Section 14.4 Tangent Planes and Linear Approximation

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Section 14.4 Tangent Planes and Linear Approximation

424 ¤ CHAPTER 14 PARTIAL DERIVATIVES

The surface and tangent plane are shown in the ﬁrst graph below. After zooming in, the surface and the tangent plane become almost indistinguishable, as shown in the second graph. (Here, the tangent plane is above the surface.) If we zoom in farther, the surface and the tangent plane will appear to coincide.

11.  ( ) = 1 +  ln( − 5). The partial derivatives are ^( ) =  · 1

 − 5() + ln( − 5) · 1 = 

 − 5 + ln( − 5) and ( ) =  · 1

 − 5() = ²

 − 5, so (2 3) = 6and (2 3) = 4. Both and are continuous functions for

  5, so by Theorem 8,  is differentiable at (2 3). By Equation 3, the linearization of  at (2 3) is given by

( ) =  (2 3) + (2 3)( − 2) + ^(2 3)( − 3) = 1 + 6( − 2) + 4( − 3) = 6 + 4 − 23.

12.  ( ) = √ = ()^12. The partial derivatives are ( ) = ¹₂()^−12() = 

2√ and

( ) = ¹₂()^−12() = 

2√, so (1 4) = 4 2√

4

= 1and (1 4) = 1 2√

4

= ¹₄. Both and are continuous functions for   0, so  is differentiable at (1 4) by Theorem 8. The linearization of  at (1 4) is

( ) =  (1 4) + (1 4)( − 1) + ^(1 4)( − 4) = 2 + 1( − 1) +¹4( − 4) =  +¹4.

13.  ( ) = ²^. The partial derivatives are ( ) = 2^and ( ) = ²^, so (1 0) = 2and (1 0) = 1. Both

and are continuous functions, so by Theorem 8,  is differentiable at (1 0). By Equation 3, the linearization of  at (1 0)is given by ( ) = (1 0) + (1 0)( − 1) + ^(1 0)( − 0) = 1 + 2( − 1) + 1( − 0) = 2 +  − 1.

14.  ( ) = 1 + 

1 +  = (1 + )(1 + )⁻¹. The partial derivatives are ( ) = (1 + )(−1)(1 + )⁻²= − 1 +  (1 + )² and

( ) = (1)(1 + )⁻¹= 1

1 + , so (1 3) = −1 and ^(1 3) =¹₂. Both and are continuous functions for

 6= −1, so  is differentiable at (1 3) by Theorem 8. The linearization of  at (1 3) is

( ) =  (1 3) + (1 3)( − 1) + ^(1 3)( − 3) = 2 + (−1)( − 1) +¹2( − 3) = − +¹2 +³₂.

15.  ( ) = 4 arctan(). The partial derivatives are ( ) = 4 · 1

1 + ()²() = 4

1 + ²², and

( ) = 4

1 + ²², so (1 1) = 2and (1 1) = 2. Both and are continuous functions, so  is differentiable at (1 1) by Theorem 8. The linearization of  at (1 1) is

( ) =  (1 1) + (1 1)( − 1) + ^(1 1)( − 1) = 4(4) + 2( − 1) + 2( − 1) = 2 + 2 +  − 4.

SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ¤ 425

16.  ( ) =  + sin(). The partial derivatives are ( ) = (1) cos()and ( ) = 1 + (−²) cos(), so

(0 3) =¹₃ and (0 3) = 1. Both and are continuous functions for  6= 0, so  is differentiable at (0 3), and the linearization of  at (0 3) is

( ) =  (0 3) + (0 3)( − 0) + ^(0 3)( − 3) = 3 +¹3( − 0) + 1( − 3) = ¹3 + .

