On the sphere zx xz and zy yz so zx zy 1 x2

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Unformatted text preview: y0 + (y1 − y0 )t) for 0 ≤ t ≤ 1; then f (x1 , y1 ) − f (x0 , y0 ) = F (1) − F (0). Apply the Mean Value Theorem to F (t) on the interval [0,1] to get [F (1) − F (0)]/(1 − 0) = F (t∗ ), F (1) − F (0) = F (t∗ ) for some t∗ in (0,1) so f (x1 , y1 ) − f (x0 , y0 ) = F (t∗ ). By the chain rule, F (t) = fx (x, y )(dx/dt) + fy (x, y )(dy/dt) = fx (x, y )(x1 − x0 ) + fy (x, y )(y1 − y0 ). Let (x∗ , y ∗ ) be the point on C for t = t∗ then f (x1 , y1 ) − f (x0 , y0 ) = F (t∗ ) = fx (x∗ , y ∗ ) (x1 − x0 ) + fy (x∗ , y ∗ ) (y1 − y0 ). Exercise Set 15.5 544 53. Let (a, b) be any point in the region, if (x, y ) is in the region then by the result of Exercise 52 f (x, y ) − f (a, b) = fx (x∗ , y ∗ )(x − a) + fy (x∗ , y ∗ )(y − b) where (x∗ , y ∗ ) is on the line segment joining (a, b) and (x, y ). If fx (x, y ) = fy (x, y ) = 0 throughout the region then f (x, y ) − f (a, b) = (0)(x − a) + (0)(y − b) = 0, f (x, y ) = f (a, b) so f (x, y ) is constant on the region. EXERCISE SET 15.5 1. At P , ∂z/∂x = 48 and ∂z/∂y = −14, tangent plane 48x − 14y − z = 64, normal line x = 1 + 48t, y = −2 − 14t, z = 12 − t. 2. At P , ∂z/∂x = 14 and ∂z/∂y = −2, tangent plane 14x − 2y − z = 16, normal line x = 2 + 14t, y = 4 − 2t, z = 4 − t. 3. At P , ∂z/∂x = 1 and ∂z/∂y = −1, tangent plane x − y − z = 0, normal line x = 1 + t, y = −t, z = 1 − t. 4. At P , ∂z/∂x = −1 and ∂z/∂y = 0, tangent plane x + z = −1, normal line x = −1 − t, y = 0, z = −t. 5. At P , ∂z/∂x = 0 and ∂z/∂y = 3, tangent plane 3y − z = −1, normal line x = π/6, y = 3t, z = 1 − t. 6. At P , ∂z/∂x = 1/4 and ∂z/∂y = 1/6, tangent plane 3x + 2y − 12z = −30, normal line x = 4 + t/4, y = 9 + t/6, z = 5 − t. 7. By implicit differentiation ∂z/∂x = −x/z , ∂z/∂y = −y/z so at P , ∂z/∂x = 3/4 and ∂z/∂y = 0, tangent plane 3x − 4z = −25, normal line x = −3 + 3t/4, y = 0, z = 4 − t. 8. By implicit differentiation ∂z/∂x = (xy )/(4z ), ∂z/∂y = x2 /(8z ) so at P , ∂z/∂x = 3/8 and ∂z/∂y = −9/16, tangent plane 6x − 9y − 16z = 5, normal line x = −3 + 3t/8, y = 1 − 9t/16, z = −2 − t. 9. The tangent plane is horizontal if the normal ∂z/∂xi + ∂z/∂y j − k is parallel to k which occurs when ∂z/∂x = ∂z/∂y = 0. (a) ∂z/∂x = 3x2 y 2 , ∂z/∂y = 2x3 y ; 3x2 y 2 = 0 and 2x3 y = 0 for all (x, y ) on the x-axis or y -axis, and z = 0 for these points, the tangent plane is horizontal at all points on the x-axis or y -axis. (b) ∂z/∂x = 2x − y − 2, ∂z/∂y = −x + 2y + 4; solve the system 2x − y − 2 = 0, −x + 2y + 4 = 0, to get x = 0, y = −2. z = −4 at (0, −2), the tangent plane is horizontal at (0, −2, −4). 