Section 14.6 Directional Derivatives and the Gradient Vector

(1)

Section 14.6 Directional Derivatives and the Gradient Vector

446 ¤ CHAPTER 14 PARTIAL DERIVATIVES

7.  ( ) =  = ⁻¹ (a) ∇( ) = 

i+

j= ⁻¹i+ (−⁻²) j = 1

i− 

²j (b) ∇(2 1) =1

1i− 2

1² j= i − 2 j

(c) By Equation 9, u (2 1) = ∇(2 1) · u = (i − 2 j) ·3 5i+⁴₅j

=³₅ −⁸5 = −1.

8.  ( ) = ²ln  (a) ∇( ) = 

i+

j= 2 ln  i + (²) j (b) ∇(3 1) = 0 i + (91) j = 9 j

(c) By Equation 9, u (3 1) = ∇(3 1) · u = 9 j ·

−13⁵ i+¹²₁₃j

= 0 +¹⁰⁸₁₃ =¹⁰⁸₁₃. 9.  (  ) = ² − ³

(a) ∇(  ) = h^(  ) (  ) (  )i =

2 − ³ ² − ³ ² − 3² (b) ∇(2 −1 1) = h−4 + 1 4 − 2 −4 + 6i = h−3 2 2i

(c) By Equation 14, u (2 −1 1) = ∇(2 −1 1) · u = h−3 2 2i · 0⁴₅ −³5

= 0 +⁸₅ −⁶5 = ²₅.

10.  (  ) = ²^

(a) ∇(  ) = h^(  ) (  ) (  )i =

²^() ²· ^() + ^· 2 ²^()

=

³^ (² + 2)^ ³^ (b) ∇(0 1 −1) = h−1 2 0i

(c) u (0 1 −1) = ∇(0 1 −1) · u = h−1 2 0i ·3

13₁₃⁴¹²₁₃

= −13³ +₁₃⁸ + 0 = ₁₃⁵

11.  ( ) = ^sin  ⇒ ∇( ) = h^sin  ^cos i, ∇(0 3) =√ 3 2 ¹₂

, and a unit vector in the direction of v is u =√ ¹

(−6)²+8²h−6 8i = 10¹ h−6 8i =

−³5⁴₅, so

u (0 3) = ∇(0 3) · u =√ 3 2 ¹₂

·

−³5⁴₅

= −³10^√³+₁₀⁴ =⁴⁻³₁₀^√³.

12.  ( ) = 

²+ ² ⇒ ∇( ) =

(²+ ²)(1) − (2)

(²+ ²)²  0 − (2) (²+ ²)²



=

 ²− ²

(²+ ²)² − 2

(²+ ²)²

 ,

∇(1 2) =₃

25 −25⁴

, and a unit vector in the direction of v = h3 5i is u = ^√9+25¹ h3 5i =

√3 34√⁵

34

, so

u (1 2) = ∇(1 2) · u =₃

25 −25⁴

·

√3 34√⁵

34



=₂₅√⁹

34−₂₅²⁰^√₃₄ = −₂₅¹¹^√₃₄. 13. ( ) = √

 ⇒ ∇( ) =√

 i+

(2√

 )

j, ∇(2 4) = 2 i +¹2j, and a unit vector in the direction of v is u= √ ¹

2²+(−1)²(2 i − j) = ^√¹₅(2 i − j), so ^u(2 4) = ∇(2 4) · u = (2 i +¹2j) · ^√¹₅(2 i − j) =^√¹₅ 4 −¹2

= ₂√⁷ 5 or

7√ 5 10 .

7.  ( ) =  = ⁻¹ (a) ∇( ) =

i+

j= ⁻¹i+ (−⁻²) j = 1

i− 

²j (b) ∇(2 1) = 1

1i− 2

1²j= i − 2 j

(c) By Equation 9, u (2 1) = ∇(2 1) · u = (i − 2 j) ·3 5i+⁴₅j

= ³₅−⁸5 = −1.

