Exercise

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 55−8 Find the arc length of the parametric curve.5. x = cos 3 t, y = sin 3 t, z = 2; 0 ≤ t ≤ π/26. x = 3cost, y = 3sint, z = 4t; 0 ≤ t ≤ π7. x = e t , y = e −t , z = √ 2t; 0 ≤ t ≤ 18. x = 1 t, y = 1 2 3 (1−t)3/2 , z = 1 3 (1+t)3/2 ; −1 ≤ t ≤ 19−12 Find the arc length of the graph of r(t).9. r(t) = t 3 i+tj+ 2√ 1 6t 2 k; 1 ≤ t ≤ 310. r(t) = (4+3t)i+(2−2t)j+(5+t)k; 3 ≤ t ≤ 411. r(t) = 3costi+3sintj+tk; 0 ≤ t ≤ 2π12. r(t) = t 2 i+(cost+tsint)j+(sint−tcost)k; 0 ≤ t ≤ π13−16 Calculate dr/dτ by the chain rule.13. r(t) = ti+t 2 j; t = 4τ +114. r(t) = 〈3cost,3sint〉; t = πτ15. r(t) = e t i+4e −t j; t = τ 216. r(t) = i+3t 3/2 j+tk; t = 1/τ17. (a) Find the arc length parametrization of the linex = t, y = tthat has the same orientation as the given line and has reference point (0,0).(b) Find the arc length parametrization of the linex = t, y = t, z = tthat has the same orientation as the given line and has reference point (0,0,0).

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 618. (a) Find the arc length parametrization of the linex = −5+3t, y = 2t, z = 5+tthat has the same direction as the given line and has reference point (−5,0,5).(b) Use the parametric equations obtained in part (a) to find the point on the line thatis 10 units from the reference point in the direction of increasing parameter.19−21 Find an arc length parametrization of the curve that has the sameorientation as the given curve and for which the reference point correspondsto t = 0.19. r(t) = (4+cost)i+(5+sint)j; 0 ≤ t ≤ 2π20. r(t) = 1 3 t3 i+ 1 2 t2 j; t ≥ 021. r(t) = sine t i+cose t j+ √ 3e t k; t ≥ 0Exercise 4.41−9 Find T(t) and N(t) at the given point.1. r(t) = (t 2 −1)i+tj; t = 1 2. r(t) = 1 2 t2 i+ 1 3 t3 j; t = 13. r(t) = 5costi+5sintj; t = π/3 4. r(t) = lnti+tj; t = e5. r(t) = 4costi+4sintj+tk; t = π/2 6. r(t) = ti+ 1 2 t2 j+ 1 3 t3 k; t = 07. x = e t cost, y = e t sint,z = e t ; t = 0 8. r(t) = costi+sintj+k; t = π/49. r(t) = e t i+e t costj+e t sintk; t = 0Exercise 5.11−8 These exercises are concerned with functions of two variables.1. Let f(x,y) = x 2 y +1. Find(a) f(2,1) (b) f(1,2) (c) f(0,0)(d) f(1,−3) (e) f(3a,a) (f) f(ab,a−b)2. Let f(x,y) = x+ 3√ xy. Find(a) f(t,t 2 ) (b) f(x,x 2 ) (c) f(2y 2 ,4y)

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 73. Let f(x,y) = xy +3. Find(a) f(x+y,x−y) (b) f(xy,3x 2 y 3 )4. Let g(x,y) = xsiny. Find(a) g(x/y) (b) g(xy) (c) g(x−y)5. Find F(g(x),h(x)) if F(x,y) = xe xy , g(x) = x 3 , and h(y) = 3y +1.6. Find g(u(x,y),v(x,y)) if g(x,y) = ysin(x 2 y), u(x,y) = x 2 y 3 , and v(x,y) = πxy.7. Let f(x,y) = x+3x 2 y 2 , x(t) = t 2 , and y(t) = t 3 . Find(a) f(x(t),y(t) (b) f(x(0),y(0)) (c) f(x(2),y(2))8. Let g(x,y) = ye −3x , x(t) = ln(t 2 +1), and y(t) = √ t. Find g(x(t),y(t)).9−12 These exercises involve functions of three variables.9. Let f(x,y,z) = xy 2 z 3 +3. Find(a) f(2,1,2)(c) f(0,0,0)(b) f(−3,2,1)(d) f(a,a,a)10. Let f(x,y,z) = zxy +x. Find(a) f(x+y,x−y,x 2 )(b) f(xy,y/x,xz)11. Find F(f(x),g(y),h(z)) if F(x,y,z) = ye xyz , f(x) = x 2 , g(y) = y +1, and h(z) =z 2 .12. Find g(u(x,y,z),v(x,y,z),w(x,y,z)) if g(x,y,z) = zsinxy, u(x,y,z) = x 2 z 3 ,v(x,y,z) = πxyz, and w(x,y,z) = xy/z.13−14 These exercises are concerned with functions of four or more variables.13. (a) Let f(x,y,z,t) = x 2 y 3√ z +t. Find f( √ 5,2,π,3π).n∑(b) Let f(x 1 ,x 2 ,...,x n ) = kx k . Find f(1,1,...,1).k=114. (a) Let f(u,v,λ,φ) = e u+v cosλtanφ. Find f(−2,2,0,π/4).(b) Let f(x 1 ,x 2 ,...,x n ) = x 2 1 +x2 2 +···+x2 n . Find f(1,2,...,n).

