20 a b f 15 6 5 4 3 2 1 0 0 1 2 3 4 5 6 d set

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Unformatted text preview: does not exist. = lim 2 + t 2t t→0 33. y ln(x2 + y 2 ) = r sin θ ln r2 = 2r(ln r) sin θ, so 34. x2 y 2 x2 + y 2 √ 35. 36. (r2 cos2 θ)(r2 sin2 θ) = r3 cos2 θ sin2 θ, so r = lim (x,y,z )→(0,0,0) tan−1 x2 y ln(x2 + y 2 ) = lim+ 2r(ln r) sin θ = 0 r →0 lim (x,y,z )→(0,0,0) e (x,y )→(0,0) x2 +y 2 +z 2 x2 + y 2 + z 2 x2 y 2 lim √ x2 +y 2 +z 2 eρ = , so ρ x2 + y 2 + z 2 e lim (x,y )→(0,0) = lim+ ρ→0 x2 + y 2 =0 eρ does not exist. ρ π 1 1 = lim+ tan−1 2 = 2 + z2 +y ρ 2 ρ→0 37. (a) No, since there seem to be points near (0, 0) with z = 0 and other points near (0, 0) with z ≈ 1/2. mx3 mx x4 = lim 2 =0 (c) lim 4 = lim 1/2 = 1/2 (b) lim 4 x→0 x + m2 x2 x→0 x + m2 x→0 2x x→0 (d) A limit must be unique if it exists, so f (x, y ) cannot have a limit as (x, y ) → (0, 0). Along y = mx : mx4 mx2 = lim 4 = 0; x→0 2x6 + m2 x2 x→0 2x + m2 along y = kx2 : 38. (a) lim lim x→0 2x6 kx5 kx = lim 2 = 0. x→0 2x + k 2 + k 2 x4 1 x6 1 = lim = = 0 x→0 2x6 + x6 x→0 3 3 (b) lim 39. (a) abct3 abct = lim 2 =0 t→0 a2 t2 + b4 t4 + c4 t4 t→0 a + b4 t2 + c4 t2 lim t4 = lim 1/3 = 1/3 t→0 t4 + t4 + t4 t→0 (b) lim 40. π/2 because x2 + 1 → +∞ as (x, y ) → (0, 1) x2 + (y − 1)2 41. −π/2 because x2 − 1 → −∞ as (x, y ) → (0, 1) x2 + (y − 1)2 42. with z = x2 + y 2 , lim + z →0 43. No, because sin z = 1 = f (0, 0) z x2 does not exist. (x,y )→(0,0) x2 + y 2 lim Along x = 0 : lim 0/y 2 = lim 0 = 0; along y = 0 : lim x2 /x2 = lim 1 = 1. y →0 y →0 x→0 x→0 44. Using polar coordinates with r > 0, xy = r2 sin θ cos θ and x2 + y 2 = r2 so |xy ln x2 + y 2 | = |r2 sin θ cos θ ln r2 | ≤ |2r2 ln r|, but lim+ 2r2 ln r = 0 thus r →0 lim (x,y )→(0,0) xy ln x2 + y 2 = 0; f (x, y ) will be continuous at (0,0) if we deﬁne f (0, 0) = 0. 533 Chapter 15 EXERCISE SET 15.3 1. (a) 9x2 y 2 (e) 6y (b) 6x3 y (f ) 6x3 (c) 9y 2 (g) 36 (d) 9x2 (h) 12 2. (a) 2e2x sin y (e) cos y (b) e2x cos y (f ) e2x (c) 2 sin y (g) 0 (d) 0 (h) 4 √ (b) − x cos y sin y (c) − √ 2x sin y (d) − √ 2x (b) 140x4 y 3 (c) 140x3 y 4 (d) 140x3 y 4 3. (a) − 1 cos y 4x3/2 4. (a) 8 + 84x2 y 5 5. (a) 3 3 ∂z =√ ; slope = ∂x 8 2 3x + 2y (b) ∂z 1 1 =√ ; slope = ∂y 4 3x + 2y 6. (a) ∂z = e−y ; slope = 1 ∂x (b) ∂z = −xe−y + 5; slope = 2 ∂y ∂z = −4 cos(y 2 − 4x); rate of change = −4 cos 7 ∂x ∂z = 2y cos(y 2 − 4x); rate of change = 2 cos 7 (b) ∂y 7. (a) 