# A 0 0 0 x2 4 2 0 0 b c x2 0 1x2 a2 y 2 b2 29

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: y 0.3 ≈ 1 + 0.5(x − 1) + 0.3(y − 1), so (1.05)0.5 (0.97)0.3 ≈ 1 + 0.5(0.05) + 0.3(−0.03) = 1.016, actual value ≈ 1.01537 29. df = (2x + 2y − 4)dx + 2xdy ; x = 1, y = 2, dx = 0.01, dy = 0.04 so df = 0.10 30. df = (1/3)x−2/3 y 1/2 dx + (1/2)x1/3 y −1/2 dy ; x = 8, y = 9, dx = −0.02, dy = 0.03 so df = 0.005 31. df = −x−2 dx − y −2 dy ; x = −1, y = −2, dx = −0.02, dy = −0.04 so df = 0.03 32. df = 33. z = x y dx + dy ; x = 0, y = 2, dx = −0.09, dy = −0.02 so df = −0.09 2(1 + xy ) 2(1 + xy ) x2 + y 2 , dz = x x2 + y 2 −1/2 dx + y x2 + y 2 −1/2 dy ; x = 3, y = 4, dx = 0.2, dy = −0.04 so dz = 0.088 cm. 34. dV = (2/3)πrhdr + (1/3)πr2 dh; r = 4, h = 20, dr = 0.05, dh = −0.05 so dV = 2.4π ≈ 7.54 in3 . Exercise Set 15.5 546 35. A = xy , dA = ydx + xdy , dA/A = dx/x + dy/y , |dx/x| ≤ 0.03 and |dy/y | ≤ 0.05, |dA/A| ≤ |dx/x| + |dy/y | ≤ 0.08 = 8% 36. V = (1/3)πr2 h, dV = (2/3)πrhdr + (1/3)πr2 dh, dV /V = 2(dr/r) + dh/h, |dr/r| ≤ 0.01 and |dh/h| ≤ 0.04, |dV /V | ≤ 2|dr/r| + |dh/h| ≤ 0.06 = 6%. 37. z = x x2 + y 2 , dz = x2 + y2 dx + y x2 x y x2 dz =2 dx + 2 dy = 2 z x + y2 x + y2 x + y2 + y2 dx x dy , + y2 x2 + y 2 dy , y x2 y2 dx dy dx dy dz ≤2 +2 , if ≤ r/100 and ≤ r/100 then 2 2 z x +y x x +y y x y x2 y2 r dz ≤2 so the percentage error in z is at most about r%. (r/100) + 2 (r/100) = 2 z x +y x + y2 100 x2 + y 2 , dz = x x2 + y 2 38. (a) z = 2 −1/2 −1/2 dx + y x2 + y 2 −1/2 dy , 2 −1/2 |dx| + y x + y |dy |; if x = 3, y = 4, |dx| ≤ 0.05, and |dz | ≤ x x + y |dy | ≤ 0.05 then |dz | ≤ (3/5)(0.05) + (4/5)(0.05) = 0.07 cm 2 2 (b) A = (1/2)xy , dA = (1/2)ydx + (1/2)xdy , |dA| ≤ (1/2)y |dx| + (1/2)x|dy | ≤ 2(0.05) + (3/2)(0.05) = 0.175 cm2 . 39. dR = 2 R2 2 dR1 (R1 + R2 ) + 2 R1 2 dR2 , (R1 + R2 ) R2 dR = R R1 + R2 dR1 R1 + R1 R1 + R2 dR2 , R2 R2 dR1 R1 dR2 dR ≤ + ; if R1 = 200, R2 = 400, |dR1 /R1 | ≤ 0.02, and R R1 + R 2 R1 R1 + R2 R 2 |dR2 /R2 | ≤ 0.02 then |dR/R| ≤ (400/600)(0.02) + (200/600)(0.02) = 0.02 = 2%. 40. dP = (k/V )dT − (kT /V 2 )dV , dP/P = dT /T − dV /V ; if dT /T = 0.03 and dV /V = 0.05 then dP/P = −0.02 so there is about a 2% decrease in pressure. a 1 da − √ dc; if a = 3, c = 5, |da| ≤ 0.01, and |dc| ≤ 0.01 then 2 2 − a2 −a cc |dθ| ≤ (1/4)(0.01) + (3/20)(0.01) = 0.004 radians. 41. dθ = √ c2 42. V = πr2 h, dV = 2πrhdr + πr2 dh; r = 2, h = 5, dr = 0.01, and dh = 0.01 so dV = (20π )(0.01) + (4π )(0.01) = 0.24π , or about 0.754 cm3 . 