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 815−18 Sketch the domain of f. Use solid lines for portions of the boundaryincluded in the domain and dashed lines for portions not included.15. f(x,y) = ln(1−x 2 −y 2 ) 16. f(x,y) = √ x 2 +y 2 −417. f(x,y) = 1x−y 218. f(x,y) = lnxy19−20 Describe the domain of f in words.19. (a) f(x,y) = xe −√ y+2(b) f(x,y,z) = √ 25−x 2 −y 2 −z 2(c) f(x,y,z) = e xyz√4−x220. (a) f(x,y) =y 2 +3xyz(c) f(x,y,z) =x+y +z(b) f(x,y) = ln(y −2x)21−30 Sketch the graph of f.21. f(x,y) = 3 22. f(x,y) = √ 9−x 2 −y 223. f(x,y) = √ x 2 +y 2 24. f(x,y) = x 2 +y 225. f(x,y) = x 2 −y 2 26. f(x,y) = 4−x 2 −y 227. f(x,y) = √ x 2 +y 2 +1 28. f(x,y) = √ x 2 +y 2 −129. f(x,y) = y +1 30. f(x,y) = x 2Exercise 5.21−6 Use limit laws and continuity properties to evaluate the limit.1. lim (4xy 2 −x) 2. lim (xy 2 sinxy)(x,y)→(1,3)(x,y)→(1/2,π)xy 33. lim(x,y)→(−1,2) x+y4. lim e 2x−y2(x,y)→(1,−3)5. lim(x,y)→(0,0) ln(1+x2 y 3 ) 6. lim(x,y)→(4,−2) x 3√ y 3 +2x7−10 Show that the limit does not exist by considering the limits as(x,y) → (0,0) along the coordinate axes.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 97. lim(x,y)→(0,0)9. lim(x,y)→(0,0)38. limx 2 +2y 2 (x,y)→(0,0)x−yx 2 +y 2 10. lim(x,y)→(0,0)x+y2x 2 +y 2cosxyx 2 +y 211−14 Evaluate the limit using the substitution z = x 2 +y 2 and observe thatz → 0 + if and only if (x,y) → (0,0).sin(x 2 +y 2 )1−cos(x 2 +y 2 )11. lim12. lim(x,y)→(0,0) x 2 +y 2 (x,y)→(0,0) x 2 +y 213. lim e −1/(x2 +y 2 )(x,y)→(0,0)14. lim(x,y)→(0,0)e −1/ √x 2 +y 2√x2 +y 215−22 Determine whether the limit exists. If so, find its value.x 4 −y 415. lim16. lim(x,y)→(0,0) x 2 +y 2 (x,y)→(0,0)xy17. lim(x,y)→(0,0)19. lim(x,y,z)→(2,−1,2)3x 2 +2y 2 18. lim(x,y)→(0,0)x 4 −16y 4x 2 +4y 21−x 2 −y 2x 2 +y 2xz 2√ 20. lim ln(2x+y −z)x2 +y 2 +z 2 (x,y,z)→(2,0,−1)sin(x 2 +y 2 +z 2 )sin √ x21. lim √ 22. lim2 +y 2 +z 2(x,y,z)→(0,0,0) x2 +y 2 +z 2 (x,y,z)→(0,0,0) x 2 +y 2 +z 2Exercise 5.31. Let f(x,y) = 3x 3 y 2 . Find(a) f x (x,y) (b) f y (x,y) (c) f x (1,y)(d) f x (x,1) (e) f y (1,y) (f) f y (x,1)(g) f x (1,2) (h) f y (1,2)2. Let z = e 2x siny. Find(a) ∂z/∂x (b) ∂z/∂y (c) ∂z/∂x| (0,y)(d) ∂z/∂x| (x,0) (e) ∂z/∂y| (0,y) (f) ∂z/∂y| (x,0)(g) ∂z/∂x| (ln2,0)(h) ∂z/∂y| (ln2,0)3. Let f(x,y) = √ 3x+2y. Find(a) Find the slope of the surface z = f(x,y) in the x-direction at the point (4,2).

RESULTADOSLa evaluación de nuestro trabajo la hacemos día a día, siendo las reuniones del equipointerdisciplinario esenciales para la misma. Creemos que el primer logro es la creación de nuestroequipo de trabajo, en donde en un clima afectuoso, de confianza, de respeto por las incumbencias, seanaliza al grupo de adultos mayores como totalidad y al caso por caso cuando se requiere. Elsegundo, es haber mantenido durante 4 años la actividad, con un crecimiento sostenido, y con eldeseo de incrementarla.En relación al grupo de adultos mayores, creemos que se logró:• Habilitar un espacio de participación activa.• Propiciar la recuperación y valorización de la propia historia de vida.• Incrementar la percepción de bienestar y autoestima positiva.• Desarrollar relaciones interpersonales, en un marco de libertad y de juego.• Ejercitar y entrenar funciones cognitivas, en todas las actividades.• Desarrollar y conservar la creatividad.Estos son los resultados a nivel grupal pero también están presentes los de cada persona, en susingularidad. Como ejemplo de esto último, tenemos a una señora que era analfabeta y empezó acursar su primaria; un señor que pudo pasar de una actitud de aislamiento en su vida de pareja, a laparticipación grupal; aquella señora que sólo se animaba a ser vestuarista pudo permitirse ser unatitiritera más; aquella otra que venció rutinas que parecían inmodificables para concurrir 8.30hs de lamañana. Los ejemplos recientemente mencionados son sólo algunos de los tantos que se logran díaa día. Una de las integrantes escribió un tango que identifica al grupo, titulado: “Dame un abrazo”,que en su letra dice “es un milagro estar reunidos….no extrañaremos la mocedad”, y no se extraña, ose extraña menos, porque se vive intensamente, en esta nueva experiencia de vida.CONCLUSIONESLas Naciones Unidas en su Asamblea General del 16 de diciembre de 1991 adoptó cincoPrincipios que apuntaban en una dirección que, a nuestro criterio, resume lo que entendemos porCalidad de Vida, o Vivir Activamente. Estos Principios son: -IndependenciaParticipaciónCuidadosAutorrealizaciónDignidadCada uno de estos Principios expresa la intención de procurar una mejor situación de Salud,entendida como ya vimos, para los Adultos Mayores y son el espíritu de nuestro trabajo. Creemosque el trabajo grupal, el abordaje desde múltiples disciplinas, la interpretación de la pieza comoespectáculo, permite recuperar en placer el esfuerzo de la elaboración, así como también les permite10