8.  ( ) = ²ln  (a) ∇( ) =

i+

j= 2 ln  i + (²) j (b) ∇(3 1) = 0 i + (91) j = 9 j

(c) By Equation 9, u (3 1) = ∇(3 1) · u = 9 j ·

−13⁵ i+¹²13j

= 0 +¹⁰⁸13 = ¹⁰⁸13. 9.  (  ) = ² − ³

(a) ∇(  ) = h^(  ) (  ) (  )i =

2 − ³ ² − ³ ² − 3² (b) ∇(2 −1 1) = h−4 + 1 4 − 2 −4 + 6i = h−3 2 2i

(c) By Equation 14, u (2 −1 1) = ∇(2 −1 1) · u = h−3 2 2i · 0⁴₅ −³5

= 0 + ⁸₅−⁶5 =²₅.

10.  (  ) = ²^

(a) ∇(  ) = h^(  ) (  ) (  )i =

²^() ²· ^() + ^· 2 ²^()

=

³^ (² + 2)^ ³^ (b) ∇(0 1 −1) = h−1 2 0i

(c) u (0 1 −1) = ∇(0 1 −1) · u = h−1 2 0i ·₃

13₁₃⁴¹²₁₃

= −13³ +₁₃⁸ + 0 = ₁₃⁵

11.  ( ) = ^sin  ⇒ ∇( ) = h^sin  ^cos i, ∇(0 3) =√ 3 2 ¹₂

, and a unit vector in the direction of v is u = √ ¹

(−6)²+8²h−6 8i =10¹ h−6 8i =

−³5⁴₅ , so

u (0 3) = ∇(0 3) · u =√ 3 2 ¹₂

·

−³5⁴₅

= −³10^√³ +₁₀⁴ =⁴⁻³₁₀^√³.

12.  ( ) = 

²+ ² ⇒ ∇( ) =

(²+ ²)(1) − (2)

(²+ ²)²  0 − (2) (²+ ²)²



=

 ²− ²

(²+ ²)² − 2

(²+ ²)²

 ,

∇(1 2) =3 25 −25⁴

, and a unit vector in the direction of v = h3 5i is u = ^√9+25¹ h3 5i =

√3 34√⁵

34

, so

u (1 2) = ∇(1 2) · u =3 25 −25⁴

·

√3 34√⁵

34



= ₂₅^√⁹₃₄−₂₅²⁰^√₃₄= −₂₅¹¹^√₃₄.

13. ( ) = √

 ⇒ ∇( ) =√

 i+

(2√

 )

j, ∇(2 4) = 2 i +¹2j, and a unit vector in the direction of v is u=√ ¹

2²+(−1)²(2 i − j) =^√¹₅(2 i − j), so ^u(2 4) = ∇(2 4) · u = (2 i +¹2j) ·^√¹₅(2 i − j) = ^√¹₅ 4 −¹2

=₂^√⁷₅ or

7√ 5 10 .

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ¤ 447 14. ( ) = ²^− ⇒ ∇( ) =

2^− i+

−²^−

j, ∇(3 0) = 6 i − 9 j, and a unit vector in the direction of v is u =√ ¹

3²+4²(3 i + 4 j) = ¹₅(3 i + 4 j), so u(3 0) = ∇(3 0) · u = (6 i − 9 j) ·¹5(3 i + 4 j) = ¹₅(18 − 36) = −¹⁸5. 15.  (  ) = ² + ² ⇒ ∇(  ) =

2 ²+ 2 ²

, ∇(1 2 3) = h4 13 4i, and a unit vector in the direction of v is u = √ ¹

4+1+4h2 −1 2i = ¹3h2 −1 2i, so

u (1 2 3) = ∇(1 2 3) · u = h4 13 4i ·¹3h2 −1 2i = ¹3(8 − 13 + 8) = ³3 = 1.

16.  (  ) = ²tan⁻¹ ⇒ ∇(  ) =



²tan⁻¹ 2 tan⁻¹ ² 1 + ²

 ,

∇(2 1 1) =

1 ·^4 4 ·^4₁₊₁² 

=

4  1, and a unit vector in the direction of v is u =√ ¹

1+1+1h1 1 1i = ^√¹₃h1 1 1i, so u (2 1 1) = ∇(2 1 1) · u =

4  1

·^√¹₃h1 1 1i = ^√¹₃

4 +  + 1

= √¹ 3

5

4 + 1.

