79. Let
; (a) Use a computer to graph .
(b) Find and when .
(c) Find and using Equations 2 and 3.
(d) Show that and .
(e) Does the result of part (d) contradict Clairaut’s Theorem? Use graphs of and to illustrate your answer.
fy x
fx y CAS
fy x0, 0 1 fx y0, 0 1
fy0, 0
fx0, 0 fyx, y x, y 0, 0
fxx, y
f
fx, y
0xx3y2 xy y23 ifif x, y 0, 0x, y 0, 0
76. (a) How many th-order partial derivatives does a function of two variables have?
(b) If these partial derivatives are all continuous, how many of them can be distinct?
(c) Answer the question in part (a) for a function of three variables.
77. If , find .
[Hint: Instead of finding first, note that it’s easier to use Equation 1 or Equation 2.]
78. If , fx, y s3x3 y3 find fx0, 0. fxx, y
fx1, 0
fx, y xx2 y232esinx2y n
TANGENT PLANES AND LINEAR APPROXIMATIONS
One of the most important ideas in single-variable calculus is that as we zoom in toward a point on the graph of a differentiable function, the graph becomes indistin- guishable from its tangent line and we can approximate the function by a linear func- tion. (See Section 2.8.) Here we develop similar ideas in three dimensions. As we zoom in toward a point on a surface that is the graph of a differentiable function of two variables, the surface looks more and more like a plane (its tangent plane) and we can approximate the function by a linear function of two variables. We also extend the idea of a differential to functions of two or more variables.
TANGENT PLANES
Suppose a surface has equation , where has continuous first partial derivatives, and let be a point on . As in the preceding section, let and be the curves obtained by intersecting the vertical planes and with the surface . Then the point lies on both and . Let and be the tangent lines to the curves and at the point . Then the tangent plane to the surface at the point is defined to be the plane that contains both tangent lines and . (See Figure 1.)
We will see in Section 11.6 that if is any other curve that lies on the surface and passes through , then its tangent line at also lies in the tangent plane. There- fore, you can think of the tangent plane to at as consisting of all possible tangent lines at to curves that lie on and pass through . The tangent plane at is the plane that most closely approximates the surface near the point .
We know from Equation 10.5.7 that any plane passing through the point has an equation of the form
By dividing this equation by and letting and , we can write it in the form
If Equation 1 represents the tangent plane at , then its intersection with the plane P z z0 ax x0 by y0
1
b BC a AC
C
Ax x0 By y0 Cz z0 0
Px0, y0,z0 P
S
P P
S P
P S
P P
S C
T2
T1
P
S P
C2
C1
T2
T1
C2
C1
P S
x x0 y y0
C2
C1
S
PSx0, y0,z0 z f x, y f
11.4
z
FIGURE 1
The tangent plane contains the tangent lines T¡TT and T™TT .
y x
must be the tangent line . Setting in Equation 1 gives
and we recognize these as the equations (in point-slope form) of a line with slope . But from Section 11.3 we know that the slope of the tangent is . Therefore,
.
Similarly, putting in Equation 1, we get , which must represent the tangent line , so .
Suppose has continuous partial derivatives. An equation of the tangent plane to the surface at the point is
EXAMPLE 1 Find the tangent plane to the elliptic paraboloid at the point .
SOLUTION Let . Then
Then (2) gives the equation of the tangent plane at as
or ■
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 . Notice that the more we zoom in, the flatter the graph appears and the more it resembles its tangent plane.
FIGURE 2 The elliptic paraboloid z=2≈+¥ appears to coincide with its tangent plane as we zoom in toward (1, 1, 3).