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 1119. f(x,y) = x 2 ye xy ; ∂f/∂x(1,1), ∂f/∂y(1,1)20. z = √ x 2 +4y 2 ; ∂z/∂x(1,2), ∂z/∂y(1,2)21. w = x 2 cosxy; ∂w/∂x ( 12 ,π) , ∂w/∂y ( 12 ,π)22. Let f(x,y,z) = x 2 y 4 z 3 +xy +z 2 +1. Find(a) f x (x,y,z) (b) f y (x,y,z) (c) f z (x,y,z)(d) f x (1,y,z) (e) f y (1,2,z) (f) f z (1,2,3)23. Let w = x 2 ycosz. Find(a) ∂w/∂x(x,y,z) (b) ∂w/∂y(x,y,z) (c) ∂w/∂z(x,y,z)(d) ∂w/∂x(2,y,z) (e) ∂w/∂y(2,1,z) (f) ∂w/∂z(2,1,0)24−26 Find f x , f y , and f z .24. f(x,y,z) = zln(x 2 ycosz) 25. f(x,y,z) = y −3/2 sec( ) 126. f(x,y,z) = tan −1 xy 2 z 327−30 Find ∂w/∂x, ∂w/∂y, and ∂w/∂z.27. w = ye z sinxz 28. w = x2 −y 2y 2 +z 229. w = √ x 2 +y 2 +z 2 30. w = y 3 e 2x+3z31. Let f(x,y,z) = y 2 e xz . Find( ) xzy(a) ∂f/∂x| (1,1,1) (b) ∂f/∂y| (1,1,1) (c) ∂f/∂z| (1,1,1)32. Let w = √ x 2 +4y 2 −z 2 . Find(a) ∂w/∂x| (2,1,−1) (b) ∂w/∂y| (2,1,−1) (c) ∂w/∂z| (2,1,−1)33. The volume V of a right circular cylinder is given by the formula V = πr 2 h, wherer is the radius and h is the height.(a) Find the formula for the instantaneous rate of change of V with respect to r ifr changes and h remains constant.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 12(b) Find the formula for the instantaneous rate of change of V with respect to h ifh changes and r remains constant.(c) Suppose that h has a constant value of 4 in, but r varies. Find the rate ofchange of V with respect to r at the point where r = 6 in.(d) Suppose that r has a constant value of 8 in, but h varies. Find the instantaneousrate of change of V with respect to h at the point where h = 10 in.34. The volume V of a right circular cone is given byV = π 24 d2√ 4s 2 −d 2where s is the slant height and d is the diameter of the base.(a) Find the formula for the instantaneous rate of change of V with respect to s ifd remains constant.(b) Find the formula for the instantaneous rate of change of V with respect to d ifs remains constant.(c) Suppose that d has a constant value of 16 cm, but s varies. Find the rate ofchange of V with respect to s when s = 10 cm.(d) Suppose that s has a constant value of 10 cm, but d varies. Find the rate ofchange of V with respect to d when d = 16 cm.35. According to the ideal gas law, the pressure, temperature, and volume of a gas arerelated by P = kT/V , where k is a constant of proprotionality. Suppose that V ismeasured in cubic inches (in 3 ), T is measured in Kelvin (K), and that for a certaingas the constant of proprotionality is k = 10 in-lb/K.(a) Find the instantaneous rate of change of pressure with respect to temperature ifthe temperature is 80 K and the volume remain fixed at 50 in 3 .(b) Find the instantaneous rate of change of volume with respect to pressure if thevolume is 50 in 3 and the temperature remain fixed at 80 K.36. The length, width, and height of a rectangular box are l = 5, w = 2, and h = 3,respectively.(a) Find the instantaneous rate of change of volume of the box with respect to thelength if w and h are held constant.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 13(b) Find the instantaneous rate of change of volume of the box with respect to thewidth if l and h are held constant.(c) Find the instantaneous rate of change of volume of the box with respect to theheight if l and w are held constant.37. The area A of a triangle is given by A = 1 absinθ, where a and b are the lengths of2two sides and θ is the angle between these sides. Suppose that a = 5, b = 10, andθ = π/3.(a) Find the rate at which A changes with respect to a if b and θ are held constant.(b) Find the rate at which A changes with respect to θ if a and b are held constant.(c) Find the rate at which b changes with respect to a if A and θ are held constant.38. (a) By differentiating implicitly, find the slope of the hyperboloid x 2 +y 2 −z 2 = 1 inthe x-direction at the points (3,4,2 √ 6) and (3,4,−2 √ 6).(b) Check the result in part (a) by solving for z and differentiating the resultingfunctions directly.39. (a) By differentiating implicitly, find the slope of the hyperboloid x 2 +y 2 −z 2 = 1 inthe y-direction at the points (3,4,2 √ 6) and (3,4,−2 √ 6).(b) Check the result in part (a) by solving for z and differentiating the resultingfunctions directly.40−43 Calculate ∂z/∂x and ∂z/∂y using implicit differentiation. Leave youranswers in terms of x, y, and z.40. (x 2 +y 2 +z 2 ) 3/2 = 1 41. ln(2x 2 +y −z 3 ) = x42. x 2 +zsinxyz = 0 43. e xy sinhz −z 2 x+1 = 044−47 Find ∂w/∂x, ∂w/∂y, and ∂w/∂z using implicit differentiation. Leaveyour answers in terms of x, y, z, and w.44. (x 2 +y 2 +z 2 +w 2 ) 3/2 = 4 45. ln(2x 2 +y −z 3 +3w) = z46. w 2 +wsinxyz = 1 47. e xy sinhw−z 2 w +1 = 048. Let z = √ xcosy. Find(a) ∂ 2 z/∂x 2 (b) ∂ 2 z/∂y 2 (c) ∂ 2 z/∂x∂y (d) ∂ 2 z/∂y∂x

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 1449. Let f(x,y) = 4x 2 −2y +7x 4 y 5 . Find(a) f xx (b) f yy (c) f xy (d) f yx50. Given f(x,y) = x 3 y 5 −2x 2 y +x, find(a) f xxy (b) f yxy (c) f yyy51. Given z = (2x−y) 5 , find(a)∂ 3 z∂y∂x∂y52. Given f(x,y) = y 3 e −5x , find(b)∂ 3 z∂x 2 ∂y(c)∂ 4 z∂x 2 ∂y 2(a) f xxy (0,1) (b) f xxx (0,1) (c) f yyxx (0,1)53. Given w = e y cosx, find∂ 3 w(a) ∣∂y 2 ∂x∣(π/4,0)54. Let f(x,y,z) = x 3 y 5 z 7 +xy 2 +y 3 z. Find(b)∂ 3 w∣∂x 2 ∂y∣(π/4,0)(a) f xy (b) f yz (c) f xz (d) f zz(e) f zyy (f) f xxy (g) f zyx (h) f xxyz55. Let w = (4x−3y +2z) 5 , find(a)∂ 2 w∂x∂z(b)∂ 3 w∂x∂y∂z(c)∂ 4 w∂z 2 ∂y∂xExercise 5.41−6 Compute the differential dz or dw of the specified function.1. z = 7x−2y 2. z = e xy3. z = x 3 y 2 4. z = 5x 2 y 5 −2x+4y +75. z = tan −1 xy 6. z = sec 2 (x−3y)7. w = 8x−3y +4z 8. w = e xyz9. w = x 3 y 2 z 10. 4x 2 y 3 z 7 −3xy +z −511. w = tan −1 (xyz) 12. w = √ x+ √ y + √ z

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 1513−18 Use a total differential to approximate the change in the values of f fromP to Q. Compare your estimate with the actual change in f.13. f(x,y) = x 2 +2xy −4x; P(1,2), Q(1.01,2.04)14. f(x,y) = x 1/3 y 1/2 ; P(8,9), Q(7.78,9.03)15. f(x,y) = x+y ; P(−1,−2), Q(−1.02,−2.04)xy16. f(x,y) = ln √ 1+xy; P(0,2), Q(−0.09,1.98)17. f(x,y,z) = 2xy 2 z 3 ; P(1,−1,2), Q(0.99,−1.02,2.02)xyz18. f(x,y,z) = ; P(−1,−2,4), Q(−1.04,−1.98,3.97)x+y +z19−26 (a) Find the local linear approximation L to the specified function f atthe designated point P. (b) Compare the error in approximating f by L at thespecified point Q with the distance between P and Q.19. f(x,y) =1√ ; P(4,3), Q(3.92,3.01)x2 +y2 20. f(x,y) = x 0.5 y 0.3 ; P(1,1), Q(1.05,0.97)21. f(x,y) = xsiny; P(0,0), Q(0.003,0.004)22. f(x,y) = lnxy; P(1,2), Q(1.01,2.02)23. f(x,y,z) = xyz; P(1,2,3), Q(1.001,2.002,3.003)24. f(x,y,z) = x+y ; P(−1,1,1), Q(−0.99,0.99,1.01)y +z25. f(x,y,z) = xe yz ; P(1,−1,−1), Q(0.99,−1.01,−0.99)26. f(x,y,z) = ln(x+yz); P(2,1,−1), Q(2.02,0.97,−1.01)Exercise 5.51−6 Use an appropriate form of the chain rule to find dz/dt.1. z = 3x 2 y 3 ; x = t t , y = t 2 2. z = ln(2x 2 +y); x = √ t, y = t 2/33. z = 3cosx−sinxy; x = 1/t, y = 3t 4. z = √ 1+x−2xy 4 ; x = lnt, y = t5. z = e 1−xy ; x = t 1/3 , y = t 3