17. (  ) = ln(3 + 6 + 9) ⇒ ∇(  ) = h3(3 + 6 + 9) 6(3 + 6 + 9) 9(3 + 6 + 9)i,

∇(1 1 1) =1

6¹₃¹₂, and a unit vector in the direction of v = 4 i + 12 j + 6 k is u = √16+144+36¹ (4 i + 12 j + 6 k) = ²₇i+⁶₇j+³₇k, so

u(1 1 1) = ∇(1 1 1) · u =₁

6¹₃¹₂

·₂

7⁶₇³₇

=₂₁¹ +²₇ +₁₄³ =²³₄₂. 18. u (2 2) = ∇(2 2) · u, the scalar projection of ∇(2 2) onto u, so we draw a

perpendicular from the tip of ∇(2 2) to the line containing u. We can use the point (2 2) to determine the scale of the axes, and we estimate the length of the projection to be approximately 3.0 units. Since the angle between ∇(2 2) and u is greater than 90^◦, the scalar projection is negative. Thus u (2 2) ≈ −3.

19.  ( ) =

 ⇒ ∇( ) =

1

2()^−12()¹₂()^−12()

=

 

2

  2





, so ∇(2 8) = 1¹₄.

The unit vector in the direction of−−→

  = h5 − 2 4 − 8i = h3 −4i is u =3 5 −⁴5

, so

u (2 8) = ∇(2 8) · u = 1¹₄

·₃

5 −⁴5

=²₅.

20.  (  ) = ²³ ⇒ ∇(  ) =

²³ 2³ 3²²

, so ∇(2 1 1) = h1 4 6i. The unit vector in the direction of−−→

  = h−2 −4 4i is u = ^√4+16+16¹ h−2 −4 4i =¹6h−2 −4 4i, so

u (2 1 1) = ∇(2 1 1) · u = h1 4 6i ·¹6h−2 −4 4i = ¹6(−2 − 16 + 24) = 1.

21.  ( ) = 4√ ⇒ ∇( ) =

4 ·¹2^−12 4√

= h2√ 4√

 i.

∇(4 1) = h1 8i is the direction of maximum rate of change, and the maximum rate is |∇(4 1)| =√

1 + 64 =√65.

22.  ( ) = ^ ⇒ ∇( ) =

^() ^() + ^(1)

=

²^ ( + 1)^.

16 + 1 =√ 17.

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ¤ 447 14. ( ) = ²^− ⇒ ∇( ) =

2^− i+

−²^−

j, ∇(3 0) = 6 i − 9 j, and a unit vector in the direction of v is u =√ ¹

3²+4²(3 i + 4 j) = ¹₅(3 i + 4 j), so u(3 0) = ∇(3 0) · u = (6 i − 9 j) ·¹5(3 i + 4 j) = ¹₅(18 − 36) = −¹⁸5. 15.  (  ) = ² + ² ⇒ ∇(  ) =

2 ²+ 2 ²

, ∇(1 2 3) = h4 13 4i, and a unit vector in the direction of v is u = √4+1+4¹ h2 −1 2i = ¹3h2 −1 2i, so

u (1 2 3) = ∇(1 2 3) · u = h4 13 4i ·¹3h2 −1 2i = ¹3(8 − 13 + 8) = ³3 = 1.

16.  (  ) = ²tan⁻¹ ⇒ ∇(  ) =



²tan⁻¹ 2 tan⁻¹ ² 1 + ²

 ,

∇(2 1 1) =

1 ·^4 4 ·^4₁₊₁² 

=_

4  1

, and a unit vector in the direction of v is u =√1+1+1¹ h1 1 1i = ^√¹₃h1 1 1i, so u (2 1 1) = ∇(2 1 1) · u =

4  1

·^√¹₃h1 1 1i = ^√¹₃

4 +  + 1

= ^√¹₃5

4 + 1.

17. (  ) = ln(3 + 6 + 9) ⇒ ∇(  ) = h3(3 + 6 + 9) 6(3 + 6 + 9) 9(3 + 6 + 9)i,

∇(1 1 1) =₁

6¹₃¹₂, and a unit vector in the direction of v = 4 i + 12 j + 6 k is u = √ ¹

16+144+36 (4 i + 12 j + 6 k) = ²₇i+⁶₇j+³₇k, so

u(1 1 1) = ∇(1 1 1) · u =₁

6¹₃¹₂

·₂

7⁶₇³₇

=₂₁¹ +²₇ +₁₄³ =²³₄₂. 18. u (2 2) = ∇(2 2) · u, the scalar projection of ∇(2 2) onto u, so we draw a

perpendicular from the tip of ∇(2 2) to the line containing u. We can use the point (2 2) to determine the scale of the axes, and we estimate the length of the projection to be approximately 3.0 units. Since the angle between ∇(2 2) and u is greater than 90^◦, the scalar projection is negative. Thus u (2 2) ≈ −3.

