www.elsolucionario.org 13.6. DIRECTIONAL DERIVATIVE 107 (c) xfx + yfy = x(8xy 3 − 3y 4 + 5x4 ) + y(12x2 y 2 − 12xy 3 ) = 8x2 y 3 − 3xy 4 + 5x5 + 12x2 y 3 − 12xy 4 = 20x2 − 15xy 4 + 5x5 = 5(4x2 y 3 − 3xy 4 + x5 ) = 5f (x, y) y y λy = λ0 f , we see that z = f (d) By observing that f =f λx x x of degree zero. 13.6 y x is homogeneous Directional Derivative 1. ∇f = (2x − 3x2 y 2 )i + (4y 3 − 2x3 y)j 2 2 2. ∇f = 4xye−2x y i + (1 + 2x2 e−2x y )j 3. ∇F = y2 2xy 3xy 2 i + j − k z3 z3 z4 4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j − xy 2 sin yzk 5. ∇f = 2xi − 8yj; ∇f (2, 4) = 4i − 32j x3 − 4y 3 27 3x2 5 6. ∇f = p i+ p j; ∇f (3, 2) = √ i − √ j 3 4 3 4 38 2 38 2 x y−y 2 x y−y 7. ∇F = 2xz 2 sin 4yi + 4x2 z 2 cos 4yj + 2x2 z sin 4yk √ √ 4π 4π 4π i + 16 cos j + 8 sin k = 2 3i − 8j − 4 3k ∇F (−2, π/3, 1) = −4 sin 3 3 3 2x 2y 2z 4 3 1 i+ 2 j+ 2 k; ∇F (−4, 3, 5) = − i + j + k x2 + y 2 + z 2 x + y2 + z2 x + y2 + z2 25 25 25 √ √ f (x + h 3/2, y + h/2) − f (x, y) (x + h 3/2)2 + (y + h/2)2 )2x − y 2 9. Du f (x, y) = lim = lim h→0 h→0 h h √ 2 2 √ √ h 3x + 3h /4 = hy + h /4 = lim = lim ( 3x + 3h/4 + y + h/4) = 3x + y h→0 h→0 h √ √ f (x + h 2/2, y + h 2/2) − f (x, y) 10. Du f (x, y) = lim h→0 h √ √ 3x + 34 2/2 − (y + h 2/2)2 − 3x + y 2 = lim h→0 h √ √ √ √ √ √ 3h 2/2 − h 2yh2 /2 = lim = lim (3 2/2 − 2y − h/2) = 3 2/2 − 2y h→0 h→0 h √ √ 3 1 15 3 2 6 3 5 11. u = i j; ∇f = 15x y i + 30x y j; ∇f (−1, 1) = 15i − 30j; Du f (−1, 1) = − 15 = 2 2 2 √ 15 ( 3 − 2) 2 8. ∇F = 108 CHAPTER 13. PARTIAL DERIVATIVES √ √ 2 2 12. u = i j; ∇f = (4 + y 2 )i(2xy − 5)j; ∇f (3, −1) = 5i − 11j; 2 2 √ √ √ 5 2 11 2 − = −3 2 Du f (3, −1) = 2 2 √ √ x 1 3 10 −y 1 10 13. u = i+ 2 j; ∇f (2, −2) = i + j; i− j; ∇f = 2 2 2 10 10 y x +y 4 4 √ √ x +√ 10 3 10 10 Du f (2, −2) = − =− 40 40 20 x2 6 8 y2 i + j; ∇f (2, −1) = i + 4j; i + j; ∇f = 10 10 (x + y)2 (x + y)2 3 16 19 Du f (2, −1) = + = 5 5 5 √ 15. u = (2i + j)/ 5; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f (3, 2) = 28i + 42j; 2(28) 42 98 Du f (3, 2) = √ + √ = √ 5 5 5 √ √ 16. u = −i; ∇F = 2x tan yi + x2 sec2 yj; ∇f (1/2, π/3) = 3i + j; Du f (1/2, π/3) = − 3 14. u = 1 1 17. u = √ j + √ k; ∇f = 2xy 2 (2z + 1)2 i2x2 y(2z + 1)2 j + 4x2 y 2 (2z + 1)k; ∇f (1, −1, 1) = 2 2 √ 18 12 6 18i − 18j + 12k; Du f (1, −1, 1) = − √ + √ = − √ = −3 2 2 2 2 2 1 2x 1 2y 2y 2 − 2x2 18. u = √ i − √ j + √ k; ∇f = 2 i − 2 j + k; ∇f (2, 4, −1) = 4i − 8j − 24k; z z z3 6 6 6 √ 4 16 24 Du f (2, 4, −1) = √ − √ − √ = −6 6 6 6 6 x2 + 4z y2 i+ p j+ p k; x2 y + 2y 2 z 2 x2 y + 2y 2 z x2 y + 2y 2 z ∇f (−2, 2, 1) = −i + j + k; Du f (−2, 2, 1) = −1 19. u = −k; ∇f = p xy √ 2 1 2 20. u = −(4i−4j+2k)/ 36 = − i+ j− k; ∇f = 2i−2yj+2zk; ∇f (4, −4, 2) = 2i+8j+4k; 3 3 3 4 16 4 8 Du f (4, −4, 2) = − + − = 3 3 3 3 √ 21. u = (−4i − j/ 17; ∇f = 2(x − y)i − 2(x − y)j; ∇f (4, 2) = 4i − 4j; 16 4 12 Du f (4, 2) = − √ + √ = − √ 17 17 17 √ 22. u = (−2i + 5j/ 29; ∇f = (3x2 − 5y)i − (5x − 2y)j; ∇f (1, 1) = −2i − 3j; 4 15 11 Du f (1, 1) = √ − √ = − √ 29 29 29 √ √ 2 2x 2x 23. ∇f = 2e sin yi + e cos yj; ∇f (0, π/4) = 2i + j 2 p √ √ √ √ 2 1/2 The maximum Du is ( 2) + ( 2/2)2 = 5/2 in the direction 2i + ( 2/2)j. 13.6. DIRECTIONAL DERIVATIVE 109 24. ∇f = (xyex−y + yex−y i + (−xyex−y + xex−y j; ∇f (5, 5) = 30i − 20j √ 1/2 The maximum Du is 302 + (−20)2 = 10 13 in the direction 30i − 20j. 25. ∇f = (2x + 4z)i + 2z 2 j + (4x + 4yz)k; ∇f (1, 2, −1) = −2i + 2j − 4k √ 1/2 The maximum Du is (−2)2 + (2)2 + (−4)2 = 2 6 in the direction −2i + 2j − 4k. 26. ∇f = yzi + xzj + xyk; ∇f (3, 1, −5) = −5i − 15j + 3k 1/2 √ The maximum Du is (−5)2 + (−15)2 + (3)2 = 259 in the direction −5i − 15j + 3k. 2 2 2 2 2 2 27. ∇f = p2x secp(x + y )ip+ 2y sec2 (x + y )j; p ∇f ( π/6, π/6) = 2 p π/6 sec (π/3)(i + j) = 8p π/6(i + j) The minimum Du is −8 π/6(12 + 12 )1/2 = −8 π/3 in the direction −(i + j). 28. ∇f = 3x2 i − 3y 2 j; ∇f (2, −2) = 12i − 12j = 12(i − j) √ 1/2 The minimum Du is −12 12 + (−1)2 = −12 2 in the direction −(i − j) = −i + j. √ y √ √ ze x 3 2 y √ 29. ∇f = i + xze j + √ k; ∇f (16, 0, 9) = i + 12j + k. The minimum Du is 8 3 2 x 2 z √ 3 2 2 2 2 1/2 − (3/8) + 12 + (2/3) = − 83281/24 in the direction − i − 12j − k. 8 3 1 1 1 i + j − k; ∇f (1/2, 1/6, 1/3) = 2i + 6j − 3k x y z 1/2 The minimum Du is − 22 + 62 (−3)2 = −7 in the direction −2i − 6j + 3k. 30. ∇f = 31. Using implicit differentiation on 2x2 + y 2 = 9 we find y 0 =√−2x/y. At (2, 1) the slope of the tangent line is −2(2)/1 = −4. √Thus, u√= ±(i − 4j)/ √ 17. Now, ∇f = i + 2yj and ∇f (3, 4) = i + 8j. Thus, Du = ±(1/ 17 − 32 17) = ±31/ 17. 2x + y − 1 x + 2y 3x + 3y − 1 √ √ 32. ∇f = (2x + y − 1)i + (x + 2y)j; Du f (x, y) = + √ = Solving 2 2 2 √ (3x + 3y − 1)/ 2 = 0 we see that Du is 0 for all points on the line 3x + 3y = 1. 33. (a) Vectors perpendicular to 4i + 3j are ±(3i − 4j). Take u = ± 3 4 i− j . 5 5 √ 4 3 (b) u = (4i + 3j)/ 16 + 9 = i + j 5 5 4 3 (c) u = − i − j 5 5 34. D−u f (a, b) = ∇f (a, b) · (−u) = −∇f (a, b) · u = −Du f (a, b) = −6 35. (a) ∇f = (3x2 − 6xy 2 )i + (−6x2 y + 3y 2 )j 3(3x2 − 6xy 2 ) −6x2 y + 3y 2 9x2 − 18xy 2 − 6x2 y + 3y 2 √ √ √ Du f (x, y) = + = 10 10 10 www.elsolucionario.org 110 CHAPTER 13. PARTIAL DERIVATIVES 3 3 (b) F (x, y) = √ (3x2 − 3xy 2 − 2x2 y + y 2 ); ∇F = √ [(6x − 6y 2 − 4xy)i + (−12xy − 10 10 2x2 + 2y)j] 3 3 3 1 2 √ √ (6x − 6y − 4xy) + √ (−12xy − 2x2 + 2y) Du F (x, y) = √ 10 10 10 10 9 3 1 = (3x − 3y 2 − 2xy) + (−6xy − x2 + y) = (27x − 27y 2 − 36xy − 3x2 + 3y) 5 5 5 12 5 α − β = 7 and Dv f (a, b) = 36. Let ∇f (a, b) = αi + βj. Then Du f (a, b) = ∇f (a, b) · u = 13 13 5 12 ∇f (a, b) · v = α − β = 3. Solving for α and β, we obtain α = 13 and β = −13/6. Thus, 13 13 ∇f (a, b) = 13i − (13/6)j. 37. 