17. Let ( ) = ^cos(). Then ( ) = ^[− sin()]() + ^cos() = ^[cos() −  sin()] and

( ) = ^[− sin()]() = −^sin(). Both and are continuous functions, so by Theorem 8,  is differentiable at (0 0). We have (0 0) = ⁰(cos 0 − 0) = 1, ^(0 0) = 0and the linear approximation of  at (0 0) is

 ( ) ≈ (0 0) + ^(0 0)( − 0) + ^(0 0)( − 0) = 1 + 1 + 0 =  + 1.

18. Let ( ) =  − 1

 + 1. Then ( ) = ( − 1)(−1)( + 1)⁻²= 1 − 

( + 1)² and ( ) = 1

 + 1. Both and are continuous functions for  6= −1, so by Theorem 8,  is differentiable at (0 0). We have ^(0 0) = 1, (0 0) = 1and the linear approximation of  at (0 0) is ( ) ≈ (0 0) + ^(0 0)( − 0) + ^(0 0)( − 0) = −1 + 1 + 1 =  +  − 1.

19. We can estimate (22 49) using a linear approximation of  at (2 5), given by

 ( ) ≈ (2 5) + ^(2 5)( − 2) + ^(2 5)( − 5) = 6 + 1( − 2) + (−1)( − 5) =  −  + 9. Thus

 (22 49) ≈ 22 − 49 + 9 = 63.

20.  ( ) = 1 −  cos  ⇒ ^( ) = − cos  and

( ) = −[(− sin ) + (cos )(1)] =  sin  −  cos , so ^(1 1) = 1, (1 1) = 1. Then the linear approximation of  at (1 1) is given by

 ( ) ≈ (1 1) + ^(1 1)( − 1) + ^(1 1)( − 1)

= 2 + (1)( − 1) + (1)( − 1) =  +  Thus (102 097) ≈ 102 + 097 = 199. We graph  and its

tangent plane near the point (1 1 2) below. Notice near  = 1 the surfaces are almost identical.

21.  (  ) =

²+ ²+ ² ⇒ ^(  ) = 

²+ ²+ ², (  ) = 

²+ ²+ ², and

(  ) = 

²+ ²+ ², so (3 2 6) = ³₇, (3 2 6) = ²₇, (3 2 6) = ⁶₇. Then the linear approximation of 

16.  ( ) =  + sin(). The partial derivatives are ( ) = (1) cos()and ( ) = 1 + (−²) cos(), so

(0 3) =¹₃ and (0 3) = 1. Both and are continuous functions for  6= 0, so  is differentiable at (0 3), and the linearization of  at (0 3) is

( ) =  (0 3) + (0 3)( − 0) + ^(0 3)( − 3) = 3 +¹3( − 0) + 1( − 3) = ¹3 + .

17. Let ( ) = ^cos(). Then ( ) = ^[− sin()]() + ^cos() = ^[cos() −  sin()] and

( ) = ^[− sin()]() = −^sin(). Both and are continuous functions, so by Theorem 8,  is differentiable at (0 0). We have (0 0) = ⁰(cos 0 − 0) = 1, ^(0 0) = 0and the linear approximation of  at (0 0) is

 ( ) ≈ (0 0) + ^(0 0)( − 0) + ^(0 0)( − 0) = 1 + 1 + 0 =  + 1.

18. Let ( ) =  − 1

 + 1. Then ( ) = ( − 1)(−1)( + 1)⁻²= 1 − 

( + 1)² and ( ) = 1

 + 1. Both and are continuous functions for  6= −1, so by Theorem 8,  is differentiable at (0 0). We have ^(0 0) = 1, (0 0) = 1and the linear approximation of  at (0 0) is ( ) ≈ (0 0) + ^(0 0)( − 0) + ^(0 0)( − 0) = −1 + 1 + 1 =  +  − 1.

19. We can estimate (22 49) using a linear approximation of  at (2 5), given by

 ( ) ≈ (2 5) + ^(2 5)( − 2) + ^(2 5)( − 5) = 6 + 1( − 2) + (−1)( − 5) =  −  + 9. Thus

 (22 49) ≈ 22 − 49 + 9 = 63.

