14 Partial Derivatives

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 Tangent Planes and Linear
14.4 Approximations

Copyright © Cengage Learning. All rights reserved.

Tangent Planes

3
Tangent Planes
Suppose a surface S has equation z = f (x, y), where f has
continuous first partial derivatives, and let P(x0, y0, z0) be a
point on S.

Let C1 and C2 be the curves obtained by intersecting the

vertical planes y = y0 and x = x0 with the surface S. Then
the point P lies on both C1 and C2.

Let T1 and T2 be the tangent lines to the curves C1 and C2

at the point P.

4
Tangent Planes
Then the tangent plane to the surface S at the point P is
defined to be the plane that contains both tangent lines
T1 and T2. (See Figure 1.)

The tangent plane contains the tangent lines T1 and T2.

Figure 1

5
Tangent Planes
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 approximates the surface S near
the point P. We know that any plane passing through the
point P(x0, y0, z0) has an equation of the form

A (x – x0) + B (y – y0) + C (z – z0) = 0

6
Tangent Planes
By dividing this equation by C and letting a = –A/C and
b = –B/C, we can write it in the form

z – z0 = a(x – x0) + b(y – y0)

If Equation 1 represents the tangent plane at P, then its

intersection with the plane y = y0 must be the tangent
line T1. Setting y = y0 in Equation 1 gives

z – z0 = a(x – x0) where y = y0

and we recognize this as the equation (in point-slope form)

of a line with slope a.
7
Tangent Planes
But we know that the slope of the tangent T1 is fx(x0, y0).

Therefore a = fx(x0, y0).

Similarly, putting x = x0 in Equation 1, we get

z – z0 = b(y – y0), which must represent the tangent line T2,
so b = fy(x0, y0).

8
Example 1
Find the tangent plane to the elliptic paraboloid z = 2x2 + y2
at the point (1, 1, 3).

Solution:
Let f (x, y) = 2x2 + y2.
Then
fx(x, y) = 4x fy(x, y) = 2y
fx(1, 1) = 4 fy(1, 1) = 2

Then gives the equation of the tangent plane at

(1, 1, 3) as
z – 3 = 4(x – 1) + 2(y – 1)
or z = 4x + 2y – 3 9
Tangent Planes
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 restricting
the domain of the function f (x, y) = 2x2 + y2.

The elliptic paraboloid z = 2x2 + y2 appears to coincide with its

tangent plane as we zoom in toward (1, 1, 3).

Figure 2
10
Tangent Planes
Notice that the more we zoom in, the flatter the graph
appears and the more it resembles its tangent plane.
In Figure 3 we corroborate this impression by zooming in
toward the point (1, 1) on a contour map of the function
f (x, y) = 2x2 + y2.

Zooming in toward (1, 1) on a contour map of f (x, y) = 2x2 + y2

Figure 3
11
Tangent Planes
Notice that the more we zoom in, the more the level curves
look like equally spaced parallel lines, which is
characteristic of a plane.

12
Linear Approximations

13
Linear Approximations
In Example 1 we found that an equation of the tangent
plane to the graph of the function f (x, y) = 2x2 + y2 at the
point (1, 1, 3) is z = 4x + 2y – 3. Therefore, the linear
function of two variables

L(x, y) = 4x + 2y – 3

is a good approximation to f (x, y) when (x, y) is near (1, 1).

The function L is called the linearization of f at (1, 1) and
the approximation

f (x, y)  4x + 2y – 3

is called the linear approximation or tangent plane

approximation of f at (1, 1).
14
Linear Approximations
For instance, at the point (1.1, 0.95) the linear
approximation gives

f (1.1, 0.95)  4(1.1) + 2(0.95) – 3 = 3.3

which is quite close to the true value of

f (1.1, 0.95) = 2(1.1)2 + (0.95)2 = 3.3225.

But if we take a point farther away from (1, 1), such as

(2, 3), we no longer get a good approximation.

In fact, L(2, 3) = 11 whereas f (2, 3) = 17.

15
Linear Approximations
In general, we know from that an equation of the
tangent plane to the graph of a function f of two variables at
the point (a, b, f (a, b)) is
z = f (a, b) + fx(a, b)(x – a) + fy(a, b)(y – b)

The linear function whose graph is this tangent plane,

namely

L (x, y) = f (a, b) + fx(a, b)(x – a) + fy(a, b)(y – b)

is called the linearization of f at (a, b)

16
Linear Approximations
The approximation

f (x, y)  f (a, b) + fx(a, b)(x – a) + fy(a, b)(y – b)

is called the linear approximation or the tangent plane

approximation of f at (a, b).

17
Linear Approximations
We have defined tangent planes for surfaces z = f (x, y),
where f has continuous first partial derivatives. What
happens if fx and fy are not continuous? Figure 4 pictures
such a function; its equation is

You can verify that its partial

derivatives exist at the origin
and, in fact, fx(0, 0) = 0 and Figure 4

fy(0, 0) = 0, but fx and fy are

not continuous. 18
Linear Approximations
The linear approximation would be f (x, y)  0, but f (x, y) =
at all points on the line y = x.