10. ∂z/∂x = 6x, ∂z/∂y = −2y , so 6x0 i − 2y0 j − k is normal to the surface at a point (x0 , y0 , z0 ) on the surface. 6i + 4j − k is normal to the given plane. The tangent plane and the given plane are parallel if their normals are parallel so 6x0 = 6, x0 = 1 and −2y0 = 4, y0 = −2. z = −1 at (1, −2), the point on the surface is (1, −2, −1). 11. ∂z/∂x = −6x, ∂z/∂y = −4y so −6x0 i − 4y0 j − k is normal to the surface at a point (x0 , y0 , z0 ) on the surface. This normal must be parallel to the given line and hence to the vector −3i + 8j − k which is parallel to the line so −6x0 = −3, x0 = 1/2 and −4y0 = 8, y0 = −2. z = −3/4 at (1/2, −2). The point on the surface is (1/2, −2, −3/4). 12. (3,4,5) is a point of intersection because it satisfies both equations. Both surfaces have (3/5)i + (4/5)j − k as a normal so they have a common tangent plane at (3,4,5). 545 Chapter 15 13. df = 2xydx + x2 dy = 0.6 + 0.2 = 0.8, ∆f = (x + ∆x)2 (y + ∆y ) − x2 y = (1.1)2 (3.2) − 12 · 3 = 0.872 14. dz = 6xdx − 2dy = −12(0.02) − 2(−0.03) = −0.18, ∆z = 3(−2 + 0.02)2 − 2(4 − 0.03) − (3(−2)2 − 2(4)) = −0.1788 16. 2(4)(−2) − (−2)3 − (0 − 13 ) = −7 15. 3/1 − (−1)/2 = 7/2 17. dz = 3x2 y 2 dx + 2x3 ydy, ∆z = (x + ∆x)3 (y + ∆y )2 − x3 y 2 18. dz = yexy dx + xexy dy, ∆z = e(x+∆x)(y+∆y) − exy 19. dz = 7dx − 2dy 21. dz = y / 1 + x2 y 2 20. dz = (10xy 5 − 2)dx + (25x2 y 4 + 4)dy dx + x/ 1 + x2 y 2 dy 22. dz = 2 sec2 (x − 3y ) tan(x − 3y )dx − 6 sec2 (x − 3y ) tan(x − 3y )dy 23. (a) Let f (x, y ) = ex sin y ; f (0, 0) = 0, fx (0, 0) = 0, fy (0, 0) = 1, so ex sin y ≈ y (b) Let f (x, y ) = 2x + 1 2x + 1 ; f (0, 0) = 1, fx (0, 0) = 2, fy (0, 0) = −1, so ≈ 1 + 2x − y y+1 y+1 24. f (1, 1) = 1, fx (x, y ) = αxα−1 y β , fx (1, 1) = α, fy (x, y ) = βxα y β −1 , fy (1, 1) = β , so xα y β ≈ 1 + α(x − 1) + β (y − 1) 25. dT = Tx dx + Ty dy ≈ 2(−0.02) − (0.02) = −0.06, T ≈ T (1, 3) + dT ≈ 93 − 0.06 = 92.94◦ 26. p(104, 103) ≈ p(100, 98) − px (100, 98)(104 − 100) − py (100, 98)(103 − 98) = 1008 + (−2)4 + (1)5 = 1005 mb 27. f (x, y ) = (x2 + y 2 )−1/2 , fx (4, 3) = 1 (3.92)2 + (3.01)2 ≈√ −x 4 −y 3 , fy (4, 3) = 2 , =− =− 125 125 (x2 + y 2 )3/2 (x + y 2 )3/2 3 1 4 (−0.08) − (0.01) = 0.20232; actual value ≈ 0.202334. − 2 125 125 +3 42 28. From Exercise 24, x0.5...
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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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