20.  ( ) = 1 −  cos  ⇒ ^( ) = − cos  and

( ) = −[(− sin ) + (cos )(1)] =  sin  −  cos , so ^(1 1) = 1, (1 1) = 1. Then the linear approximation of  at (1 1) is given by

 ( ) ≈ (1 1) + ^(1 1)( − 1) + ^(1 1)( − 1)

= 2 + (1)( − 1) + (1)( − 1) =  +  Thus (102 097) ≈ 102 + 097 = 199. We graph  and its

tangent plane near the point (1 1 2) below. Notice near  = 1 the surfaces are almost identical.

21.  (  ) =

²+ ²+ ² ⇒ ^(  ) = 

²+ ²+ ², (  ) = 

²+ ²+ ², and

(  ) = 

²+ ²+ ², so (3 2 6) = ³₇, (3 2 6) = ²₇, (3 2 6) = ⁶₇. Then the linear approximation of 

at (3 2 6) is given by

 (  ) ≈ (3 2 6) + ^(3 2 6)( − 3) + ^(3 2 6)( − 2) + ^(3 2 6)( − 6)

= 7 +³₇( − 3) +²7( − 2) +⁶7( − 6) =³7 +²₇ +⁶₇ Thus

(302)²+ (197)²+ (599)²=  (302 197 599) ≈ ³7(302) +²₇(197) +⁶₇(599) ≈ 69914.

22. From the table, (40 20) = 28. To estimate (40 20)and (40 20)we follow the procedure used in Exercise 14.3.4. Since

(40 20) = lim

→0

 (40 +  20) − (40 20)

 , we approximate this quantity with  = ±10 and use the values given in the table:

(40 20) ≈  (50 20) − (40 20)

10 =40 − 28

10 = 12, (40 20) ≈ (30 20) − (40 20)

−10 = 17 − 28

−10 = 11 Averaging these values gives (40 20) ≈ 115. Similarly, ^(40 20) = lim

→0

 (40 20 + ) − (40 20)

 , so we use  = 10

and  = −5:

(40 20) ≈  (40 30) − (40 20)

10 =31 − 28

10 = 03, (40 20) ≈ (40 15) − (40 20)

−5 = 25 − 28

−5 = 06 Averaging these values gives (40 15) ≈ 045. The linear approximation, then, is

 ( ) ≈ (40 20) + ^(40 20)( − 40) + ^(40 20)( − 20) ≈ 28 + 115( − 40) + 045( − 20)

When  = 43 and  = 24, we estimate (43 24) ≈ 28 + 115(43 − 40) + 045(24 − 20) = 3325, so we would expect the wave heights to be approximately 3325 ft.

23. From the table, (94 80) = 127. To estimate (94 80)and (94 80)we follow the procedure used in Section 14.3. Since

(94 80) = lim

→0

 (94 +  80) − (94 80)

(94 80) ≈  (96 80) − (94 80)

2 =135 − 127

2 = 4, (94 80) ≈  (92 80) − (94 80)

−2 =119 − 127

−2 = 4

Averaging these values gives (94 80) ≈ 4. Similarly, ^(94 80) = lim

→0

 (94 80 + ) − (94 80)

 , so we use  = ±5:

(94 80) ≈ (94 85) − (94 80)

5 = 132 − 127

5 = 1, (94 80) ≈  (94 75) − (94 80)

−5 =122 − 127

−5 = 1

Averaging these values gives (94 80) ≈ 1. The linear approximation, then, is

 ( ) ≈ (94 80) + ^(94 80)( − 94) + ^(94 80)( − 80)

≈ 127 + 4( − 94) + 1( − 80) [or 4 +  − 329]

Thus when  = 95 and  = 78, (95 78) ≈ 127 + 4(95 − 94) + 1(78 − 80) = 129, so we estimate the heat index to be approximately 129^◦F.