(c) 2 1 0
2
1 0 40
20 0 _20
y z
x (b)
2 0 _2
2 0
_2 40
20 0 _20
y z
x (a)
40 20 0 _20
y z
4 2 0 _2 _4
4 2 0x _2 _4
fx, y 2x2 y2 z 4x 2y 3
z 3 4x 1 2y 1
1, 1, 3
fx1, 1 4 fy1, 1 2 fxx, y 4x fyx, y 2y fx, y 2x2 y2
1, 1, 3 z 2x2 y2
V
z z0 fxx0, y0x x0 fyx0, y0y y0 Px0, y0,z0 z f x, y
f
2
b fyx0, y0 T2
z z0 by y0 x x0
a fxx0, y0 T1 fxx0, y0
a y y0
z z0 ax x0 y y0 T1
y y0
■ Note the similarity between the equa- tion of a tangent plane and the equation of a tangent line:
y y0 fx0x x0
Visual 11.4 shows an animation of Figure 2.
In Figure 3 we corroborate this impression by zooming in toward the point (1, 1) on a contour map of the function . Notice that the more we zoom in, the more the level curves look like equally spaced parallel lines, which is charac- teristic of a plane.
LINEAR APPROXIMATIONS
In Example 1 we found that an equation of the tangent plane to the graph of the func-
tion at the point (1, 1, 3) is . Therefore, in view
of the visual evidence in Figures 2 and 3, the linear function of two variables
is a good approximation to when is near (1, 1). The function L is called the linearization of f at (1, 1) and the approximation
is called the linear approximation or tangent plane approximation of f at (1, 1).
For instance, at the point (1.1, 0.95) the linear approximation gives
which is quite close to the true value of .
But if we take a point farther away from (1, 1), such as (2, 3), we no longer get a good
approximation. In fact, whereas .
In general, we know from (2) that an equation of the tangent plane to the graph of a function f of two variables at the point is
if and are continuous. The linear function whose graph is this tangent plane, namely
is called the linearization of f at and the approximation
is called the linear approximation or the tangent plane approximation of atf a, b.
fx, y f a, b fxa, bx a fya, by b
4
a, b
Lx, y f a, b fxa, bx a fya, by b
3
fy
fx
z f a, b fxa, bx a fya, by b
a, b, f a, b
f2, 3 17 L2, 3 11
f1.1, 0.95 21.12 0.952 3.3225 f1.1, 0.95 41.1 20.95 3 3.3
fx, y 4x 2y 3
x, y
fx, y
Lx, y 4x 2y 3
z 4x 2y 3 fx, y 2x2 y2
FIGURE 3
Zooming in toward (1, 1) on a contour map of
f(x, y)=2≈+¥ 0.95 1.05
1.05
0.8 1.2
1.2
0.5 1.5
1.5
fx, y 2x2 y2
We have defined tangent planes for surfaces , where has continuous first partial derivatives. What happens if and are not continuous? Figure 4 pic- tures such a function; its equation is
You can verify (see Exercise 36) that its partial derivatives exist at the origin and, in fact, and , but and are not continuous. The linear approx- imation would be , but at all points on the line . 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, , if x changes from a to we defined the increment of as
In Chapter 2 we showed that if is differentiable at a, then
Now consider a function of two variables, , and suppose x changes from a to and y changes from b to . Then the corresponding increment of
is
Thus the increment represents the change in the value of when changes from to . By analogy with (5) we define the differentiability of a function of two variables as follows.
DEFINITION If , then is differentiable at if can be expressed in the form
where and as .
Definition 7 says that a differentiable function is one for which the linear approxi- mation (4) is a good approximation when is near . In other words, the tan- gent plane approximates the graph of f well near the point of tangency.
It’s sometimes hard to use Definition 7 directly to check the differentiability of a function, but the following theorem provides a convenient sufficient condition for differentiability.
THEOREM If the partial derivatives and exist near and are con- tinuous at 8 a, b, then is differentiable at f fx a, bfy. a, b
a, b
x, y
x, y l 0, 0
2l 0
1
z fxa, b x fya, b y 1x 2y
z
a, b
z f x, y f
7
a x, b y
a, b z f x, y
z f a x, b y f a, b
6
z a x b yz f x, y
where l 0 as x l 0
y f a x x
5
f
y f a x f a
y
a x, y f x
y x fx, y 12
fx, y 0fy0, 0 0 fx fy
fx0, 0 0
fx, y
0x2xy y2 ifif x, y 0, 0x, y 0, 0
fy
fx
z f x, y f
z y
x
xy
≈+¥ if (x, y)≠(0, 0), f(0, 0)=0
FIGURE 4 f(x, y)=
■ This is Equation 2.5.5.
