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 Theorem15.7.2 [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 pointsabove 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 thelim
)

(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),we have

(6 4)

(3 5)

+ ( 3 0)

=

1 2 2

0 35. This means that as we

move through the point (6 4) in the direction of u, the temperature increases at a rate of approximately 0 35 C m. 6+ Alternatively, we can use DeÞnition 15.6.2 [ ET 14.6.2]: which we can estimate with (6 4) (4 2) 2 [ ( = ±2 2. Then
u

(6 4) = lim

1 2

4+

1 2

(6 4) ,

0

u

(6 4)

(8 6) (64) 80 = 2 2 2

80 = 0, 2
u

u

(6 4) 74 80 3 = = . Averaging these values, we have 2 2 2 2 )] = lim ( (6 4) + ) ( ) , so (6 4) = lim

(6 4)

3 2 2

1 1 C m.

(c)

(

)=

(6 4 + )

(6 4) (6 6) and

0

0

which we can (6 2). If we

estimate with use

= ±2. We have

3 5 from part (a), but we will also need values for

= ±2 and the values given in the table, we have(8 6) 2 (6 6) = 80 2 75 = 2 5, (6 6) (4 6) 2 (6 6) = 68 2 75 = 3 5.

(6 6)

Averaging these values, we estimate

(6 6)

3 0. Similarly,

F.
CHAPTER 15 REVIEW ET CHAPTER 14 ¤ 215

(6 2)

(8 2) 2

(6 2)

=

90 2

87 (6 2)

= 1 5,

(6 2)

(4 2) 2

(6 2)

=

74 2

87

= 6 5.

Averaging these values, we estimate (6 4) (6 6) 2 (6 4) =

4 0. Finally, weestimate = 0 25. 0 25, (6 4)

(6 4): (6 2) 2 (6 4) = 40 2 35 = 0 25.

30 2 (6 4)
2

35

Averaging these values, we have
13.

(

)=

2 +
1

2

= 1 (2 + 2 = tan
1

)

1 2

(2) =

1 2 +

2

,

= 1 (2 + 2

2

)

1 2

(2 ) =

2 +

2

15.

(

)=

tan

,

=

1+

2

17.

( ( ( = ( =

) = ln( + )=4 )= 1) = + + =
2 1 3 2

)

= ln( + = 12 =2 2

), = ,

= 2 =
2

+ ,
1

,

= = 24 , , = =

+ = 2 ,
1 1 1

19. 21.

,

= ,

= = (

2 1)
2 1

1

,
1

,
1

=

(

1)

,

=

,

=

=

,

23.

=

= + (1 +

+ + =

,

= = 2

+

and + (1 + + = + + = + .

25. (a)

=6 +2 1 = 8(

2) = 8 and

2) = 4, so an equation of the tangent plane is

1) + 4( + 2) or

= 8 + 4 + 1. 2...
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