24. From the table, (−15 50) = −29. To estimate ^(−15 50) and ^(−15 50) we follow the procedure used in Section 14.3.

Since (−15 50) = lim

→0

 (−15 +  50) − (−15 50)

(−15 50) ≈  (−10 50) − (−15 50)

5 = −22 − (−29)

5 = 14

(−15 50) ≈ (−20 50) − (−15 50)

−5 = −35 − (−29)

−5 = 12

Averaging these values gives (−15 50) ≈ 13. Similarly ^(−15 50) = lim

→0

 (−15 50 + ) − (−15 50)

 ,

so we use  = ±10:

(−15 50) ≈ (−15 60) − (−15 50)

10 = −30 − (−29)

10 = −01

(−15 50) ≈ (−15 40) − (−15 50)

−10 = −27 − (−29)

−10 = −02

Averaging these values gives (−15 50) ≈ −015. The linear approximation to the wind-chill index function, then, is

 ( ) ≈ (−15 50) + ^(−15 50)( − (−15)) + ^(−15 50)( − 50) ≈ −29 + (13)( + 15) − (015)( − 50).

Thus when  = −17^◦Cand  = 55 kmh, (−17 55) ≈ −29 + (13)(−17 + 15) − (015)(55 − 50) = −3235, so we estimate the wind-chill index to be approximately −3235^◦C.

25.  = ^−2cos 2 ⇒

 =

 +

 = ^−2(−2) cos 2  + ^−2(− sin 2)(2)  = −2^−2cos 2  − 2^−2sin 2 

26.  =

²+ 3² = (²+ 3²)^12 ⇒

 =

 +

 = ¹₂(²+ 3²)^−12(2)  +¹₂(²+ 3²)^−12(6)  = 

²+ 3²  + 3

²+ 3² 

27.  = ⁵³ ⇒  = 

  +

  = 5⁴³ + 3⁵²

28.  = 

1 +  ⇒

 = 

 +

  +



= (−1)(1 + )⁻²()  +1(1 + ) − ()

(1 + )²  + (−1)(1 + )⁻²() 

= − ²

(1 + )² + 1

(1 + )² − ² (1 + )²

29.  = ²cos  ⇒  = 

 +

  +

  = ²cos   + 2 cos   − ²sin  

30.  = ^−²^−² ⇒

 = 

 +

 +

  = ^−²^−² + ^−²^−²(−2)  + [ · ^−²^−²(−2) + ^−²^−²· 1] 

= ^−²^−² − 2^−²^−² + (1 − 2²)^−²^−²

31.  = ∆ = 005,  = ∆ = 01,  = 5²+ ², = 10, = 2. Thus when  = 1 and  = 2,

 = (1 2)  + (1 2)  = (10)(005) + (4)(01) = 09while

∆ =  (105 21) − (1 2) = 5(105)²+ (21)²− 5 − 4 = 09225.

32.  = ∆ = −004,  = ∆ = 005,  = ²−  + 3², = 2 − , ^= 6 − . Thus when  = 3 and  = −1,

 = (7)(−004) + (−9)(005) = −073 while ∆ = (296)²− (296)(−095) + 3(−095)²− (9 + 3 + 3) = −07189.

33.  = 

 +

 =   +  and |∆| ≤ 01, |∆| ≤ 01. We use  = 01,  = 01 with  = 30,  = 24; then the maximum error in the area is about  = 24(01) + 30(01) = 54 cm².

34. Let  be the volume. Then  = ²and ∆ ≈  = 2  + ²is an estimate of the amount of metal. With

 = 005and  = 02 we get  = 2(2)(10)(005) + (2)²(02) = 280 ≈ 88 cm³.

35. The volume of a can is  = ²and ∆ ≈  is an estimate of the amount of tin. Here  = 2  + ², so put

 = 004,  = 008 (004 on top, 004 on bottom) and then ∆ ≈  = 2(48)(004) + (16)(008) ≈ 1608 cm³. Thus the amount of tin is about 16 cm³.