38. ∇f = h2x, −5yi, |∇f | = p 10x2 + 25y 2 = 10, 4x2 + 25y 2 = 100, x2 y2 + =1 25 4 y x 39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, the maximum decrease in temperature) is in the direction −∇T (4, 3) = −16i − 4j. 40. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be −∇T (x, y), we have x0 (t)i + y 0 (t)j = −∇T (x, y) = 4xi + 2yj. Separating variables in dx/dt = 4x, we obtain dx/x = 4dt, ln x = 4t + c1 , and x = C1 e4t . Separating variables in dy/dt = 2y, we obtain dy/y = 2dt, ln y = 2t + c2 , and y = C2 e2t . Since x(0) = 4 and y(0) = 2, we have x = 4e4t and y = 2e2t . The equation of the path is 4e4t i + 2e2t j for t ≥ 0, or eliminating the parameter, x = y 2 , y ≥ 0. 41. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be ∇T (x, y), we have x0 (t)i + y 0 (t)j = ∇T (x, y) = −4xi − 2yj. Separating variables in dx/dt = −4x, we obtain dx/x = −4dt, ln x = −4t + c1 , and x = C1 e−4t . Separating variables in dy/dt = −2y, we obtain dy/y = −2dt, ln y = −2t + c2 , and y = C2 e−2t . Since x(0) = 3 and y(0) = 4, we have x = 3e−4t and y = 4e−2t . The equation of the path is 3e−4t i + 4e−2t j, or eliminating the parameter, 16x = 3y 2 , y ≥ 0. 13.6. DIRECTIONAL DERIVATIVE 111 42. Substituing x = 0, y = 0, z = 1, and T = 500 into t = and T (x, y, z) = k we see that k = 500 x2 + y 2 + z 2 500 . x2 + y 2 + z 2 1 1 2 2 h1, −2, −2i = i − j − k 3 3 3 3 1000y 1000z 1000x i− 2 j− 2 k ∇T = − 2 2 2 2 2 2 2 (x + y + z ) (x + y + z ) (x + y 2 + z 2 )2 500 750 750 ∇T (2, 3, 3) = − i− j− k 121 121 121 1 500 2 750 2 750 2500 Du T (2, 3, 3) = − − − − − = 3 121 3 121 3 121 363 (a) u = (b) The direction of maximum increase is ∇T (2, 3, 3) = − 500 750 750 252 i− j− k= (−2i − 3j − 3k). 121 121 121 121 (c) The maximum rate of change of T is |∇T (2, 3, 3)| = √ 250 √ 250 22 4+9+9= . 121 121 Gmx Gmy Gm i+ 2 j= 2 (xi + yj) 2 3/2 2 3/2 +y ) (x + y ) (x + y 2 )3/2 The maximum and minimum values of Du U (x, y) are obtained when u is in the directions ∇U and −∇U , respectively. Thus, at a point (x,y), not (0,0), the directions of maximum and minimum increase in U are xi+yj and −xi−yj, respectively. A vector at (x, y) in the direction ±(xi+yj) lies on a line through the origin. 43. ∇U = (x2 44. Since ∇f = fx (x, y)i + fy (x, y)j, we have ∂f /∂x = 3x2 + y 3 + yexy . Integrating, we obtain f (x, y) = x3 + xy 3 + exy + g(y). Then fy = 3xy 2 + xexy + g 0 (y) = −2y 2 + 3xy 2 + xexy . Thus, g 0 (y) = −2y 2 , g(y) = − 32 y 3 + c, and f (x, y) = x3 + xy 3 + exy − 23 + C. 45. ∇(cf ) = ∂ ∂ (cf )i + j = cfx i + cfy j = c(fx i + fy j) = c∇f ∂x ∂y 46. ∇(f + g) = (fx + gx )i + (fy + gy )j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g 47. ∇(f g) = (f gx + fx g)i + (f gy + fy g)j = f (gx