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

SectION 14.3 Partial Derivatives 915
x
z
0
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S
FIGURE 11
The partial derivatives of f at (a, sa, b) bd are
the slopes of the tangents to C¡ 1 and C.
2.
T¡
C¡
P(a, b, c)
(a, b, 0)
T
C
y
Interpretations of Partial Derivatives
To give a geometric interpretation of partial derivatives, we recall that the equation
z − f sx, yd represents a surface S (the graph of f ). If f sa, bd − c, then the point Psa, b, cd
lies on S. By fixing y − b, we are restricting our attention to the curve C 1 in which the
ver tical plane y − b intersects S. (In other words, C 1 is the trace of S in the plane y − b.d
Likewise, the vertical plane x − a intersects S in a curve C 2 . Both of the curves C 1 and
C 2 pass through the point P. (See Figure 1.)
Note that the curve C 1 is the graph of the function tsxd − f sx, bd, so the slope of its tangent
T 1 at P is t9sad − f x sa, bd. The curve C 2 is the graph of the function Gsyd − f sa, yd,
so the slope of its tangent T 2 at P is G9sbd − f y sa, bd.
Thus the partial derivatives f x sa, bd and f y sa, bd can be interpreted geometrically as
the slopes of the tangent lines at Psa, b, cd to the traces C 1 and C 2 of S in the planes y − b
and x − a.
As we have seen in the case of the heat index function, partial derivatives can also be
interpreted as rates of change. If z − f sx, yd, then −zy−x represents the rate of change of
z with respect to x when y is fixed. Similarly, −zy−y represents the rate of change of z with
respect to y when x is fixed.
EXAMPLE 2 If f sx, yd − 4 2 x 2 2 2y 2 , find f x s1, 1d and f y s1, 1d and interpret these
numbers as slopes.
SOLUTION We have
f x sx, yd − 22x
f y sx, yd − 24y
f x s1, 1d − 22 f y s1, 1d − 24
The graph of f is the paraboloid z − 4 2 x 2 2 2y 2 and the vertical plane y − 1 intersects
it in the parabola z − 2 2 x 2 , y − 1. (As in the preceding discussion, we label it
C 1 in Figure 2.) The slope of the tangent line to this parabola at the point s1, 1, 1d is
f x s1, 1d − 22. Similarly, the curve C 2 in which the plane x − 1 intersects the paraboloid
is the parabola z − 3 2 2y 2 , x − 1, and the slope of the tangent line at s1, 1, 1d is
f y s1, 1d − 24. (See Figure 3.)
z
z
z=4-≈-2¥
z=4-≈-2¥
z
z
z=4-≈-2¥
z=4-≈-2¥
C¡ C¡
C C
y=1 y=1
x=1 x=1
(1, 1, (1, 1) 1, 1)
(1, 1, (1, 1) 1, 1)
y y
(1, 1) (1, 1)
2 2
x x
x
FIGURE
FIGURE 22 2
FIGURE 3 33 ■
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y y
2 2
x
(1, 1) (1, 1)
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