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 166−9 Use an appropriate form of the chain rule to find dw/dt.6. w = 5x 2 y 3 z 4 ; x = t 2 , y = t 3 , z = t 57. w = ln(3x 2 −2y +4z 3 ); x = t 1/2 , y = t 2/3 , z = t −28. w = 5cosxy −sinxz; x = 1/t, y = t, z = t 39. w = √ 1+x−2yz 4 x; x = lnt, y = t, z = 4t10−15 Use an appropriate forms of the chain rule to find ∂z/∂u and ∂z/∂v.10. z = 8x 2 y −2x+3y; x = uv, y = u−v11. z = x 2 −ytanx; x = u/v, y = u 2 v 212. z = x/y; x = 2cosu, y = 3sinv13. z = 3x−2y; x = y +vlnu, y = u 2 −vlnv14. z = e x2y ; x = √ uv, y = 1/v15. z = cosxsiny; x = u−v, y = u 2 +v 216−23 Use an appropriate forms of the chain rule to find the derivatives.16. Let T = x 2 y −xy 3 +2; x = rcosθ, y = rsinθ. Find ∂T/∂r and ∂T/∂θ.17. Let R = e 2s−t2 ; s = 3φ, t = φ 1/2 . Find dR/dφ.18. Let t = u/v; u = x 2 −y 2 , v = 4xy 3 . Find ∂t/∂x and ∂t/∂y.19. Let w = rs/(r 2 +s 2 ); r = uv, s = u−2v. Find ∂w/∂u and ∂w/∂v.20. Let z = ln(x 2 +1), where x = rcosθ. Find ∂z/∂r and ∂z/∂θ.21. Let u = rs 2 lnt; r = x 2 , s = 4y +1, t = xy 3 . Find ∂u/∂x and ∂u/∂y.22. Let w = 4x 2 + 4y 2 + z 2 , x = ρsinφcosθ, y = ρsinφsinθ, z = ρcosφ. Find∂w/∂ρ, ∂w/∂φ, and ∂w/∂θ.23. Let w = 3xy 2 z 3 , y = 3x 2 +2, z = √ x−1. Find dw/dx.24. Use a chain rule to find the value of dwds∣ if w = r 2 −rtanθ; r = √ s, θ = πs.s=1/4

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 1725. Use a chain rule to find the value of∂f∣∂u∣u=1,v=−2and∂f∣∂v∣u=1,v=−2if f(x,y) = x 2 y 2 −x+2y; x = √ u, y = uv 3 .26. Use a chain rule to find the value of∂z∣∂r∣r=2,θ=π/6and∂z∣∂θ∣r=2,θ=π/6if z = xye x/y ; x = rcosθ, y = rsinθ.27. Use a chain rule to find dzdt∣ if z = x 2 y; x = t 2 , y = t+7.t=328−31 Use Theorem14.4 to find dy/dx and check your result usingimplicit differentiation.28. x 2 y 3 +cosy = 0 29. x 3 −3xy 2 +y 3 = 530. e xy +ye y = 1 31. x− √ xy +3y = 432. Two straight roads intersect at right angles. Car A, moving on one of the roads, approachesthe intersection at 25 mi/h and car B, moving on the other road, approachesthe intersection at 30 mi/h. At what rate is the distance between the cars changingwhen A is 0.3 mile from the intersection and B is 0.4 mile from the intersection?33. Use the ideal gas law P = kT/V with V in cubic inches (in 3 ), T in kelvins (K),and k = 10 in · lb/K to find the rate at which the temperature of a gas is changingwhen the volume is 200 in 3 and increasing at the rate of 4 in 3 /s, while the pressureis 5 lb/in 2 and decreasing at the rate of 1 lb/in 2 /s.34. Two sides of a triangle have lengths a = 4 cm and b = 3 cm but are increasing atthe rate of 1 cm/s. If the area of the triangle remains constant, at what rate is theangle θ between a and b changing when θ = π/6?35. Two sides of a triangle have lengths a = 5 cm and b = 10 cm, and the includedangle is θ = π/3. If a is increasing at a rate of 2 cm/s, b is increasing at a rate of 1cm/s, and θ remains constant, at what rate is the third side changing? Is it increasingor decreasing? [Hint: Use the law of cosines.]

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 1836. The length, width, and height of a rectangular box are increasing at rates of 1 in/s, 2in/s, and 3 in/s, respectively.(a) At what rate is the volume increasing when the length is 2 in, the width is 3 in,and the height is 6 in?(b) At what rate is the length of the diagonal increasing at that instant?37. Consider the box in Exercise36. At what rate is the surface area of the box increasingat the given instant?Exercise 5.61−8 Find D u f at P.1. f(x,y) = (1+xy) 3/2 ; P(3,1); u = 1 √2i+ 1 √2j2. f(x,y) = e xy ; P(4,0); u = − 3 5 i+ 4 5 j3. f(x,y) = ln(1+x 2 +y); P(0,0); u = − 1 √10i− 3 √10j4. f(x,y) = cx+dyx−y ; P(3,4); u = 4 5 i+ 3 5 j5. f(x,y,z) = 4x 5 y 2 z 3 ; P(2,−1,1); u = 1 3 i+ 2 3 j− 2 3 k6. f(x,y,z) = ye xz +z 2 ; P(0,2,3); u = 2 7 i− 3 7 j+ 6 7 k7. f(x,y,z) = ln(x 2 +2y 2 +3z 2 ); P(−1,2,4); u = − 3 i− 4 12j− k13 13 138. f(x,y,z) = sinxyz; P ( 12 , 1 3 ,π) ; u = 1 √3i− 1 √3j+ 1 √3k9−18 Find the directional derivative of f at P in the direction of a.9. f(x,y) = 4x 3 y 2 ; P(2,1); a = 4i−3j10. f(x,y) = x 2 −3xy +4y 3 ; P(−2,0); a = i+2j11. f(x,y) = y 2 lnx; P(1,4); a = 3i+3j12. f(x,y) = e x cosy; P(0,π/4); a = 5i−2j