8. (a) ∂z 1 1 =− ; rate of change = − ∂x (x + y )2 4 (b) ∂z 1 1 =− ; rate of change = − ∂y (x + y )2 4 9. ∂z/∂x = slope of line parallel to xz -plane = −4; ∂z/∂y = slope of line parallel to yz -plane = 1/2 10. The slope at P in the positive x-direction is negative, the slope in the positive y -direction is negative, thus ∂z/∂x < 0, ∂z/∂y < 0; the curve through P which is parallel to the x-axis is concave down, so ∂ 2 z/∂x2 < 0; the curve parallel to the y -axis is concave down, so ∂ 2 z/∂y 2 < 0. 23 11. ∂z/∂x = 8xy 3 ex y 23 , ∂z/∂y = 12x2 y 2 ex y 12. ∂z/∂x = −5x4 y 4 sin(x5 y 4 ), ∂z/∂y = −4x5 y 3 sin x5 y 4 13. ∂z/∂x = x3 /(y 3/5 + x) + 3x2 ln(1 + xy −3/5 ), ∂z/∂y = −(3/5)x4 /(y 8/5 + xy ) 14. ∂z/∂x = yexy sin(4y 2 ), ∂z/∂y = 8yexy cos(4y 2 ) + xexy sin(4y 2 ) 15. y (x2 − y 2 ) ∂z x(x2 − y 2 ) ∂z =− 2 =2 , 2 )2 ∂x (x + y ∂y (x + y 2 )2 17. fx (x, y ) = (3/2)x2 y 5x2 − 7 fy (x, y ) = (1/2)x3 3x2 − 7 16. 3x5 y − 7x3 y 3x5 y − 7x3 y xy 3 (3x + 4y ) ∂z x2 y 2 (6x + 5y ) ∂z = = , 3/2 ∂x ∂y 2(x + y ) 2(x + y )3/2 −1/2 −1/2 18. fx (x, y ) = −2y/(x − y )2 , fy (x, y ) = 2x/(x − y )2 19. fx (x, y ) = y −1/2 xy −3/2 3 , fy (x, y ) = − 2 − y −5/2 tan−1 (x/y ) 2 + x2 2 y y +x 2 Exercise Set 15.3 534 20. fx (x, y ) = 3x2 e−y + (1/2)x−1/2 y 3 sec 21. fx (x, y ) = −(4/3)y 2 sec2 x y 2 tan x √ √ √ x tan x, fy (x, y ) = −x3 e−y + 3y 2 sec x −7/3 , fy (x, y ) = −(8/3)y tan x y 2 tan x −7/3 √ 1 x sinh xy 2 cosh xy 2 + x−1/2 sinh x sinh2 xy 2 2 √ fy (x, y ) = 4xy cosh x sinh xy 2 cosh xy 2 22. fx (x, y ) = 2y 2 cosh √ 23. fx (x, y ) = −2x, fx (3, 1) = −6; fy (x, y ) = −21y 2 , fy (3, 1) = −21 24. ∂f /∂x = x2 y 2 exy + 2xyexy , ∂f /∂x 25. ∂z/∂x = x(x2 +4y 2 )−1/2 , ∂z/∂x 26. ∂w/∂x = −x2 y sin xy + 2x cos xy, (1,1) (1,2) = 3e; ∂f /∂y = x3 yexy + x2 exy , ∂f /∂y (1,1) √ = 1/ 17 ; ∂z/∂y = 4y (x2 +4y 2 )−1/2 , ∂z/∂y = 2e (1,2) √ = 8/ 17 ∂w ∂w (1/2, π ) = −π/4; ∂w/∂y = −x3 sin x, (1/2, π ) = −1/8 ∂x ∂y 27. fx = 8x − 8y 4 , fy = −32xy 3 + 35y 4 , fxy = fyx = −32y 3 28. fx = x/ x2 + y 2 , fy = y/ x2 + y 2 , fxy = fyx = −xy (x2 + y 2 )−3/2 29. fx = ex cos y, fy = −ex sin y, fxy = fyx = −ex sin y 30. fx = ex−y , fy = −2yex−y , fxy = fyx = −2yex−y 2 2 2 31. fx = 4/(4x − 5y ), fy = −5/(4x − 5y ), fxy = fyx = 20/(4x − 5y )2 32. fx = 4xy 2 /(x2 + y 2 )2 , fy = −4x2 y/(x2 + y 2 )2 , fxy...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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