■ Theorem 8 is proved in Appendix B.
EXAMPLE 2 Show that is differentiable at (1, 0) and find its lin- earization there. Then use it to approximate .
SOLUTION The partial derivatives are
Both and are continuous functions, so is differentiable by Theorem 8. The linearization is
The corresponding linear approximation is
so
Compare this with the actual value of . ■
DIFFERENTIALS
For a differentiable function of one variable, , 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 is then defined as
(See Section 2.8.) Figure 6 shows the relationship between the increment and the differential : represents the change in height of the curve and rep- resents the change in height of the tangent line when changes by an amount
For a differentiable function of two variables, , we define the differen- tials and to be independent variables; that is, they can be given any values. Then the differential , also called the total differential, is defined by
(Compare with Equation 9.) Sometimes the notation is used in place of .
If we take and in Equation 10, then the dif-
ferential of is
So, in the notation of differentials, the linear approximation (4) can be written as fx, y f a, b dz
dz fxa, bx a fya, by b
z
dy y y b dx x x a
dz d f
dz fxx, y dx fyx, y dy &z
&x dx &z
&y dy
10
dz dy dx
z f x, y
dx x.
x
dy y f x
y dy
y dy f x dx
9
y
y f x
f1.1, 0.1 1.1e0.11 0.98542 f1.1, 0.1 1.1 0.1 1
xexy x y
1 1x 1 1 y x y
Lx, y f 1, 0 fx1, 0x 1 fy1, 0y 0
f fy
fx
fy1, 0 1 fx1, 0 1
fyx, y x2exy fxx, y exy xyexy
f1.1, 0.1
fx, y xexy
V
FIGURE 5
1 0 _1
6 4 2 0
x y z
1
0
■ Figure 5 shows the graphs of the function and its linearization in Example 2.
L f
a a+Îx x y
0
dx=Îx y=ƒ
dy Îy
FIGURE 6 tangent line y=f(a)+fª(a)(x-a)
Figure 7 is the three-dimensional counterpart of Figure 6 and shows the geometric interpretation of the differential and the increment : represents the change in height of the tangent plane, whereas represents the change in height of the surface
when changes from to .
EXAMPLE 3
(a) If , find the differential .
(b) If changes from 2 to and changes from 3 to , compare the values of and .
SOLUTION
(a) Definition 10 gives
(b) Putting , , , and , we get
The increment of is
Notice that but is easier to compute. ■
EXAMPLE 4 The base radius and height of a right circular cone are measured as 10 cm and 25 cm, respectively, with a possible error in measurement of as much as
cm in each. Use differentials to estimate the maximum error in the calculated volume of the cone.
0.1
dz
z dz
0.6449
2.052 32.052.96 2.962 22 323 32
z f 2.05, 2.96 f 2, 3
z
0.65
dz 22 33 0.05 32 23 0.04
dy y 0.04 y 3
dx x 0.05 x 2
dz &z
&x dx &z
&y dy 2x 3y dx 3x 2y dy dz
z x 2.05 y 2.96
dz z f x, y x2 3xy y2
V
y x
z
Îx=d x 0
{a, {{ b, f(a, b)}
(a, b, 0)
(a+Îx, b+Îy, 0) {a+Îx, b+Îy, f (a+Îx, b+Îy)}
f(a, b)
f(a, b)
Îy=dy
tangent plane
z-f(a, b)=fff (a, b)(x-a)+fx f (a, b)(y-b)y
surface z=f(x, y)
dz Îz
FIGURE 7
a x, b y
a, b
x, y
z f x, y dz z z dz
FIGURE 8 60
0
5 3 1 2
x y
z 20
4 40
4 2 _20
0
0
■ In Example 3, is close to because the tangent plane is a good approximation
to the surface near
. (See Figure 8.)