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 1913. f(x,y) = tan −1 (y/x); P(−2,2); a = −i−j14. f(x,y) = xe y −ye x ; P(0,0); a = 5i−2j15. f(x,y,z) = x 3 z −yx 2 +z 2 ; P(2,−1,1); a = 3i−j+2k16. f(x,y,z) = y − √ x 2 +z 2 ; P(−3,1,4); a = 2i−2j−k17. f(x,y,z) = z −x ; P(1,0,−3); a = −6i+3j−2kz +y18. f(x,y,z) = e x+y+3z ; P(−2,2,−1); a = 20i−4j+5k19−21 Find the directional derivative of f at P in the direction of a vectormaking the counterclockwise angle θ with the positive x-axis.19. f(x,y) = √ xy; P(1,4); θ = π/320. f(x,y) = x−y ; P(−1,−2); θ = π/2x+y21. f(x,y) = tan(2x+y); P(π/6,π/3); θ = 7π/422. Find the directional derivative of f(x,y) = xx+yQ(−1,−1).at P(1,0) in the direction of23. Find the directional derivative of f(x,y) = e −x secy at P(0,π/4) in the direction ofthe origin.24. Find the directional derivative of f(x,y) = √ xye y at P(1,1) in the direction of the25. Letnegative y-axis.f(x,y) = yx+yFind a unit vector u for which D u f(2,3) = 0.26. Find the directional derivative off(x,y,z) = yx+zat P(2,1,−1) in the direction form P to Q(−1,2,0).

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2027. Find the directional derivative of the functionf(x,y,z) = x 3 y 2 z 5 −2xz +yz +3xat P(−1,−2,1) in the direction of the negative z-axis.28−31 Find ∇z or ∇w.28. z = 4x−8y 29. z = e −3y cos4x30. w = ln √ x 2 y 2 +z 2 31. w = e −5x secx 2 yz32−35 Find the gradient of f at the indicate point.32. f(x,y) = (x 2 +xy) 3 ; (−1,−1) 33. f(x,y) = (x 2 +y 2 ) −1/2 ; (3,4)34. f(x,y,z) = yln(x+y +z); (−3,4,0) 35. f(x,y,z) = y 2 ztan 3 x; (π/4,−3,1)Exercise 6.11−12 Evaluate the iterated integral.1.3.5.7.9.∫ 1 ∫ 20 0∫ 4∫ 12 0∫ ln3 ∫ ln20 0∫ 0∫ 5−1 2∫ 1 ∫ 1011.(x+3)dydx 2.x 2 ydxdy 4.e x+y dydx 6.dxdy 8.0∫ ln2∫ 10x(xy +1) dydx 20xye y2x dydx 12.∫ 3 ∫ 11 −1∫ 0∫ 2−2 −1∫ 2 ∫ 10 0∫ 6∫ 7410. ∫ π−3∫ 2π/2 1∫ 4∫ 231(2x−4y)dydx(x 2 +y 2 )dxdyysinxdydxdydxxcosxydydx1(x+y) 2 dydx13−16 Evaluate the double integral over the rectangular region R.13.∫∫R4xy 3 dA; R = {(x,y) : −1 ≤ x ≤ 1, −2 ≤ y ≤ 2}

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2114.15.16.∫∫R∫∫R∫∫Rxy√ dA; R = {(x,y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}x2 +y 2 +1x √ 1−x 2 dA; R = {(x,y) : 0 ≤ x ≤ 1, 2 ≤ y ≤ 3}(xsiny −ysinx)dA; R = {(x,y) : 0 ≤ x ≤ π/2, 0 ≤ y ≤ π/3}17−20 Use a double integral to find the volume.17. The volume under the plane z = 2x+y and over the rectangleR = {(x,y) : 3 ≤ x ≤ 5, 1 ≤ y ≤ 2}.18. The volume under the surface z = 3x 3 +3x 2 y and over the rectangleR = {(x,y) : 1 ≤ x ≤ 3, 0 ≤ y ≤ 2}.19. The volume of the solid enclosed by the surface z = x 2 and the planes x = 0,x =2,y = 3,y = 0, and z = 0.20. The volume in the first octant bounded by the coordinate planes, the plane y = 4,and the plane (x/3)+(z/5) = 1.21. Evaluate the integral by choosing a convenient order of integration:∫∫xcos(xy)cos 2 πxdA; R = [0, 1]×[0,π]2R22. (a) Sketch the solid in the first octant that is enclosed by the planes x = 0, z = 0,x = 5, z −y = 0, and z = −2y +6.(b) Find the volume of the solid by breaking it into two parts.Exercise 6.21−10 Evaluate the iterated integral.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 221.3.5.7.9.∫ 1 ∫ x0∫ 30xy 2 dydx 2.x 2∫ √ 9−y 2ydxdy 4.0∫ √ 2π√ π∫ x 20∫ π ∫ x 2π/2 0∫ 1 ∫ x00∫ 3/2 ∫ 3−y1∫ 1sin y x dydx 6. ∫ 11x cos y ∫ 1x dydx 8.y √ x 2 −y 2 dydx 10.y∫ x1/4∫ x−1∫ x0 0∫ 2 ∫ y 21ydxdyx 2 √ xy dxdy−x 2 (x 2 −y)dydx0e x2 dydxe x/y2 dxdy11. Let R be the region shown in the accompanying figure. Fill in the missing limits ofintegration.∫∫(a) f(x,y)dA =(b)R∫∫Rf(x,y)dA =∫ ✷∫ ✷✷ ✷∫ ✷ ∫ ✷✷✷f(x,y)dydxf(x,y)dxdyyy = x 2R2x12. Let R be the region shown in the accompanying figure. Fill in the missing limits ofintegration.∫∫(a) f(x,y)dA =R∫∫(b) f(x,y)dA =∫ ✷ ∫ ✷✷ ✷∫ ✷ ∫ ✷✷ ✷Rf(x,y)dydxf(x,y)dxdy

••MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 23yy = √ xRy = x 2x13. Let R be the region shown in the accompanying figure. Fill in the missing limits ofintegration.∫∫(a) f(x,y)dA =R∫∫(b)f(x,y)dA =∫ 2 ∫ ✷1 ✷∫ 5 ∫ ✷+4 ✷∫ ✷∫ ✷✷ ✷Rf(x,y)dydx+f(x,y)dydxf(x,y)dxdyy∫ 4 ∫ ✷R2 ✷(1,3) (5,3)f(x,y)dydx• •(2,1) (4,1)x14. Let R be the region shown in the accompanying figure. Fill in the missing limits ofintegration.∫∫(a) f(x,y)dA =(b)R∫∫Rf(x,y)dA =∫ ✷ ∫ ✷✷ ✷∫ ✷ ∫ ✷✷ ✷f(x,y)dydxf(x,y)dxdyy1−1R1x−1

