Exercicios resolvidos calculo 2
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F.TX.10
CHAPTER 15 REVIEW ET CHAPTER 14 ¤ 213
1. True.
(
) = lim
(
+ ) (
(
) )
0
from Equation 15.3.3 [ET 14.3.3]. Let ( ) .
=
. As
0,
. Then by
substituting, we get
2
(
) = lim
3. False.
=
.
5. False. See Example 15.2.3 [ET 14.2.3]. 7. True. If
has a local minimum and ( )=h i. (
is differentiable at ( ) (
) then by Theorem 15.7.2 [ET 14.7.2],
(
) = 0 and
(
9. False. 11. True. u ) = 0, so (
)i = h0 0i = 0.
) = h0 1
= hcos ( )=
cos i, so | ·u=|
|=
cos2
+ cos2 . But |cos | u 1, so | ( )|
|
2. Now 2.
| |u| cos , but u is a unit vector, so |
2·1·1=
1. ln( +
+ 1) is deÞned only when is {( )|
+ +1
0
1,
so the domain of line = 1.
1}, all those points above the
3.
= (
)=1
2
, a parabolic cylinder
5. The level curves are
4
2
+
2
=
or 4
2
+
2
=
2
,
0, a family of ellipses.
F.
214 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
7.
9.
is a rational function, so it is continuous on its domain. Since limit:
(
is deÞned at (1 1), we use direct substitution to evaluate the
lim
)
(1 1)
2
2 +2
2
=
2(1)(1) 2 = . 12 + 2(1)2 3 4) (6 4) , so we can approximate (8 4) 2 (6 4) (6 4) by considering = 86 2 80 = 3, (6 4) to be approximately = ±2: 3 5. Averaging these = ±2 and
11. (a)
(6 4) = lim
(6 +
0
using the values given in the table: (6 4) (4 4) 2 (6 4) = 72
(6 4)
80 = 4. Averaging these values, we estimate 2 (6 4 + ) 80 2 (6 4)
3 5 C m. Similarly, (6 4) (6 6) 2
(6 4) = lim (6 4) = 75
0
, which we can approximate with (6 2) 2 (6 4) = 87
=
2 5,
(6 4)
80 = 2
values, we estimate (b) Here u =
1 2 1 2
(6 4) to be approximately
3 0 C m. u , so by Equation 15.6.9 [ ET 14.6.9], u (6 4) =
1 2
(6 4) · u =
1 2
(6 4)
1 2
+
(6 4)
1 . 2
Using our estimates from part (a),