19.  ( ) =

 ⇒ ∇( ) =

1

2()^−12()¹₂()^−12()

=

 

2

  2





, so ∇(2 8) = 1¹₄.

The unit vector in the direction of−−→

  = h5 − 2 4 − 8i = h3 −4i is u =3 5 −⁴5

, so

u (2 8) = ∇(2 8) · u = 1¹₄

·3 5 −⁴5

=²₅.

20.  (  ) = ²³ ⇒ ∇(  ) =

²³ 2³ 3²²

, so ∇(2 1 1) = h1 4 6i. The unit vector in the direction of−−→

  = h−2 −4 4i is u = ^√_4+16+16¹ h−2 −4 4i =¹6h−2 −4 4i, so

u (2 1 1) = ∇(2 1 1) · u = h1 4 6i ·¹6h−2 −4 4i = ¹6(−2 − 16 + 24) = 1.

21.  ( ) = 4√

 ⇒ ∇( ) =

4 ·¹2^−12 4√



= h2√

 4√

 i.

1 + 64 =√ 65.

22.  ( ) = ^ ⇒ ∇( ) =

^() ^() + ^(1)

=

²^ ( + 1)^.

16 + 1 =√ 17.

23.  ( ) = sin() ⇒ ∇( ) = h cos()  cos()i, ∇(1 0) = h0 1i. Thus the maximum rate of change is

|∇(1 0)| = 1 in the direction h0 1i.

24.  (  ) =  ln() ⇒ ∇(  ) =



ln()  · 

  · 





=



ln()







, ∇(1 2¹2) =

0¹₂ 2. Thus

the maximum rate of change is

∇(1 2¹2) =

0 +¹₄+ 4 =

17

4 =^√₂¹⁷ in the direction

0¹₂ 2or equivalently h0 1 4i.

25.  (  ) = ( + ) = ( + )⁻¹ ⇒

∇(  ) =

1( + ) −( + )⁻²(1) −( + )⁻²(1)

=

 1

 +  − 

( + )² −  ( + )²

 ,

∇(8 1 3) =1

4 −4⁸² −4⁸²

=1

4 −¹2 −¹2

. Thus the maximum rate of change is

|∇(8 1 3)| =

1

16+¹₄+¹₄ =

9

16= ³₄in the direction₁

4 −¹2 −¹2

or equivalently h1 −2 −2i.

26.  (  ) = arctan() ⇒ ∇(  ) =

 

1 + ()² 

1 + ()² 

1 + ()²



, ∇(1 2 1) =₂

5¹₅²₅. Thus the maximum rate of change is |∇(1 2 1)| =

4

25+₂₅¹ +₂₅⁴ =

9

25 =³₅ in the direction₂

5¹₅²₅

or equivalently h2 1 2i.

27. (a) As in the proof of Theorem 15, u = |∇| cos . Since the minimum value of cos  is −1 occurring when  = , the minimum value of uis − |∇| occurring when  = , that is when u is in the opposite direction of ∇

(assuming ∇ 6= 0).

(b) ( ) = ⁴ − ²³ ⇒ ∇( ) =

4³ − 2³ ⁴− 3²²

, so  decreases fastest at the point (2 −3) in the direction −∇(2 −3) = − h12 −92i = h−12 92i.

28.  ( ) = ²+ ³ ⇒ ∇( ) =

2 + ³ 3²

so ∇(2 1) = h5 6i. If u = h i is a unit vector in the desired direction then u (2 1) = 2 ⇔ h5 6i · h i = 2 ⇔ 5 + 6 = 2 ⇔  = ¹3−⁵6. But ²+ ²= 1 ⇔

²+1 3−⁵62

= 1 ⇔ ⁶¹36²−⁵9 +¹₉ = 1 ⇔ 61²− 20 − 32 = 0. By the quadratic formula, the solutions are

 = −(−20) ±

(−20)²− 4(61)(−32)

2(61) = 20 ±√

8208

122 = 10 ± 6√ 57

61 . If  =10 + 6√ 57

61 ≈ 09065 then

 =1 3−5

6

10 + 6√ 57 61



≈ −04221, and if  = 10 − 6√ 57

61 ≈ −05787 then  = 1 3−5

6

10 − 6√ 57 61



≈ 08156.