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2415. Evaluate∫∫xydA, where R is the region inR(a) Exercise11(b) Exercise13∫∫16. Evaluate (x+y)dA, where R is the region inR(a) Exercise12(b) Exercise1417−20 Evaluate the double integral in two ways using iterated integrals: (a) viewing Ras a type I region, and (b) viewing R as a type II region.∫∫17. x 2 dA; R is the region bounded by y = 16/x, y = x, and x = 8.18.19.20.R∫∫R∫∫R∫∫xy 2 dA; R is the region bounded by y = 1, y = 2, x = 0, and y = x.(3x−2y)dA; R is the region enclosed by the circle x 2 +y 2 = 1.ydA; R is the region in the first quadrant enclosed between the circle x 2 +y 2 = 25Rand the line x+y = 5.21−26 Evaluate the double integral.21.∫∫x(1+y 2 ) −1/2 dA; R is the region in the first quadrant enclosed by y = x 2 , y = 4,22.23.Rand x = 0.∫∫xcosydA; R is the triangular region bounded by the lines y = x, y = 0, andRx = π.∫∫xydA; R is the region enclosed by y = √ x, y = 6−x, and y = 0.24.R∫∫RxdA; R is the region enclosed by y = sin −1 x, x = 1/ √ 2, and y = 0.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2525.∫∫(x − 1)dA; R is the region in the first quadrant enclosed between y = x and26.Ry = x 3 .∫∫x 2 dA; R is the region in the first quadrant enclosed by xy = 1, y = x, andRy = 2x.27−29 Use a double integration to find the area of the plane region enclosed by thegiven curves.27. y = sinx and y = cosx, for 0 ≤ x ≤ π/4.28. y 2 = −x and 3y −x = 4.29. y 2 = 9−x and y 2 = 9−9x.30−31 Use double integration to find the volume of the solid.30.z33x+2y +4z = 12x46y31.z2x 2 +z 2 = 4x22x 2 +y 2 = 4y32−40 Use double integration to find the volume of the solid.32. The solid bounded by the cylinder x 2 +y 2 = 9 and the planes z = 0 and z = 3−x.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2633. The solid in the first octant bounded above by the paraboloid z = x 2 + 3y 2 , belowby the plane z = 0, and laterally by y = x 2 and y = x.34. The solid bounded above by the paraboloid z = 9x 2 +y 2 , below by the plane z = 0,and laterally by the planes x = 0, y = 0, x = 3, and y = 2.35. The solid enclosed by y 2 = x, z = 0 and x+z = 1.36. The solid in the first octant bounded above by z = 9 − x 2 , below by z = 0, andlaterally by y 2 = 3x.37. The solid that is common to the cylinders x 2 +y 2 = 25 and x 2 +z 2 = 25.38. The solid bounded above by the paraboloid z = x 2 +y 2 , below by the xy-plane, andlaterally by the circular cylinder x 2 +(y −1) 2 = 1.39−44 Express the integral as an equivalent integral with the order of integrationreversed.39.41.43.∫ 2 ∫ √ x0 0∫ 2∫ e y0 1∫ 1 ∫ π/20 sin −1 yf(x,y)dydx 40.f(x,y)dxdy 42.f(x,y)dxdy 44.∫ 4 ∫ 80 2y∫ e∫ lnx1 0∫ 1 ∫ √ y0 y 2f(x,y)dxdyf(x,y)dydxf(x,y)dxdy45−48 Evaluate the integral by first reversing the order of integration.45.47.∫ 1 ∫ 40 4x∫ 4∫ 20e −y2 dydx 46.√ ye x3 dxdy 48.∫ 2 ∫ 10 y/2∫ 3∫ lnx1 0cos(x 2 )dxdyxdydxExercise 6.31−6 Evaluate the iterated integral.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 271.3.5.∫ π/2 ∫ sinθ0 0∫ π/2 ∫ asinθ0 0∫ π∫ 1−sinθ0 0rcosθdrdθ 2.r 2 drdθ 4.r 2 cosθdrdθ 6.∫ π ∫ 1+cosθ0 0∫ π/6 ∫ cos3θ0 0∫ π/2∫ cosθ0 0rdrdθrdrdθr 3 drdθ7−10 Use a double integral in polar coordinates to find the area of the regiondescribed.7. The region enclosed by the cardioid r = 1−cosθ.8. The region enclosed by the rose r = sin2θ.9. The region in the first quadrant bounded by r = 1 and r = sin2θ, with π/4 ≤ θ ≤π/2.10. The region inside the circle x 2 +y 2 = 4 and to the right of the line x = 1.11−12 Let R be the region described. Sketch the region R and fill in themissing limits of integration.∫∫f(r,θ)dA =∫ ✷ ∫ ✷✷ ✷Rf(r,θ)drdθ11. The region inside the circle r = 4sinθ and outside the circle r = 2.12. The region inside the circle r = 1 and outside the cardioid r = 1+cosθ.13−16 Use polar coordinates to evaluate the double integral.13.14.∫∫R∫∫e −(x2 +y 2) dA, where R is the region enclosed by the circle x 2 +y 2 = 1.√9−x2 −y 2 dA, where R is the region in the first quadrant within the circleRx 2 +y 2 = 9.∫∫115. dA, where R is the sector in the first quadrant bounded by y = 0,1+x 2 +y2 Ry = x, and x 2 +y 2 = 4.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2816.∫∫2ydA, where R is the region in the first quadrant bounded above by the circleR(x−1) 2 +y 2 = 1 and below by the line y = x.17−24 Evaluate the iterated integral by converting to polar coordinates.17.19.21.23.∫ 1 ∫ √ 1−x 20 0∫ 2∫ √ 2x−x 20 0∫ 1 ∫ √ y0 y∫ 4∫ √ 25−x 20 3(x 2 +y 2 )dydx 18.√x2 +y 2 dydx 20.√x2 +y 2 dxdy 22.dydx 24.∫ 2−2∫ 1∫ √ 4−y 2e −(x2 +y 2) dydx− √ 4−y y∫ √ 1−y 2cos(x 2 +y 2 )dxdy0 0∫ a ∫ √ a 2 −x 20 0∫ √ 20∫ √ 4−y 2y1dydx (a > 0)(1+x 2 +y 2 )3/21√1+x2 +y 2 dxdyExercise 6.41−8 Evaluate the iterated integral.1.3.5.7.∫ 1 ∫ 2 ∫ 1−1 0 0∫ 2 ∫ y 2 ∫ z0 −1 −1∫ 3∫ √ 9−z∫ 2 x0 0 0∫ 2 ∫ √ 4−x ∫ 2 3−x 2 −y 20 0(x 2 +y 2 +z 2 )dxdydz 2.yzdxdzdy 4.xydydxdz 6.−5+x 2 +y 2 xdzdydx 8.∫ 1/2 ∫ π ∫ 11/3 0 0∫ π/4 ∫ 1 ∫ x 20 0 0∫ 3∫ x 2 ∫ lnz1 x 0∫ 2 ∫ 2 ∫ √ 3y1z0zxsinxydzdydxxcosydzdxdyxe y dydzdxyx 2 +y 2 dxdydz9−12 Evaluate the triple integral.9.∫∫∫xysinyzdV , where G is the rectangular box defined by the inequalities 0 ≤10.Gx ≤ π, 0 ≤ y ≤ 1, 0 ≤ z ≤ π/6.∫∫∫ydV , where G is the solid enclosed by the plane z = y, the xy-plane, and theGparabolic cylinder y = 1−x 2 .