i + gy j) + g(fx i + fy j) = f ∇g + g∇f 48. ∇(f /g) = (gfx − f gx )/g 2 i + (gfy − f gy )/g 2 j = g(fx i + fy j)/g 2 − f (gx i + gy j)/g 2 = g∇f /g 2 − f ∇g/g 2 = (g∇f − f ∇g)/g 2 p ∂r x x ∂r y y x2 + y 2 so =p = and =p = 2 2 2 2 ∂x r ∂y r x +y x +y Dx yE 1 r This gives ∇r = , = hx, yi = r r r r 49. r(x, y) = 112 50. CHAPTER 13. PARTIAL DERIVATIVES ∂ (f (r)) df ∂r ∂ (f (r)) df ∂r ∂ (f (r)) ∂ (f (r)) = and = so that ∇f (r) = h , i ∂x dr ∂x ∂y dr ∂y ∂x ∂y df ∂r df ∂r df ∂r ∂r =h , i= h , i = f 0 (r)∇r = f 0 (r)r/r dr ∂x dr ∂y dr ∂x ∂y 51. Let u = u1 i + u2 j and v = v1 i + v2 j. Dv f = (fx i+ fy j) · v = v1 fx + v2 fy ∂ ∂ Du Dv f = (v1 fx + v2 fy )i + (v1 fx + v2 fy )j · u = [(v1 fxx + v2 fyz )i + (v1 fxy + v2 fyy )j] · u ∂x ∂y = u1 v1 fxx + u1 v2 fyx + u2 v1 fxy + u2 v2 fyy D − uf = (fx i + fy j) · u = u1 fx + u2 fy ∂ ∂ (u1 fx + u2 fy )i + (u1 fx + u2 fy )j · v = [(u1 fxx + u2 fyx )i + (u1 fxy + u2 fyy )j] · v Dv Du f = ∂x ∂y = u1 v1 fxx + u2 v1 fyx + u1 v2 fxy + u2 v2 fyy Since the second partial derivatives are continuous, fxy = fyx and Du Dv f = Dv Du f. [Note that this result is a generalization fxy = fyx since Di Dj f = fyx and Dj Di f = fxy ] i ∂ ∂x f1 k ∂ 52. ∇ × F = ∂z f3 ∂f3 ∂f2 ∂f3 ∂f1 ∂f2 ∂f1 = − i− − j+ − k ∂y ∂z ∂x ∂z ∂x ∂y 13.7 j ∂ ∂y f2 Tangent Planes and Normal Lines 1. Since f (6, 1) = 4, the level curve is x − 2y = 4. ∇f = i − 2j; ∇f (6, 1) = i − 2j y x 2. Since f (1, 3) = 5, the level curve is y+2x = 5x or y = 3x, x 6= 0. y 1 ∇f = − 2 i + j; ∇f (1, 3) = −3i + j x x y x www.elsolucionario.org 13.7. TANGENT PLANES AND NORMAL LINES 113 3. Since f (2, 5) = 1, the level curve is y = x2 + 1. ∇f = −2xi + j; ∇f (2, 5) = −4i + j y x 4. Since f (−1, 3) = 10, the level curve is x2 + y 2 = 10. ∇f = 2xi + 2yj; ∇f (−1, 3) = −2i + 6j y x 5. Since f (−2, −3) = 2, the level curve is x2 /4 + y 2 /0 = 2 x 2y x2 /8 + y 2 /18 = 1. ∇f = i + j; ∇f (−2, −3) = −i − 2 9 y or 2 j 3 x 6. Since f (2, 2) = 2, the level curve is y 2 = 2x, x 6= 0. 2y y2 ∇f = − 2 i + j; ∇f (2, 2) = −i + 2j x x y x 114 CHAPTER 13. PARTIAL DERIVATIVES 7. Since f (1, 1) = −1, the level curve is (x − 1)2 − y 2 = −1 or y 2 − (x − 1)2 = 1. ∇f = 2(x − 1)i − 2yj; ∇f (1, 1) = −2j y x 8. Since f (π/6, 3/2) = 1, the level curve is y − 1 = sin x or −(y − 1) cos x 1 y = 1 + sin x, sin x 6= 0. ∇f = i+ j; 2 sin x sin x √ ∇f (π/6, 3/2) = − 3i + 2j y x 9. Since f (3, 1, 1) = 2, the level curve is y + z = 2 ∇f = j + k; ∇f (3, 1, 1) = j + k z 2 2 y x 10. Since f (1, 1, 3) = −1, the level curve is x2 + y 2 − z = −1 or z = 1 + x2 + y 2 . ∇f = 2xi + 2yj − k; ∇f (1, 1, 3) = 2i + 2j − k z y x 11. Since F (3, 4, 0) = 5, the level curve is x2 + y 2 + z 2 = 25. x y z ∇F = p i+ p j+ p k; 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 3 4 ∇F (3, 4, 0) = i + j 4 5 z 5 5 y x 