43. dT = π g L/g dL − πL g2 L/g dg , 1 dL 1 dg dT = − ; |dL/L| ≤ 0.005 and |dg/g | ≤ 0.001 so T 2L 2g |dT /T | ≤ (1/2)(0.005) + (1/2)(0.001) = 0.003 = 0.3% 44. Let h be the height of the building, x the distance to the building, and θ the angle of elevation, then h = x tan θ, dh = tan θdx + x sec2 θdθ; if x = 100, θ = 60◦ , |dx| ≤ 1/6 ft, and √ 3 (1/6) + (100)(4)(π/900) < 1.7 ft. |dθ| ≤ (0.2)(π/180) = π/900 radians, then |dh| ≤ 45. (a) z = xy , dz = ydx + xdy , dz/z = dx/x + dy/y ; (r + s)%. (b) z = x/y , dz = dx/y − xdy/y 2 , dz/z = dx/x − dy/y ; (r + s)%. 547 Chapter 15 (c) z = x2 y 3 , dz = 2xy 3 dx + 3x2 y 2 dy , dz/z = 2dx/x + 3dy/y ; (2r + 3s)%. (d) z = x3 y 1/2 , dz = 3x2 y 1/2 dx + x3 dy/ 2y 1/2 , dz/z = 3dx/x + (1/2)dy/y ; (3r + s/2)%. k k k k ; at a point a, b, on the surface, − 2 , − 2 , −1 and hence bk, ak, a2 b2 is xy ab a b ab normal to the surface so the tangent plane is bkx + aky + a2 b2 z = 3abk . The plane cuts the x, 3k y , and z -axes at the points 3a, 3b, and , respectively, so the volume of the tetrahedron that is ab 9 1 3k 1 (3a)(3b) = k , which does not depend on a and b. formed is V = 3 ab 2 2 46. z = 47. (a) 2t + 7 = (−1 + t)2 + (2 + t)2 , t2 = 1, t = ±1 so the points of intersection are (−2, 1, 5) and (0, 3, 9). (b) ∂z/∂x = 2x, ∂z/∂y = 2y so at (−2, 1, 5) the vector n = −4i + 2j − k is normal to the surface. v = i + j+2k is parallel to the line; n · v = −4 so the cosine of the acute angle is √ √√ [n · (−v)]/( n − v ) = 4/ 21 6 = 4/ 3 14 . Similarly, at (0,3,9) the vector n = 6j − k is normal to the surface, n · v = 4 so the cosine of the acute angle is √ √√ 4/ 37 6 = 4/ 222. 48. z = xf (u) where u = x/y , ∂z/∂x = xf (u)∂u/∂x + f (u) = (x/y )f (u) + f (u) = uf (u) + f (u), ∂z/∂y = xf (u)∂u/∂y = −(x2 /y 2 )f (u) = −u2 f (u). If (x0 , y0 , z0 ) is on the surface then, with u0 = x0 /y0 , [u0 f (u0 ) + f (u0 )] i − u2 f (u0 ) j − k is normal to the surface so the tangent plane is 0 [u0 f (u0 ) + f (u0 )] x − u2 f (u0 )y − z = [u0 f (u0 ) + f (u0 )]x0 − u2 f (u0 )y0 − z0 0 0 x2 x0 0 f (u0 ) + f (u0 ) x0 − 2 f (u0 ) y0 − z0 y0 y0 = x0 f (u0 ) − z0 = 0 so all tangent planes pass through the origin. = 49. Use implicit diﬀerentiation to get ∂z/∂x = −c2 x/ a2 z , ∂z/∂y = −c2 y/ b2 z . At (x0 , y0 , z0 ), z0 = 0, a normal to the surface is − c2 x0 / a2 z0 − 2 2 c y0 c x0 x− 2 y−z = 2z a0 b z0 c2 x2 −20 a z0 − 2 c...
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