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 2911.∫∫∫xyzdV , where G is the solid in the first octant that is bounded by the parabolic12.Gcylinder z = 2−x 2 and the planes z = 0, y = x, and y = 0.∫∫∫cos(z/y)dV , where G is the solid defined by the inequalities π/6 ≤ y ≤ π/2,Gy ≤ x ≤ π/2, 0 ≤ z ≤ xy.13−15 Use a triple integral to find the volume of the solid.13. The solid in the first octant bounded by the coordinate planes and the plane 3x +6y +4z = 12.14. The solid bounded by the surface z = √ y and the planes x + y = 0, x = 0, andz = 0.15. The solid bounded by the surface y = x 2 and the planes y +z = 4 and z = 0.16−17 Set up (but do not evaluate) an iterated triple integral for the volume of thesolid enclosed between the given surfaces.16. The elliptic cylinder x 2 +9y 2 = 9 and the planes z = 0 and z = x+3.17. The cylinders x 2 +y 2 = 1 and x 2 +z 2 = 1.Exercise 7.11−2 Confirm that φ is a potential function for F(r) on some region.1. (a) φ(x,y) = tan −1 xyyF(x,y) =1+x 2 y i+ x2 1+x 2 y j 2(b) φ(x,y,z) = x 2 −3y 2 +4z 2F(x,y,z) = 2xi−6yj+8zk2. (a) φ(x,y) = 2y 2 +3x 2 y −xy 3F(x,y) = (6xy −y 3 )i+(4y +3x 2 −3xy 2 )j(b) φ(x,y,z) = xsinz +ysinx+zsinyF(x,y,z) = (sinz +ycosx)i+(sinx+zcosy)j+(siny +xcosz)k

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 303−8 Find divF and curlF.3. F(x,y,z) = x 2 i−2j+yzk 4. F(x,y,z) = xz 3 i+2y 4 x 2 j+5z 2 yk5. F(x,y,z) = 7y 3 z 2 i−8x 2 z 5 j−3xy 4 k 6. F(x,y,z) = e xy i−cosyj+sin 2 zk7. F(x,y,z) =1√x2 +y 2 +z 2(xi+yj+zk)8. F(x,y,z) = lnxi+e xyz j+tan −1 (z/x)k9−10 Find ∇·(F×G).9. F(x,y,z) = 2xi+j+4ykG(x,y,z) = xi+yj−zk10. F(x,y,z) = yzi+xzj+xykG(x,y,z) = xyj+xyzk11−12 Find ∇·(∇×F).11. F(x,y,z) = sinxi+cos(x−y)j+zk 12. F(x,y,z) = e xz i+3xe y j−e yz k13−14 Find ∇×(∇×F).13. F(x,y,z) = xyj+xyzk 14. F(x,y,z) = y 2 xi−3yzj+xykExercise 7.21. Let C be the line segment from (0,0) to (0,1). In each part, evaluate the line integralalong C by inspection, and explain your reasoning.∫∫(a) ds(b) sinxydyCC2. Let C be the line segment from (0,2) to (0,4). In each part, evaluate the line integralalong C by inspection, and explain your reasoning.∫∫(a) ds(b) e xy dxCC

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 313. Let C be the curve represented by the equationsx = 2t, y = 3t 2 (0 ≤ t ≤ 1)In each part, evaluate the line integral along C.∫∫(a) (x−y)ds(b) (x−y)dxCC∫(c) (x−y)dyC4. Let C be the curve represented by the equationsx = t, y = 3t 2 , z = 6t 3 (0 ≤ t ≤ 1)In each part, evaluate the line integral along C.∫∫(a) xyz 2 ds(b) xyz 2 dxCC∫∫(c) xyz 2 dy(d) xyz 2 dzCC5. In each part, evaluate the integral∫(3x+2y)dx+(2x−y)dyalong the stated curve.(a) The line segment from (0,0) to (1,1).(b) The parabolic arc y = x 2 from (0,0) to (1,1).(c) The curve y = sin(πx/2) from (0,0) to (1,1).(d) The curve x = y 3 from (0,0) to (1,1).C6. In each part, evaluate the integral∫ydx+zdy −xdzalong the stated curve.(a) The line segment from (0,0,0) to (1,1,1).(b) The twisted cubic x = t, y = t 2 , z = t 3 from (0,0,0) to (1,1,1).(c) The helix x = cospit, y = sinπt, z = t from (1,0,0) to (−1,0,1).

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 327.7−10 Evaluate the line integral with respect to s along the curve C.∫C11+x dsC : r(t) = ti+ 2 3 t3/2 j (0 ≤ t ≤ 3)∫x8.1+y ds 2CC : x = 1+2t, y = t (0 ≤ t ≤ 1)∫9. 3x 2 yzdsCC : x = t, y = t 2 , z = 2 3 t3 (0 ≤ t ≤ 1)∫e −z10.x 2 +y ds 211.CC : r(t) = 2costi+2sintj+tk (0 ≤ t ≤ 2π)11−18 Evaluate the line integral along the curve C.∫C(x+2y)dx+(x−y)dyC : x = 2cost, y = 4sint (0 ≤ t ≤ π/4)∫12. (x 2 −y 2 )dx+xdyCC : x = t 2/3 , y = t (−1 ≤ t ≤ 1)∫13. −ydx+xdyCC : y 2 = 3x from (3,3) to (0,0).∫14. (y −x)dx+x 2 ydyCC : y 2 = x 3 from (1,−1) to (1,1).∫15. (x 2 +y 2 )dx−xdyCC : x 2 +y 2 = 1, counterclockwise from (1,0) to (0,1).∫16. (y −x)dx+xydyCC : the line segment from (3,4) to (2,1).