13.7. TANGENT PLANES AND NORMAL LINES 115 12. Since F (0, −1, 1) = 0, the level curve is x2 − y 2 + z = 0 or z = y 2 − x2 . ∇F = 2xi − 2yj + k; ∇F (0, −1, 1) = 2i + k z y x 13. F (x, y, z) = x2 + y 2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c 4i + j + 21 k or 2x = 4c, 2y = c, −1 = c/2. From the third equation c = −2. Thus, x = −4 and y = −1. Since z = x2 + y 2 = 16 + 1 = 17, the point on the surface is (−4, −1, −17). 14. F (x, y, z) = x3 + y 3 + z; ∇F = 3x2 i + 2yj + k. We want ∇F = c(27i + 8j + k) or 3x2 = 27c, 2y = 8c, 1 = c. From c = 1 we obtain x = ±3 and y = 4. Since z = 15 − x3 − y 2 = 15 − (±3)3 − 16 = −1 ∓ 27, the points on the surface are (3, 4, −28) and (−3, 4, 26). 15. F (x, y, z0 = x2 + y 2 + z 2 ; ∇F = 2xi + 2yj + 2zk. ∇F (−2, 2, 1) = −4i + 4j + 2k. The equation of the tangent plane is −4(x + 2) + 4(y − 2) + 2(z − 1) = 0 or −2x + 2y + z = 9. 16. F (x, y, z) = 5x2 − y 2 + 4z 2 ; ∇F = 10xi − 2yj + 8zk; ∇F (2, 4, 1) = 20i − 8j + 8k.The equation of the tangent plane is 20(x − 2) − 8(y − 4) + 8(z − 1) = 0 or 5x − 2y + 2z = 4. 17. F (x, y, z) = x2 − y 2 − 3z 2 ; ∇F = 2xi − 2yj − 6zk; ∇F (6, 2, 3) = 12i − 4j − 18k. The equation of the tangent plane is 12(x − 6) − 4(y − 2) − 18(z − 3) = 0 or 6x − 2y − 9z = 5. 18. F (x, y, z) = xy + yz + zx; ∇F = (y + z)i + (x + z)j + (y + x)k; ∇F (1, −3, −5) = −8i − 4j − 2k. The equation of the tangent plane is −8(x − 1) − 4(y + 3) − 2(z + 5) = 0 or 4x + 2y + z = −7. 19. F (x, y, z) = x2 + y 2 + z; ∇F = 2xi + 2yj + k; ∇F (3, −4, 0) = 6i − 8j + k. The equation of the tangent plane is 6(x − 3) − 8(y + 4) + z = 0 or 6x − 8y + z = 50. 20. F (x, y, z) = xz; ∇F = zi + xk; ∇F (2, 0, 3) = 3i + 2k. The equation of the tangent plane is 3(x − 2) + 2(z − 3) = 0 or 3x + 2z = 12. √ 21. F (x, y,√z) = cos(2x+y)−z; ∇F = −2 sin(2x+y)i−sin(2x+y)j−k; ∇F (π/2, π/4, −1 2) = √ √ √ 2 π 2 π 1 2i+ j−k. The equation of the tangent plane is 2 x − + y− − z+√ = 2 2 2 4 2 √ π π √ 1 5π 0, 2 x − + y− − 2 z+√ = 0, or 2x + y − 2z = + 1. 2 4 4 2 22. F (x, y, z) = x2 y 3 + 6z; ∇F = 2xy 3 i + 3x2 y 2 j + 6k; ∇F (2, 1, 1) = 4i + 12j + 6k. The equation of the tangent plane is 4(x − 2) + 12(y − 1) + 6(z − 1) = 0 or 2x + 6y + 3z = 13. √ √ √ 2x 2y 23. F (x, y, z) = ln(x2 + y 2 ) − z; ∇F = 2 i+ 2 j − k; ∇F (1/ 2, 1/ 2, 0) = 2i + 2 2 x +y x +y √ √ √ 1 1 √ √ 2j − k. The equation of the tangent plane is 2 x − + 2 y− − (z − 0) = 2 2 √ √ √ 1 1 +2 y− √ − 2z = 0, or 2x + 2y − 2z = 2 2. 0, 2 x − √ 2 2 www.elsolucionario.org 116 CHAPTER 13. PARTIAL DERIVATIVES −2y 24. F (x, sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) = √ y, z) = 8e 16 3i − 8j − k. The equation of the tangent plane is √ √ √ 2 3π − 4. 