2, 3, 13 z x2 3xy y2
z dz
SOLUTION The volume of a cone with base radius and height is . So the differential of is
Since each error is at most cm, we have , . To find the largest error in the volume we take the largest error in the measurement of and of
. Therefore, we take and along with , . This gives
Thus the maximum error in the calculated volume is about cm cm . ■ 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
and the linearization is the right side of this expression.
If , then the increment of is
The differential is defined in terms of the differentials , , and of the inde- pendent variables by
EXAMPLE 5 The dimensions of a rectangular box are measured to be 75 cm, 60 cm, and 40 cm, and each measurement is correct to within 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 , , and , its volume is and so
We are given that , , and . To find the largest error in the volume, we therefore use , , and together with
, , and :
Thus an error of only cm in measuring each dimension could lead to an error of as much as 1980 cm in the calculated volume! This may seem like a large error, but
it’s only about 1% of the volume of the box. ■
3
0.2
V dV 60400.2 75400.2 75600.2 1980 z 40
y 60 x 75
dz 0.2 dy 0.2
dx 0.2
z 0.2 y 0.2 x 0.2dV &V
&x dx &V
&y dy &V
&z dz yz dx xz dy xy dz V xyz z
y x
0.2 dw &w
&x dx &w
&y dy &w
&z dz
dz dy dx dw
w f x x, y y, z z f x, y, z
w w f x, y, z Lx, y, z
fx, y, z f a, b, c fxa, b, cx a fya, b, cy b fza, b, cz c
3 3 63 20
dV 500
3 0.1 100
3 0.1 20
h 25 r 10
dh 0.1 dr 0.1
h
h 0.1 r r 0.10.1 dV &V
&r dr &V
&h dh 2rh
3 dr r2 3 dh V
Vr2h3 h
r V
18 –22 ■ Find the differential of the function.
18.
19. 20.
21. 22.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
If and changes from to
compare the values of and .
24. If and changes from to
, compare the values of and .
25. The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measurement of at most cm in each. Use differentials to estimate the maxi- mum error in the calculated area of the rectangle.
26. The dimensions of a closed rectangular box are measured as 80 cm, 60 cm, and 50 cm, respectively, with a possible error of cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box.
Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is
cm thick.
28. Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is cm thick and the metal in the sides is cm thick.
29. A model for the surface area of a human body is given by , where is the weight (in pounds), is the height (in inches), and is measured in square feet.
If the errors in measurement of and are at most 2%, use differentials to estimate the maximum percentage error in the calculated surface area.
30. The pressure, volume, and temperature of a mole of an ideal gas are related by the equation , where is measured in kilopascals, in liters, and in kelvins. Use differentials to find the approximate change in the pressure if the volume increases from 12 L to 12.3 L and the temper- ature decreases from 310 K to 305 K.
31. If is the total resistance of three resistors, connected in parallel, with resistances , , , then
If the resistances are measured in ohms as , , and , with a possible error of in each case, estimate the maximum error in the calculated value of .R
0.5%
R3 50 R2 40
R1 25 1
R 1 R1 1
R2 1 R3
R3
R2
R1
R
T V
P PV 8.31T
h w S h
S 0.1091w0.425h0.725 w 0.05
0.1 0.04
27.
0.2 0.1
dz
z
2.96, 0.95z x2 xy 3y2 x, y 3, 1
dz
z 1, 2 1.05, 2.1,
x, y
z 5x2 y2 23.
w xyexz R 2 cos '
u etsins 2t
z x3 lny2 v y cos xy 1–6 ■ Find an equation of the tangent plane to the given
surface at the specified point.
1. ,
2. ,
3. ,
4. ,
5. ,
6. ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
;7– 8 ■ Graph the surface and the tangent plane at the given point. (Choose the domain and viewpoint so that you get a good view of both the surface and the tangent plane.) Then zoom in until the surface and the tangent plane become indistinguishable.
