and so ∇f
(1,1,1)
=
(2,4,6)
. Therefore, from this problem, the equation of the plane is
(2,4,6)
⋅
(x − 1,y − 1,z − 1)
= 0 or in other words, 2x − 12 + 4y + 6z = 0.
Example 21.10.2The point
(√ - )
3,1,4
is on both the surfaces, z = x^{2} + y^{2}andz = 8 −
( )
x2 + y2
. Find the cosine of the angle between the two tangent planes atthis point.
Recall this is the same as the angle between two normal vectors. Of course there is
some ambiguity here because if n is a normal vector, then so is −n and replacing n with
−n in the formula for the cosine of the angle will change the sign. We agree to look for
the acute angle and its cosine rather than the obtuse angle. The normals are
(2√3, 2,− 1)
and
(2√3, 2,1)
. Therefore, the cosine of the angle desired is
√- 2
(2-3)+4−1
17
=
15
17
.
Example 21.10.3The point
( √ - )
1, 3,4
is on the surface z = x^{2} + y^{2}. Find theline perpendicular to the surface at this point.
All that is needed is the direction vector of this line. The surface is the level surface
x^{2} + y^{2}− z = 0. The normal to this surface is given by the gradient at this point. Thus
the desired line is