Thus the two directions giving a directional derivative of 2 are approximately h09065 −04221i and h−05787 08156i.

29. The direction of fastest change is ∇( ) = (2 − 2) i + (2 − 4) j, so we need to ﬁnd all points ( ) where ∇( ) is parallel to i + j ⇔ (2 − 2) i + (2 − 4) j =  (i + j) ⇔  = 2 − 2 and  = 2 − 4. Then 2 − 2 = 2 − 4 ⇒

 =  + 1so the direction of fastest change is i + j at all points on the line  =  + 1.

23.  ( ) = sin() ⇒ ∇( ) = h cos()  cos()i, ∇(1 0) = h0 1i. Thus the maximum rate of change is

|∇(1 0)| = 1 in the direction h0 1i.

24.  (  ) =  ln() ⇒ ∇(  ) =



ln()  · 

  · 





=



ln()







, ∇(1 2¹2) =

0¹₂ 2. Thus

the maximum rate of change is

∇(1 2¹2) =

0 +¹₄+ 4 =

17

4 =^√₂¹⁷ in the direction 0¹₂ 2

or equivalently h0 1 4i.

25.  (  ) = ( + ) = ( + )⁻¹ ⇒

∇(  ) =

1( + ) −( + )⁻²(1) −( + )⁻²(1)

=

 1

 +  − 

( + )² −  ( + )²

 ,

∇(8 1 3) =1

4 −4⁸² −4⁸²

=1

4 −¹2 −¹2

. Thus the maximum rate of change is

|∇(8 1 3)| =

1

16+¹₄+¹₄ =

9

16= ³₄in the direction₁

4 −¹2 −¹2

or equivalently h1 −2 −2i.

26.  (  ) = arctan() ⇒ ∇(  ) =

 

1 + ()² 

1 + ()² 

1 + ()²



, ∇(1 2 1) =₂

5¹₅²₅ . Thus

the maximum rate of change is |∇(1 2 1)| =

4

25+₂₅¹ +₂₅⁴ =

9

25 =³₅ in the direction2

5¹₅²₅or equivalently h2 1 2i.

27. (a) As in the proof of Theorem 15, u = |∇| cos . Since the minimum value of cos  is −1 occurring when  = , the minimum value of uis − |∇| occurring when  = , that is when u is in the opposite direction of ∇

(assuming ∇ 6= 0).

(b) ( ) = ⁴ − ²³ ⇒ ∇( ) =

4³ − 2³ ⁴− 3²²

, so  decreases fastest at the point (2 −3) in the direction −∇(2 −3) = − h12 −92i = h−12 92i.

28.  ( ) = ²+ ³ ⇒ ∇( ) =

2 + ³ 3²

so ∇(2 1) = h5 6i. If u = h i is a unit vector in the desired direction then u (2 1) = 2 ⇔ h5 6i · h i = 2 ⇔ 5 + 6 = 2 ⇔  = ¹3−⁵6. But ²+ ²= 1 ⇔

²+₁

3−⁵62

= 1 ⇔ ⁶¹36²−⁵9 +¹₉ = 1 ⇔ 61²− 20 − 32 = 0. By the quadratic formula, the solutions are

 = −(−20) ±

(−20)²− 4(61)(−32)

2(61) = 20 ±√

8208

122 = 10 ± 6√ 57

61 . If  =10 + 6√ 57

61 ≈ 09065 then

 =1 3−5

6

10 + 6√ 57 61



≈ −04221, and if  = 10 − 6√ 57

61 ≈ −05787 then  = 1 3−5

6

10 − 6√ 57 61



≈ 08156.

Thus the two directions giving a directional derivative of 2 are approximately h09065 −04221i and h−05787 08156i.

