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

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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...
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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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