7. ,
8. ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
9–10 ■ Draw the graph of and its tangent plane at the given point. (Use your computer algebra system both to compute the partial derivatives and to graph the surface and its tangent plane.) Then zoom in until the surface and the tangent plane become indistinguishable.
9.
10.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
11–14 ■ Explain why the function is differentiable at the given point. Then find the linearization of the function at that point.
,
12. ,
13. ,
14. ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
15. Find the linear approximation of the function
at and use it to approxi- mate .
;16. Find the linear approximation of the function
at and use it to approximate . Illustrate by graphing and the tangent plane.
Find the linear approximation of the function at and use it to approximate the number fx, y, z sx2 y2 zs3.022 3, 2, 62 1.972 5.992. 17.
f f6.9, 2.06 7, 2
fx, y lnx 3y
f1.95, 1.08 2, 1
fx, y s20 x2 7y2
3, 0
fx, y sx e4y
1, 0
fx, y tan1x 2y
6, 3
fx, y xy
1, 4
fx, y xsy 11.
Lx, y
fx, y exy10
(
sx sy sxy)
, 1, 1, 3e0.1 fx, y xy sinx y1 x2 y2, 1, 1, 0
CAS f
1, 1, 4
z arctanxy2
1, 1, 5
z x2 xy 3y2
1, 1, 1
z ex2y2
2, 2, 2
z y cosx y
1, 4, 0
z y ln x
1, 1, 1
z s4 x2 2y2
1, 2, 18
z 9x2 y2 6x 3y 5
1, 2, 4
z 4x2 y2 2y
EXERCISES
11.4
Prove that if is a function of two variables that is differen- tiable at , then is continuous at .
Hint: Show that
36. (a) The function
was graphed in Figure 4. Show that and both exist but is not differentiable at . [Hint: Use the result of Exercise 35.]
(b) Explain why fxand fyare not continuous at 0, 0.
0, 0
f fy0, 0
fx0, 0
fx, y
0x2xy y2 ifif x, y 0, 0x, y 0, 0
x, y l 0, 0lim fa x, b y f a, b
a, b
a, b f 35. f
32. Suppose you need to know an equation of the tangent plane to a surface at the point . You don’t have an equation for but you know that the curves
both lie on . Find an equation of the tangent plane at . 33–34 ■ Show that the function is differentiable by finding values of and that satisfy Definition 7.
34.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
fx, y xy 5y2 fx, y x2 y2 33.
2
1
P S
r2u 1 u2, 2u3 1, 2u 1
r1t 2 3t, 1 t2, 3 4t t2 S
P2, 1, 3
S
THE CHAIN RULE
Recall that the Chain Rule for functions of a single variable gives the rule for differ- entiating a composite function: If and , where and are differen- tiable functions, then is indirectly a differentiable function of and
For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for differentiating a composite function. The first version (The- orem 2) deals with the case where and each of the variables and is, in turn, a function of a variable . This means that is indirectly a function of , , and the Chain Rule gives a formula for differentiating as a function of . We assume that is differentiable (Definition 11.4.7). Recall that this is the case when and are continuous (Theorem 11.4.8).
THE CHAIN RULE (CASE 1) Suppose that is a differentiable function of and , where and are both differentiable func- tions of . Then is a differentiable function of and
PROOF A change of in produces changes of in and in . These, in turn, produce a change of in , and from Definition 11.4.7 we have
where and as . [If the functions and are not
defined at 1l 00, 0, we can define them to be 0 there.] Dividing both sides of this 2l 0 x, y l 0, 0 1 2
z &f
&x x &f
&yy 1x 2y z
zt x x y y
t
dz dt &f
&x dx dt &f
&y dy dt z t
t
y ht
x tt
y x
z f x, y
2
fy
fx
f t
z
z f tt, ht t z t
y z f x, y x
dy dt dy
dx dx
1 dt
t y
t f x tt
y f x