29. The direction of fastest change is ∇( ) = (2 − 2) i + (2 − 4) j, so we need to ﬁnd all points ( ) where ∇( ) is parallel to i + j ⇔ (2 − 2) i + (2 − 4) j =  (i + j) ⇔  = 2 − 2 and  = 2 − 4. Then 2 − 2 = 2 − 4 ⇒

 =  + 1so the direction of fastest change is i + j at all points on the line  =  + 1.

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ¤ 449 30. The ﬁsherman is traveling in the direction h−80 −60i. A unit vector in this direction is u =100¹ h−80 −60i =

−⁴5 −³5

,

and if the depth of the lake is given by ( ) = 200 + 002²− 0001³, then ∇( ) =

004 −0003² .

u (80 60) = ∇(80 60) · u = h32 −108i ·

−⁴5 −³5

= 392. Since u (80 60)is positive, the depth of the lake is

increasing near (80 60) in the direction toward the buoy.

31.  = 

²+ ²+ ² and 120 =  (1 2 2) = 

3so  = 360.

(a) u = h1 −1 1i√

3 ,

u (1 2 2) = ∇ (1 2 2) · u =

−360

²+ ²+ ²_−32

h  i

(122)· u = −⁴⁰3h1 2 2i ·^√¹₃h1 −1 1i = −₃⁴⁰^√₃ (b) From (a), ∇ = −360

²+ ²+ ²_−32

h  i, and since h  i is the position vector of the point (  ), the vector − h  i, and thus ∇ , always points toward the origin.

32. ∇ = −400^−²^−3²^−9²h 3 9i (a) u = √¹

6h1 −2 1i, ∇ (2 −1 2) = −400⁻⁴³h2 −3 18i and

u (2 −1 2) =



−400⁻⁴³

√6



(26) = −5200√ 6 3⁴³

◦Cm.

(b) ∇ (2 −1 2) = 400⁻⁴³h−2 3 −18i or equivalently h−2 3 −18i.

(c) |∇ | = 400^−²^{− 3}²^{− 9}²

²+ 9²+ 81^{2 ◦}Cm is the maximum rate of increase. At (2 −1 2) the maximum rate of increase is 400⁻⁴³√

337 ^◦Cm.

33. ∇ (  ) = h10 − 3 +   − 3 i, ∇ (3 4 5) = h38 6 12i (a) u (3 4 5) = h38 6 12i ·^√¹₃h1 1 −1i = ^√³²₃

(b) ∇ (3 4 5) = h38 6 12i, or equivalently, h19 3 6i.

(c) |∇ (3 4 5)| =√

38²+ 6²+ 12²=√

1624 = 2√ 406

34.  =  ( ) = 1000 − 0005²− 001² ⇒ ∇( ) = h−001 −002i and ∇(60 40) = h−06 −08i.

(a) Due south is in the direction of the unit vector u = −j and

u (60 40) = ∇ (60 40) · h0 −1i = h−06 −08i · h0 −1i = 08. Thus, if you walk due south from (60 40 966) you will ascend at a rate of 08 vertical meters per horizontal meter.

(b) Northwest is in the direction of the unit vector u =√¹

2h−1 1i and

u (60 40) = ∇ (60 40) ·^√¹₂h−1 1i = h−06 −08i ·^√¹₂h−1 1i = −^02^√₂ ≈ −014. Thus, if you walk northwest from (60 40 966) you will descend at a rate of approximately 014 vertical meters per horizontal meter.

1

(2)

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ¤ 451

38. If we place the initial point of the gradient vector ∇(4 6) at (4 6), the vector is perpendicular to the level curve of  that includes (4 6), so we sketch a portion of the level curve through (4 6) (using the nearby level curves as a guideline) and draw a line perpendicular to the curve at (4 6). The gradient vector is

parallel to this line, pointing in the direction of increasing function values, and with length equal to the maximum value of the directional derivative of  at (4 6). We can estimate this length by ﬁnding the average rate of change in the direction of the gradient. The line intersects the contour lines corresponding to

−2 and −3 with an estimated distance of 05 units. Thus the rate of change is approximately −2 − (−3)

05 = 2, and we sketch the gradient vector with length 2.