36.  = 1312 + 06215 − 1137^016+ 03965 ^016, so the differential of  is

 = 

  +

  = (06215 + 03965^016)  +

−1137(016)^−084+ 03965 (016)^−084



= (06215 + 03965^016)  + (−18192 + 006344 )^−084

Here we have |∆ | ≤ 1, |∆| ≤ 2, so we take  = 1,  = 2 with  = −11,  = 26. The maximum error in the calculated value of  is about  = (06215 + 03965(26)^016)(1) + (−18192 + 006344(−11))(26)^−084(2) ≈ 096.

37.  = 

2²+ ², so the differential of  is

 = 

 +

  = (2²+ ²)() − (2)

(2²+ ²)²  +(2²+ ²)(0) − (4) (2²+ ²)² 

=(2²− ²)

(2²+ ²)²  − 4

(2²+ ²)² 

Here we have ∆ = 01 and ∆ = 01, so we take  = 01,  = 01 with  = 3,  = 07. Then the change in the tension  is approximately

 = [2(07)²− (3)²]

[2(07)²+ (3)²]² (01) − 4(3)(07) [2(07)²+ (3)²]²(01)

= −0802

(998)² −084

(998)² = − 1642

996004 ≈ −00165

Because the change is negative, tension decreases.

SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ¤ 429 38. Here  = ∆ = 03,  = ∆ = −5,  = 831

, so

 =

831





 −831 · 

²  = 831



−5 12−310

144· 3 10



≈ −883. Thus the pressure will drop by about 883 kPa.

39. First we ﬁnd 

1

implicitly by taking partial derivatives of both sides with respect to 1:



1

1





=  [(11) + (12) + (13)]

1 ⇒ −⁻² 

1 = −⁻²1 ⇒ 

1

=²

²₁. Then by symmetry,



2

=²

²₂, 

3

= ²

²₃. When 1= 25, 2= 40and 3= 50, 1

= 17

200 ⇔  = ²⁰⁰17 Ω. Since the possible error for each is 05%, the maximum error of  is attained by setting ∆= 0005. So

∆ ≈  = 

1

∆1+ 

2

∆2+ 

3

∆3 = (0005)²

 1

1

+ 1

2

+ 1

3



= (0005) =₁₇¹ ≈ 0059 Ω.

40. The errors in measurement are at most 2%, so



∆





 ≤ 002 and



∆





 ≤ 002. The relative error in the calculated surface

area is

∆

 ≈ 

 =01091(0425^0425⁻¹)^0725 + 01091^0425(0725^0725⁻¹) 

01091^0425^0725 = 0425

 + 0725

 To estimate the maximum relative error, we use

 =



∆





 = 002 and



 =



∆





 = 002 ⇒



 = 0425 (002) + 0725 (002) = 0023. Thus the maximum percentage error is approximately 23%.

41. (a) ( ) = ² ⇒ ( ) = 1² and ( ) = −2³. Since

  0, both and are continuous functions, so  is differentiable at (23 110). We have (23 110) = 23(110)²≈ 1901, ^(23 110) = 1(110)² ≈ 08264, and

(23 110) = −2(23)(110)³≈ −3456, so the linear approximation of  at (23 110) is

 ( ) ≈ (23 110)+^(23 110)(−23)+^(23 110)(−110) ≈ 1901+08264(−23)−3456(−110) or ( ) ≈ 08264 − 3456 + 3802.

(b) From part (a), for values near  = 23 and  = 110, ( ) ≈ 08264 − 3456 + 3802. If  increases by 1 kg to 24 kg and  increases by 003 m to 113 m, we estimate the BMI to be

(24 113) ≈ 08264(24) − 3456(113) + 3802 ≈ 18801. This is very close to the actual computed BMI:

(24 113) = 24(113)²≈ 18796.