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 3317.18.∫yzdx−xzdy +xydzCC : x = e t , y = e 3t , z = e −t (0 ≤ t ≤ 1)∫x 2 dx+xydy +z 2 dzCC : x = sint, y = cost, z = t 2 (0 ≤ t ≤ π/2)19−20 Evaluate ∫ ydx−xdy along the curve C shown in the figure.C19. (a) y(0,1)(1,0)x(b)y(1,1)(1,0)x20. (a)y(1,1)(2,0)x(b)y(0,5)(−5,0)(5,0)x

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 3421−24 Evaluate ∫ F·dr along the curve C.C21. F(x,y) = x 2 i+xyjC : r(t) = 2costi+2sintj (0 ≤ t ≤ π)22. F(x,y) = x 2 yi+4jC : r(t) = e t i+e −t j (0 ≤ t ≤ 1)23. F(x,y) = (x 2 +y 2 ) −3/2 (xi+yj)C : r(t) = e t sinti+e t costj (0 ≤ t ≤ 1)24. F(x,y,z) = zi+xj+ykC : r(t) = sinti+3sintj+sin 2 tk (0 ≤ t ≤ π/2)25−28 Find the work done by the force field F on a particle that moves alongthe curve C.25. F(x,y) = xyi+x 2 jC : x = y 2 from (0,0) to (1,1).26. F(x,y) = (x 2 +xy)i+(y −x 2 y)jC : x = t, y = 1/t (1 ≤ t ≤ 3)27. F(x,y,z) = xyi+yzj+xzkC : r(t) = ti+t 2 j+t 3 k (0 ≤ t ≤ 1)28. F(x,y,z) = (x+y)i+xyj−z 2 kC : along line segments from (0,0,0) to (1,3,1) to (2,−1,4).28−29 Find the work done by the force field1F(x,y) =x 2 +y i+ 42 x 2 +y j 2on a particle that moves along the curve C show in the figure.28.

•MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 35y(0,4)(4,0)x29.y•(6,3)x

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 36Exercise 7.31−6 Determine whether F is a conservative vector field. If so, find a potentialfunction for it.1. F(x,y) = xi+yj 2. F(x,y) = 3y 2 i+6xyj3. F(x,y) = x 2 yi+5xy 2 j 4. F(x,y) = e x cosyi−e x sinyj5. F(x,y) = xlnyi+ylnxj 6. F(x,y) = (cosy +ycosx)i+(sinx−xsiny)j7. (a) Show that the line integral ∫ C y2 dx+2xydy is independent of the path.(b) Evaluate the integral in part (a) along the line segment from (−1,2) to (1,3).(c) Evaluate the integral ∫ (1,3)(−1,2) y2 dx+2xydy using Theorem16.1, and confirm thatthe value is the same as that obtained in part (b).8. (a) Show that the line integral ∫ ysinxdx−cosxdy is independent of the path.C(b) Evaluate the integral in part (a) along the line segment from (0,1) to (π,−1).(c) Evaluate the integral ∫ (π,−1)(0,1)ysinxdx−cosxdy using Theorem16.1, and confirmthat the value is the same as that obtained in part (b).9−14 Show that the line integral is independent of the path, and useTheorem16.1 to find its value.9.11.∫ (4,0)(1,2)∫ (3,2)(0,0)3ydx+3xdy 10.2xe y dx+x 2 e y dy 12.∫ (1,π/2)(0,0)∫ (−1,0)(2,−2)e x sinydx+e x cosydy2xy 3 dx+3y 2 x 2 dy13.14.∫ (0,1)(−1,2)∫ (3,3)(1,1)(3x−y +1)dx−(x+4y +2)dy) ( )(e x lny − ey exdx+x y −ey lnx dy, where x and y are positive.15−18 Confirm that the force field F is conservative in some open connectedregion containing the points P and Q, and then find the work done by the forcefield on a particle moving along an arbitrary smooth curve in the region fromP to Q.

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 3715. F(x,y) = xy 2 i+x 2 yj; P(1,1), Q(0,0)16. F(x,y) = 2xy 3 i+3x 2 y 2 j; P(−3,0), Q(4,1)17. F(x,y) = ye xy i+xe xy j; P(−1,1), Q(2,0)18. F(x,y) = e −y cosxi−e −y sinxj; P(π/2,1), Q(−π/2,0)19−20 Find the exact value of ∫ F·dr using any method.C19. F(x,y) = (e y +ye x )i+(xe y +e x )jC : r(t) = sin(πt/2)i+lntj (1 ≤ t ≤ 2)20. F(x,y) = 2xyi+(x 2 +cosy)jC : r(t) = ti+tcos(t/3)j (0 ≤ t ≤ π)Exercise 7.41−2 Evaluate the line integral using Green’s Theorem and check the answer byevaluating it directly.∮1. y 2 dx+x 2 dy, where C is the square with vertices (0,0), (1,0), (1,1), and (0,1)Coriented counterclockwise.∮2. ydx+xdy, where C is the unit circle oriented counterclockwise.C3−13 Use Green’s Theorem to evaluate the integral. In each exercise, assumethat the curve C is oriented counterclockwise.∮3. 3xydx+2xydy, where C is the rectangle bounded by x = −2, x = 4, y = 1, andCy = 2.∮4. (x 2 −y 2 )dx+xdy, where C is the circle x 2 +y 2 = 9.C∮5. xcosydx−ysinxdy, where C is the square with vertices (0,0), (π/2,0), (π/2,π/2),Cand (0,π/2).∮6. ytan 2 xdx+tanxdy, where C is the circle x 2 +(y +1) 2 = 1.C

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 387.8.9.10.11.12.13.∮(x 2 −y)dx+xdy, where C is the circle x 2 +y 2 = 4.∮CC(e x +y 2 )dx+(e y +x 2 )dy, where C is the boundary of the region between y = x 2and y = x.∮ln(1 + y)dx− xy dy, where C is the triangle with vertices (0,0), (2,0), andC 1+y(0,4).∮x 2 ydx − y 2 xdy, where C is the boundary of the region in the first quadrant,Cenclosed between the coordinate axes and the circle x 2 +y 2 = 16.∮tan −1 ydx− y2 xdy, where C is the square with vertices (0,0), (1,0), (1,1),C 1+y2 and (0,1).∮cosxsinydx+sinxcosydy, where C is the triangle with vertices (0,0), (3,3),Cand (0,3).∮x 2 ydx+(y+xy 2 )dy, where C is the boundary of the region enclosed by y = x 2Cand x = y 2 .14. Use a line integral to find the area enclosed by the astroidx = acos 3 φ, y = asin 3 φ (0 ≤ φ ≤ 2π)15. Use a line integral to find the area of the triangle with vertices (0,0), (a,0), and(0,b), where a > 0 and b > 0.16−17 Use Green’s Theorem to find the work done by the force field F on aparticle that moves along the stated path.16. F(x,y) = xyi+ ( 12 x2 +xy ) j; the particle starts at (5,0), traverses the upper semicirclex 2 +y 2 = 25, and returns to its starting point along the x-axis.17. F(x,y) = √ yi + √ xj; the particle moves counterclockwise one time around theclosed curve given by the equations y = 0, x = 2, and y = x 3 /4.