39.  ( ) = ³+ 5² + ³ ⇒

u ( ) = ∇( ) · u =

3²+ 10 5²+ 3²

·₃

5⁴₅

= ⁹₅²+ 6 + 4²+¹²₅²= ²⁹₅²+ 6 +¹²₅². Then

²u ( ) = u[u ( )] = ∇ [^u ( )] · u =58

5 + 6 6 +²⁴₅

·3 5⁴₅

= ¹⁷⁴₂₅ +¹⁸₅ +²⁴₅ +⁹⁶₂₅ = ²⁹⁴₂₅ +¹⁸⁶₂₅ and ²u (2 1) = ²⁹⁴₂₅(2) +¹⁸⁶₂₅(1) = ⁷⁷⁴₂₅.

40. (a) From Equation 9 we have u = ∇ · u = h^ i · h i = ^ + and from Exercise 39 we have

u² = u[u ] = ∇ [^u ] · u = h^ +   + i · h i = ^²+  +  + ². But = by Clairaut’s Theorem, so u² = ²+ 2 + ².

(b) ( ) = ^2 ⇒  = ^2, = 2^2,  = 0, = 2^2, = 4^2and a unit vector in the direction of v is u = √ ¹

4²+6²h4 6i =

√2 13√³

13



= h i. Then

u² = ²+ 2 + ²= 0 ·

√2 13

2

+ 2 · 2^2

√2 13

  3

√13



+ 4^2

√3 13

2

=²⁴₁₃^2+³⁶₁₃^2.

41. Let  (  ) = 2( − 2)²+ ( − 1)²+ ( − 3)². Then 2( − 2)²+ ( − 1)²+ ( − 3)²= 10is a level surface of  .

(  ) = 4( − 2) ⇒ ^(3 3 5) = 4, (  ) = 2( − 1) ⇒ ^(3 3 5) = 4, and

(  ) = 2( − 3) ⇒ ^(3 3 5) = 4.

(a) Equation 19 gives an equation of the tangent plane at (3 3 5) as 4( − 3) + 4( − 3) + 4( − 5) = 0 ⇔ 4 + 4 + 4 = 44or equivalently  +  +  = 11.

(b) By Equation 20, the normal line has symmetric equations − 3

4 =  − 3

4 =  − 5

4 or equivalently

 − 3 =  − 3 =  − 5. Corresponding parametric equations are  = 3 + ,  = 3 + ,  = 5 + .

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ¤ 453

47.  (  ) =  +  + ,

∇ (  ) = h +   +   + i,

∇ (1 1 1) = h2 2 2i, so an equation of the tangent plane is 2 + 2 + 2 = 6 or  +  +  = 3, and the normal line is given by  − 1 =  − 1 =  − 1 or

 =  = . To graph the surface we solve for :

 =3 − 

 +  .

48. (  ) = , ∇ (  ) = h  i,

∇ (1 2 3) = h6 3 2i, so an equation of the tangent plane is 6 + 3 + 2 = 18, and the normal line is given by − 1

6 = − 2

3 = − 3

2 or  = 1 + 6,  = 2 + 3,

 = 3 + 2. To graph the surface we solve for :  = 6

.

49.  ( ) =  ⇒ ∇( ) = h i, ∇(3 2) = h2 3i. ∇(3 2) is perpendicular to the tangent line, so the tangent line has equation

∇(3 2) · h − 3  − 2i = 0 ⇒ h2 3i · h − 3  − 2i = 0 ⇒ 2( − 3) + 3( − 2) = 0 or 2 + 3 = 12.

50. ( ) = ²+ ²− 4 ⇒ ∇( ) = h2 − 4 2i,

∇(1 2) = h−2 4i. ∇(1 2) is perpendicular to the tangent line, so the tangent line has equation ∇(1 2) · h − 1  − 2i = 0 ⇒ h−2 4i · h − 1  − 2i = 0 ⇒ −2( − 1) + 4( − 2) = 0 ⇔

−2 + 4 = 6 or equivalently − + 2 = 3.

51. ∇ (⁰ 0 0) =

20

² 20

² 20

²



. Thus an equation of the tangent plane at (0 0 0)is

20

²  +20

²  +20

²  = 2

²0

² +0²

² +²0

²



= 2(1) = 2since (0 0 0)is a point on the ellipsoid. Hence

0

²  +0

²  +0

²  = 1is an equation of the tangent plane.