16 3(x − π/24) − 8(y − 0) − (z − 4) = 0 or 16 3x − 8y − z = 3 25. The gradient of F (x, y, z) = x2 + y 2 + z 2 is ∇F = 2xi + 2yj + 2zk, so the normal vector to the surface at (x0 , y0 , z0 ) is 2x0 i + 2y0 j + 2z0 k. A normal vector to the plane 2x + 4y + 6z = 1 is 2i + 4j + 6k. Since we want the tangent plane to be parallel to the given plane, we find c so that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c, y0 = √ 2c, z0 = 3c. Now, (x0 , y0 , z0 ) is on 2 2 2 2 2. Thus, the points on the surface the surface, so c + (2c) + (3c) = 14c = 7 and c = ±1/ √ √ √ √ √ √ are ( 2/2, 2, 3 2/2) and − 2/2, − 2, −3 2/2). 26. The gradient of F (x, y, z) = x2 − 2y 2 − 3z 2 is ∇F (x, y, z) = 2xi − 4yj − 6zk, so a normal vector to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i − 4y0 j − 6z0 k. A normal vector to the plane 8x+4y+6z = 5 is 8i+4j+6k. Since we want the tangent plane to be parallel to the given plane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c or x0 = 4c, y0 = −c, √z0 = −c. 2 2 2 Now (x0 , y0 , z0 ) is on the surface, 2(−c)2 − 3(−c) √so (4c) √ −√ √ = √ 11c √ = 33 and c = ± 3. Thus, the points on the surface are (4 3, − 3, − 3) and (−4 3, 3, 3). 27. The gradient of F (x, y, z) = x2 +4x+y 2 +z 2 −2z is ∇F = (2x+4)i+2yj+(2z−2)k, so a normal to the surface at (x0 , y0 , z0 ) is (2x0 + 4)i + 2y0 j + (2z0 − 2)k. A horizontal plane has normal ck for c 6= 0. Thus, we want 2x0 + 4 = 0, 2y0 = 0, 2z0 − 2 = c or x0 = −2, y0 = 0, z0 = c + 1. Since (x0 , y0 , z0 ) is on the surface, (−2)2 + 4(−2) + (c + 1)2 − 2(c + 1) = c2 − 5 = 11 and c = ±4. The points on the surface are (−2, 0, 5) and (−2, 0, −3). 28. The gradient of F (x, y, z) = x2 + 3y 2 + 4z 2 − 2xy is ∇F = (2x − 2y)i + (6y − 2x)j + 8zk, so a normal to the surface at (x0 , y0 , z0 ) is 2(x0 − y0 )i + 2(3y0 − x0 )j + 8z0 k. (a) A normal to the xz plane is cj for c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = c, 8z0 = 0 or x0 = y0 , 3y0 −x0 = c/2, z0 = 0. Solving the first two equations, we obtain 2 x0 = y0 = c/4. Since (x0 , y0√ , z0 ) is on the surface, (c/4)2 +3(c/4)2 +4(0)√ −2(c/4)(c/4) = √ 2 2c /16 = 16 and c = ±16/ 2. Thus, the points on the surface are (4/ 2, 4/ 2, 0) and √ √ (−4 2, −4 2, 0). (b) A normal to the yz-plane is ci for c 6= 0. Thus, we want 2(x0 − y0 ) = c, 2(3y0 − x0 ) = 0, 8z0 = 0 or x0 − y0 = c/2, x0 = 3y0 , z0 = 0. Solving the first two equations, we obtain x0 = 3c/4 and y0 = c/4. Since (x0 , y0 , z0 )√is on the surface, (3c/4)2 + 3(c/4)2 + 4(0)2√− 2(3c/4)(c/4) = 6c2 /16 √ √ = 16 and √ c = ±16 6. Thus, the points on the surface are (12/ 6, 4/ 6, 0) and (−12/ 6, −4/ 6, 0). (c) A normal to the xy-plane is ckfor c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = 0, 8z0 = c or x0 = y0 , 3y0 −x0 = 0, z0 = c/8. Solving the first two equations, we obtain x0 = y0 = 0. Since (x0 , y0 , z0 ) is on the surface, 02 +3(0)2 +4(c/8)2 −2(0)(0) = c2 /16 = 16 and c = ±16. Thus, the points on the surface are (0, 0, 2) and (0, 0, −2). 