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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online