Section 14.6 Directional Derivatives and the Gradient Vector

Section 14.6 Directional Derivatives and the Gradient Vector

12. Let f (x, y, z) = y²e^xyz, P (0, 1, −1), u =₃

13,₁₃⁴,¹²₁₃. (a) Find the gradient of f .

(b) Evaluate the gradient at the point P .

(c) Find the rate of change of f at P in the direction of the vector u.

Solution:

1412 ¤ CHAPTER 14 PARTIAL DERIVATIVES 7. ( ) = arctan() ⇒ ^( ) = 

1 + ()² and ( ) = 

1 + ()². If u is a unit vector in the direction

 = 34, then from Equation 6,

u (2 −3) = ^(2 −3) cos

3

4



+ (2 −3) sin

3

4



= − 3 37



−

√2 2

 + 2

37

 √2 2



= 5√ 2 74 . 8. ( ) = ²^

(a) ∇( ) = 

i+

j= 2^i+ ²^j (b) ∇(3 0) = 2(3)⁰i+ 3²⁰j= 6 i + 9 j

(c) By Equation 9, u (3 0) = ∇(3 0) · u = (6i + 9 j) ·3 5i−⁴5j

= ¹⁸₅ −³⁶5 = −¹⁸5. 9. ( ) =  = ⁻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.

10. ( ) = 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. 11. (  ) = 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.

12. (  ) = 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 13. ( ) = ^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.

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17. Find the directional derivative of the function at the given point in the direction of the vector v.

f (x, y, z) = x²y + y²z, (1, 2, 3), v =< 2, −1, 2 >

Solution:

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.  (  ) = ^+ ^+ ^ ⇒ ∇(  ) = h^+ ^ ^+ ^ ^+ ^i, ∇(0 0 0) = h1 1 1i, and a unit vector in the direction of v is u = √ ¹

25+1+4h5 1 −2i = ^√1₃₀h5 1 −2i, so

u (0 0 0) = ∇(0 0 0) · u = h1 1 1i ·^√1₃₀h5 1 −2i = ^√4₃₀.

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

1

 − + 

²



, ∇(1 1 −1) = h−1 −1 −2i. Thus the maximum rate of

change is |∇(1 1 −1)| =√

1 + 1 + 4 =√

6in the direction h−1 −1 −2i.

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32. Find the maximum rate of change of f at the given point and the direction in which it occurs.

f (p, q, r) = arctan(pqr), (1, 2, 1) Solution:

448 ¤ CHAPTER 14 PARTIAL DERIVATIVES

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 = ^√₂¹⁷in 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 =³₄ in 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= ³₅in 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.

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38. The temperature at a point (x, y, z) is given by

T (x, y, z) = 200e^{−x2−3y2−9z2} where T is measured in^◦C and x, y, z in meters.

(a) Find the rate of change of temperature at the point P (2, −1, 2) in the direction toward the point (3, −3, 3).

(b) In which direction does the temperature increase fastest at P ? (c) Find the maximum rate of increase at P .

Solution:

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 = −₃^40√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2}h 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.

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51. Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point.

x + y + z = e^xyz, (0, 0, 1) Solution:

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ¤ 1419

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

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

(3 1 −1) = −2.

(a) By Equation 19, an equation of the tangent plane at (3 1 −1) is (−1)( − 3) + 2( − 1) + (−2)[ − (−1)] = 0 or

− + 2 − 2 = 1 or  − 2 + 2 = −1.

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

−1 =  − 1

2 =  − (−1)

−2 or equivalently

 − 3 =  − 1

−2 = + 1

2 and parametric equations  = 3 − ,  = 1 + 2,  = −1 − 2.

49.Let  (  ) = ²³. Then ²³= 8is a level surface of  and ∇ (  ) =

²³ 2³ 3²². (a) ∇ (2 2 1) = h4 8 24i is a normal vector for the tangent plane at (2 2 1), so an equation of the tangent plane is

4( − 2) + 8( − 2) + 24( − 1) = 0 or 4 + 8 + 24 = 48 or equivalently  + 2 + 6 = 12.

(b) The normal line has direction ∇ (2 2 1) = h4 8 24i or equivalently h1 2 6i, so parametric equations are  = 2 + ,

 = 2 + 2,  = 1 + 6, and symmetric equations are  − 2 = − 2

2 = − 1 6 .

50.Let  (  ) =  +  + . Then  +  +  = 5 is a level surface of  and ∇ (  ) = h +   +   + i.

(a) ∇ (1 2 1) = h3 2 3i is a normal vector for the tangent plane at (1 2 1), so an equation of the tangent plane is 3( − 1) + 2( − 2) + 3( − 1) = 0 or 3 + 2 + 3 = 10.

(b) The normal line has direction h3 2 3i, so parametric equations are  = 1 + 3,  = 2 + 2,  = 1 + 3, and symmetric equations are  − 1

2 = − 2

1 = − 1 3 .

51.Let  (  ) =  +  +  − ^. Then  +  +  = ^is the level surface  (  ) = 0, and ∇ (  ) = h1 − ^ 1 − ^ 1 − ^i.

(a) ∇ (0 0 1) = h1 1 1i is a normal vector for the tangent plane at (0 0 1), so an equation of the tangent plane is 1( − 0) + 1( − 0) + 1( − 1) = 0 or  +  +  = 1.

(b) The normal line has direction h1 1 1i, so parametric equations are  = ,  = ,  = 1 + , and symmetric equations are

 =  =  − 1.

52.Let  (  ) = ⁴+ ⁴+ ⁴− 3²²². Then ⁴+ ⁴+ ⁴= 3²²²is the level surface  (  ) = 0, and ∇ (  ) =

4³− 6²² 4³− 6²² 4³− 6²².

(a) ∇ (1 1 1) = h−2 −2 −2i or equivalently h1 1 1i is a normal vector for the tangent plane at (1 1 1), so an equation of the tangent plane is 1( − 1) + 1( − 1) + 1( − 1) = 0 or  +  +  = 3.

(b) The normal line has direction h1 1 1i, so parametric equations are  = 1 + ,  = 1 + ,  = 1 + , and symmetric equations are  − 1 =  − 1 =  − 1 or equivalently  =  = .

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