29. If (x0 , y0 , z0 ) is on x2 /a2 + y 2 /b2 + z 2 /c2 = 1, then x20 /a2 + y02 /b2 + z02 /c2 = 1 and x0 , y0 , z0 ) is on the plane xx0 /a2 + yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is 13.7. TANGENT PLANES AND NORMAL LINES 117 ∇F (x0 , y0 , z0 ) = (2x − 0/a2 )i + (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i + (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane, the normal vectors are parallel and the plane is tangent to the surface. 30. If (x0 , y0 , z0 ) is on x2 /a2 − y 2 /b2 + z 2 /c2 = 1, then x20 /b2 − y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 ) is on the plane xx0 /a2 − yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = (2x0 /a2 )i − (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i − (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane, the normal vectors are parallel, and the plane is tangent to the surface. 31. F (x, y, z) = x2 + 2y 2 + z 2 ; ∇F = 2xi + 4yj + 2zk; ∇F (1, −1, 1) = 2i − 4j + 2k. Parametric equations of the line are x = 1 + 2t, y = −1 − 4t, z = 1 + 2t. 32. F (x, y, z) = 2x2 − 4y 2 − z; ∇F = 4xi − 8yj − k; ∇F (3, −2, 2) = 12i + 16j − k. Parametric equations of the line are x = 3 + 12t; y = −2 + 16t, z = 2 − t. 33. F (x, y, z) = 4x2 + 9y 2 − z; ∇F = 8xi + 18yj − k; ∇F (1/2, 1/3, 3) = 4i + 6j − k. Symmetric y − 1/3 z−3 x − 1/2 = = . equations of the line are 4 6 −1 34. F (x, y, z) = x2 + y 2 − z 2 ; ∇F = 2xi + 2yj − 2zk; ∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric x−3 y−4 z−5 equations of the line are = = . 6 8 −10 35. Let F (x, y, z) = x2 + y 2 − z 2 . Then ∇F = 2xi + 2yj − 2zk and a normal to the surface at (x0 , y0 , z0 ) is x0 i + y0 j − z0 k. An equation of the tangent plane at (x0 , y0 , z0 ) is x0 (x − x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0 or x0 x + y0 y − z0 z = x20 + y02 − z02 . Since (x0 , y0 , z0 ) is on the surface, z02 = x20 + y02 and x20 + y02 − z02 = 0. Thus, the equation of the tangent plane is x0 x + y0 y − z0 z = 0, which passes through the origin. 1 1 1 √ √ x+ y + z. Then ∇F = √ i+ √ j+ √ k and a normal to the surface 2 x 2 y 2 z 1 1 1 at (x0 , y0 , z0 ) is √ i + √ j + √ k. An equation of the tangent plane at (x0 , y0 , z0 ) is 2 x0 2 y0 2 z0 1 1 1 1 1 1 √ √ √ √ (x−x0 )+ √ (y−y0 )+ √ (z−z0 ) = 0 or √ x+ √ y+ √ z = x0 + y0 + z0 = 2 x0 2 y0 2 z0 x0 y0 z0 √ √ √ √ √ √ √ √ √ √ √ √a. The √ sum of the intercepts is x0 a + y0 a + z0 a = ( x0 + y0 + z0 ) a = a · a = a. 36. Let F (x, y, z) = √ 37. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i + 2y0 j + 2z0 k. Parametric equations of the normal line are x = x0 + 2x0 t, y = y0 + 2y0 t, z = z0 + 2z0 t. Letting t = −1/2, we see that the normal line passes through the origin. 