James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

928 Chapter 14 Partial Derivatives
x
T
z
0
FIGURE 1
The tangent plane contains the
tangent lines T 1 and T 2.
P
T¡
C¡
C
y
Tangent Planes
Suppose a surface S has equation z − f sx, yd, where f has continuous first partial derivatives,
and let Psx 0 , y 0 , z 0 d be a point on S. As in the preceding section, let C 1 and C 2 be
the curves obtained by intersecting the vertical planes y − y 0 and x − x 0 with the surface
S. Then the point P lies on both C 1 and C 2 . Let T 1 and T 2 be the tangent lines to the
curves C 1 and C 2 at the point P. Then the tangent plane to the surface S at the point P is
defined to be the plane that contains both tangent lines T 1 and T 2 . (See Figure 1.)
We will see in Section 14.6 that if C is any other curve that lies on the surface S and
passes through P, then its tangent line at P also lies in the tangent plane. Therefore you
can think of the tangent plane to S at P as consisting of all possible tangent lines at P to
curves that lie on S and pass through P. The tangent plane at P is the plane that most
closely approx imates the surface S near the point P.
We know from Equation 12.5.7 that any plane passing through the point Psx 0 , y 0 , z 0 d
has an equation of the form
Asx 2 x 0 d 1 Bsy 2 y 0 d 1 Csz 2 z 0 d − 0
By dividing this equation by C and letting a − 2AyC and b − 2ByC, we can write it in
the form
1 z 2 z 0 − asx 2 x 0 d 1 bsy 2 y 0 d
If Equation 1 represents the tangent plane at P, then its intersection with the plane y − y 0
must be the tangent line T 1 . Setting y − y 0 in Equation 1 gives
z 2 z 0 − asx 2 x 0 d where y − y 0
and we recognize this as the equation (in point-slope form) of a line with slope a. But
from Section 14.3 we know that the slope of the tangent T 1 is f x sx 0 , y 0 d. Therefore
a − f x sx 0 , y 0 d.
Similarly, putting x − x 0 in Equation 1, we get z 2 z 0 − bsy 2 y 0 d, which must represent
the tangent line T 2 , so b − f y sx 0 , y 0 d.
Note the similarity between the equation
of a tangent plane and the equation
of a tangent line:
y 2 y 0 − f9sx 0dsx 2 x 0d
2 Suppose f has continuous partial derivatives. An equation of the tangent
plane to the surface z − f sx, yd at the point Psx 0 , y 0 , z 0 d is
z 2 z 0 − f x sx 0 , y 0 dsx 2 x 0 d 1 f y sx 0 , y 0 dsy 2 y 0 d
EXAMPLE 1 Find the tangent plane to the elliptic paraboloid z − 2x 2 1 y 2 at the
point s1, 1, 3d.
SOLUTION Let f sx, yd − 2x 2 1 y 2 . Then
f x sx, yd − 4x
f y sx, yd − 2y
f x s1, 1d − 4 f y s1, 1d − 2
Then (2) gives the equation of the tangent plane at s1, 1, 3d as
z 2 3 − 4sx 2 1d 1 2sy 2 1d
or z − 4x 1 2y 2 3 ■
Figure 2(a) shows the elliptic paraboloid and its tangent plane at (1, 1, 3) that we
found in Example 1. In parts (b) and (c) we zoom in toward the point (1, 1, 3) by restrict-
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