So a function of two variables can behave badly even

though both of its partial derivatives exist.

To rule out such behavior, we formulate the idea of a

differentiable function of two variables.

Recall that for a function of one variable, y = f (x), if x

changes from a to a + x, we defined the increment of y as

y = f (a + x) – f (a)
19
Linear Approximations
If f is differentiable at a, then

y = f (a) x +  x where   0 as x  0

Now consider a function of two variables, z = f (x, y), and

suppose x changes from a to a + x and y changes from
b to b + y. Then the corresponding increment of z is

z = f (a + x, b + y) – f (a, b)

Thus the increment z represents the change in the value

of f when (x, y) changes from (a, b) to (a + x, b + y).
20
Linear Approximations
By analogy with we define the differentiability of a
function of two variables as follows.

Definition 7 says that a differentiable function is one for

which the linear approximation is a good approximation
when (x, y) is near (a, b). In other words, the tangent plane
approximates the graph of f well near the point of tangency.

21
Linear Approximations
It’s sometimes hard to use Definition 7 directly to check the
differentiability of a function, but the next theorem provides
a convenient sufficient condition for differentiability.

22
Differentials

23
Differentials
For a differentiable function of one variable, y = f (x), we
define the differential dx to be an independent variable; that
is, dx can be given the value of any real number.

The differential of y is then defined as

dy = f (x) dx

24
Differentials
Figure 6 shows the relationship between the increment y
and the differential dy: y represents the change in height
of the curve y = f (x) and dy represents the change in height
of the tangent line when x changes by an amount dx = x.

Figure 6

25
Differentials
For a differentiable function of two variables, z = f (x, y), we
define the differentials dx and dy to be independent
variables; that is, they can be given any values. Then the
differential dz, also called the total differential, is
defined by

Sometimes the notation df is used in place of dz.

26
Differentials
If we take dx = x = x – a and dy = y = y – b in
Equation 10, then the differential of z is

dz = fx(a, b)(x – a) + fy(a, b)(y – b)

So, in the notation of differentials, the linear approximation

can be written as

f (x, y)  f (a, b) + dz

27
Differentials
Figure 7 is the three-dimensional counterpart of Figure 6
and shows the geometric interpretation of the differential dz
and the increment z: dz represents the change in height of
the tangent plane, whereas z represents the change in
height of the surface z = f (x, y) when (x, y) changes from
(a, b) to (a + x, b + y).

Figure 6 Figure 7
28
Example 4
(a) If z = f (x, y) = x2 + 3xy – y2, find the differential dz.

(b) If x changes from 2 to 2.05 and y changes from 3 to

2.96, compare the values of z and dz.

Solution:
(a) Definition 10 gives

29
Example 4 – Solution cont’d

(b) Putting x = 2, dx = x = 0.05, y = 3, and

dy = y = –0.04, we get

dz = [2(2) + 3(3)]0.05 + [3(2) – 2(3)](–0.04) = 0.65

The increment of z is
z = f (2.05, 2.96) – f (2, 3)
= [(2.05)2 + 3(2.05)(2.96) – (2.96)2]
– [22 + 3(2)(3) – 32]
= 0.6449

Notice that z  dz but dz is easier to compute.

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Functions of Three or More Variables

31
Functions of Three or More Variables
Linear approximations, differentiability, and differentials can
be defined in a similar manner for functions of more than
two variables. A differentiable function is defined by an
expression similar to the one in Definition 7. For such
functions the linear approximation is

f (x, y, z)  f (a, b, c) + fx(a, b, c)(x – a) + fy(a, b, c)(y – b)

+ fz(a, b, c)(z – c)

and the linearization L(x, y, z) is the right side of this

expression.

32
Functions of Three or More Variables
If w = f (x, y, z), then the increment of w is

w = f (x + x, y + y, z + z) – f (x, y, z)

The differential dw is defined in terms of the differentials

dx, dy, and dz of the independent variables by

33
Example 6
The dimensions of a rectangular box are measured to be
75 cm, 60 cm, and 40 cm, and each measurement is
correct to within 0.2 cm. Use differentials to estimate the
largest possible error when the volume of the box is
calculated from these measurements.

Solution:
If the dimensions of the box are x, y, and z, its volume is
V = xyz and so

34
Example 6 – Solution cont’d

We are given that | x |  0.2, | y |  0.2, and | z |  0.2.

To estimate the largest error in the volume, we therefore

use dx = 0.2, dy = 0.2, and dz = 0.2 together with x = 75,
y = 60, and z = 40:

V  dV = (60)(40)(0.2) + (75)(40)(0.2) + (75)(60)(0.2)

= 1980

35
Example 6 – Solution cont’d

Thus an error of only 0.2 cm in measuring each dimension

could lead to an error of approximately 1980 cm3 in the
calculated volume! This may seem like a large error, but it’s
only about 1% of the volume of the box.

36