38. The normal lines to F (x, y, z) = 0 and G(x, y, z) = 0 are Fx i+Fy j+Fz k and Gx i+Gy j+Gz k, respectively. These vectors are orthogonal if and only if their dot product is 0. Thus, the surfaces are orthogonal at P if and only if Fx Gx + Fy Gy + Fz Gz = 0. 39. We have F (x, y, z) = x2 + y 2 + z 2 and G(x, y, z) = x2 + y 2 − z 2 . ∇F = h2x, 2y, 2zi = 6 0 except at the origin 118 CHAPTER 13. PARTIAL DERIVATIVES ∇G = h2x, 2y, −2zi = 6 0 except at the origin Therefore, the gradient vectors are nonzero at each of the intersection points. Now Fx Gx + Fy Gy + Fx Gz = (2x)(2x) + (2y)(2y) + (2z)(−2z) = 4x2 + 4y 2 − 4z 2 = 4(x2 + y 2 + z 2 ) = 4(0) = 0 The second to last equality follows from the fact that the intersection points lie on both surfaces and hence satisfy the second equation x2 + y 2 − z 2 = 0. 40. Let F (x, y, z) = x2 − y 2 + z 2 − 4 and G(x, y, z) = 1/xy 2 − z. Then Fx Gx + Fy Gy + Fz Gz = (2x)(−1/x2 y 2 ) + (−2y)(−2/xy 3 ) + (2z)(−1) = −2/xy 2 + 4/xy 2 − 2z = 2(1/xy 2 − z). For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 2(0) and the surfaces are orthogonal at points of intersection. 13.8 Extrema of Multivariable Functions 1. fx = 2x; fxx = 2; fxy = 0; fy = 2y; fyy = 2; D = 4. Solving fx = 0 and fy = 0, we obtain the critical point (0, 0). Since D(0, 0) = 4 > 0 and fxx (0, 0) = 2 > 0, f (0, 0) = 5 is a relative minimum. 2. fx = 8x; fxx = 8; fxy = 0; fy = 16y; fyy = 16; D = 128. Solving fx = 0 and fy = 0, we obtain the critical point (0, 0). Since D(0, 0) = 128 > 0 and fxx (0, 0) = 8 > 0, f (0, 0) = 0 is a relative minimum. 3. fx = −2x + 8; fxx = −2; f xy = 0; fy = −2y + 6; fyy = −2; D = 4. Solving fx = 0 and fy = 0 we obtain the critical point (4, 3). Since D(4, 3) = 4 > 0 and fxx (4, 3) = −2 < 0, f (4, 3, ) = 25 is a relative maximum. 4. fx = 6x − 6; f xx = 6; f xy = 0; fy = 4y + 8; fyy = 4; D = 24. Solving fx = 0 and fy = 0, we obtain the critical point (1, −2). Since D(1, −2) = 24 > 0 and fxx (1, −2) = 6 > 0, f (1, −2) = −11 is a relative minimum. 5. fx = 10x + 20; fxx = 10; fxy = 0; fy = 10y − 10; fyy = 10; D = 100. Solving fx = 0 and fy = 0, we obtain the critical point (−2, 1). Since D(−2, 1) = 100 > 0 and fxx (−2, 1) = 10 > 0, f (−2, 1) = 15 is a relative minimum. 6. fx = −8x − 8; fxx = −8; fxy = 0; fy = −4y + 12; fyy = −4; D = 32. Solving fx = 0 and fy = 0, we obtain the critical point (−1, 3). Since D(−1, 3) = 32 > 0 and fxx (−1, 3) = −8 < 0, f (−1, 3) = 27 is a relative maximum. 7. fx = 12x2 − 12; fxx = 24x; fxy = 0; fy = 3y 2 − 3; fyy = 6y; D = 144xy. Solving fx = 0 and fy = 0, we obtain the critical points (−1, −1), (−1, 1), (1, −1), and (1, 1). Since D(−1, 1) = −144 < 0 and D(1, −1) = −144 < 0, these points do not give relative extrema. Since D(−1, −1) = 144 > 0 and fxx (−1, −1) = −24 < 0, f (−1, −1) = 10 is a