42. r1() =

2 + 3 1 − ² 3 − 4 + ²

⇒ r⁰1() = h3 −2 −4 + 2i, r²() =

1 + ² 2³− 1 2 + 1

⇒ r⁰₂() =

2 6² 2

. Both curves pass through  since r1(0) = r2(1) = h2 1 3i, so the tangent vectors r⁰1(0) = h3 0 −4i and r⁰2(1) = h2 6 2i are both parallel to the tangent plane to  at  . A normal vector for the tangent plane is

r⁰₁(0) × r⁰2(1) = h3 0 −4i × h2 6 2i = h24 −14 18i, so an equation of the tangent plane is 24( − 2) − 14( − 1) + 18( − 3) = 0 or 12 − 7 + 9 = 44.

43. ∆ =  ( + ∆  + ∆) − ( ) = ( + ∆)²+ ( + ∆)²− (²+ ²)

= ²+ 2 ∆ + (∆)²+ ²+ 2 ∆ + (∆)²− ²− ²= 2 ∆ + (∆)²+ 2 ∆ + (∆)²

But ( ) = 2and ( ) = 2and so ∆ = ( ) ∆ + ( ) ∆ + ∆ ∆ + ∆ ∆, which is Deﬁnition 7 with 1= ∆and 2= ∆. Hence  is differentiable.

44. ∆ =  ( + ∆  + ∆) − ( ) = ( + ∆)( + ∆) − 5( + ∆)²− ( − 5²)

=  +  ∆ +  ∆ + ∆ ∆ − 5²− 10 ∆ − 5(∆)²−  + 5²

= ( − 10) ∆ +  ∆ + ∆ ∆ − 5 ∆ ∆,

but ( ) = and ( ) =  − 10 and so ∆ = ^( ) ∆ + ( ) ∆ + ∆ ∆ − 5∆ ∆, which is Deﬁnition 7 with 1= ∆and 2= −5 ∆. Hence  is differentiable.

45. To show that  is continuous at ( ) we need to show that lim

()→() ( ) =  ( )or equivalently lim

(∆∆)→(00) ( + ∆  + ∆) =  ( ). Since  is differentiable at ( ),

 ( + ∆  + ∆) − ( ) = ∆ = ^( ) ∆ + ( ) ∆ + 1∆ + 2∆, where 1and 2→ 0 as

(∆ ∆) → (0 0). Thus ( + ∆  + ∆) = ( ) + ^( ) ∆ + ( ) ∆ + 1∆ + 2∆. Taking the limit of both sides as (∆ ∆) → (0 0) gives lim

(∆∆)→(00) ( + ∆  + ∆) =  ( ). Thus  is continuous at ( ).

46. (a) lim

→0

 ( 0) − (0 0)

 = lim

→0

0 − 0

 = 0and lim

→0

 (0 ) − (0 0)

 = lim

→0

0 − 0

 = 0. Thus (0 0) = (0 0) = 0.

To show that  isn’t differentiable at (0 0) we need only show that  is not continuous at (0 0) and apply Exercise 45. As ( ) → (0 0) along the -axis ( ) = 0²= 0for  6= 0 so ( ) → 0 as ( ) → (0 0) along the -axis. But as ( ) → (0 0) along the line  = , ( ) = ²

2²

= ¹₂for  6= 0 so ( ) →¹2 as ( ) → (0 0) along this line. Thus lim

()→(00) ( )doesn’t exist, so  is discontinuous at (0 0) and thus not differentiable there.

(b) For ( ) 6= (0 0), ^( ) =(²+ ²) − (2)

(²+ ²)² = (²− ²)

(²+ ²)². If we approach (0 0) along the -axis, then

( ) = (0 ) = ³

⁴ = 1

, so ( ) → ±∞ as ( ) → (0 0). Thus lim

()→(00)( )does not exist and

( )is not continuous at (0 0) Similarly, ( ) =(²+ ²) − (2)

(²+ ²)² =(²− ²)

(²+ ²)² for ( ) 6= (0 0), and if we approach (0 0) along the -axis, then ( ) = ( 0) = ³

⁴ = 1

. Thus lim

()→(00)( )does not exist and

( )is not continuous at (0 0)

1

(2)

r⁰1(0) × r⁰2(1) = h3 0 −4i × h2 6 2i = h24 −14 18i, so an equation of the tangent plane is 24( − 2) − 14( − 1) + 18( − 3) = 0 or 12 − 7 + 9 = 44.

43. ∆ =  ( + ∆  + ∆) − ( ) = ( + ∆)²+ ( + ∆)²− (²+ ²)

= ²+ 2 ∆ + (∆)²+ ²+ 2 ∆ + (∆)²− ²− ²= 2 ∆ + (∆)²+ 2 ∆ + (∆)²

But ( ) = 2and ( ) = 2and so ∆ = ( ) ∆ + ( ) ∆ + ∆ ∆ + ∆ ∆, which is Deﬁnition 7 with 1= ∆and 2= ∆. Hence  is differentiable.

44. ∆ =  ( + ∆  + ∆) − ( ) = ( + ∆)( + ∆) − 5( + ∆)²− ( − 5²)

=  +  ∆ +  ∆ + ∆ ∆ − 5²− 10 ∆ − 5(∆)²−  + 5²

= ( − 10) ∆ +  ∆ + ∆ ∆ − 5 ∆ ∆,

but ( ) = and ( ) =  − 10 and so ∆ = ^( ) ∆ + ( ) ∆ + ∆ ∆ − 5∆ ∆, which is Deﬁnition 7 with 1= ∆and 2= −5 ∆. Hence  is differentiable.

45. To show that  is continuous at ( ) we need to show that lim

()→() ( ) =  ( )or equivalently lim

(∆∆)→(00) ( + ∆  + ∆) =  ( ). Since  is differentiable at ( ),

 ( + ∆  + ∆) − ( ) = ∆ = ^( ) ∆ + ( ) ∆ + 1∆ + 2∆, where 1and 2→ 0 as

(∆ ∆) → (0 0). Thus ( + ∆  + ∆) = ( ) + ^( ) ∆ + ( ) ∆ + 1∆ + 2∆. Taking the limit of both sides as (∆ ∆) → (0 0) gives lim

(∆∆)→(00) ( + ∆  + ∆) =  ( ). Thus  is continuous at ( ).

46. (a) lim

→0

 ( 0) − (0 0)

 = lim

→0

0 − 0

 = 0and lim

→0

 (0 ) − (0 0)

 = lim

→0

0 − 0

 = 0. Thus (0 0) = (0 0) = 0.

To show that  isn’t differentiable at (0 0) we need only show that  is not continuous at (0 0) and apply Exercise 45. As ( ) → (0 0) along the -axis ( ) = 0²= 0for  6= 0 so ( ) → 0 as ( ) → (0 0) along the -axis. But as ( ) → (0 0) along the line  = , ( ) = ²

2²

= ¹₂for  6= 0 so ( ) →¹2 as ( ) → (0 0) along this line. Thus lim

()→(00) ( )doesn’t exist, so  is discontinuous at (0 0) and thus not differentiable there.

(b) For ( ) 6= (0 0), ^( ) =(²+ ²) − (2)

(²+ ²)² = (²− ²)

(²+ ²)². If we approach (0 0) along the -axis, then

( ) = (0 ) = ³

⁴ = 1

, so ( ) → ±∞ as ( ) → (0 0). Thus lim

()→(00)( )does not exist and

( )is not continuous at (0 0) Similarly, ( ) =(²+ ²) − (2)

(²+ ²)² =(²− ²)

(²+ ²)² for ( ) 6= (0 0), and if we approach (0 0) along the -axis, then ( ) = ( 0) = ³

⁴ = 1

. Thus lim

()→(00)( )does not exist and

( )is not continuous at (0 0)