M1320FullCourseKnolwledgeChecklistAndPractice-2 (1)
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Meru University College of Science and Technology (MUCST)**We aren't endorsed by this school
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MGMT MISC
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Jan 13, 2025
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Math 1320 Full Course KnowledgeChecklist and PracticeMarch 31, 2024Click on the (red-numbered links) to view example questions.Chapter 61.(6.1) Areas between curves.Given two functionsf(x)≥g(x) on [a, b], be able to perform thefollowing.(a) Be able to equate the area between the curvesAas the integralA=Zba[f(x)-g(x)] dx,and be able to compute the integral exactly when necessary. (1)(b) Iff(x)> g(x) forx(a, b), andf(a) =g(a),f(b) =g(b), be able to solve for the pointsx=aandx=band compute the area as in (a). (2)(c) Be able to compute areas enclosed between two curves and other given vertical boundaries, andareas enclosed by multiple curves. (3,4)2.(6.1) Areas between parametric curves (optional depending on instructor).(a) Be able to equate the area between the bottom curveA(x(t), y1(t)) and top curve (x(t), y2(t)) ont∈[a, b] by formulating a height functionh(t) =y2(t)-y1(t) and width differentialdx=x0(t)dt,to be formA=Zbah(t)x0(t)dt.A similar method can be applied on the vertical axis withdyin lieu ofdx.(b) Be able to compute the aforementioned integrals.3.(6.2) Volumes by stacked disks and washers.(a) Given a radius functionr(u)≥0 on [a, b] about theu-axis, whereu=xory, depending, describinga solid of revolution, be able to identify the volume of the solid with the integral. (11)(b) Be able to equate the area between the curvesAas the integralA=Zbaπr2(u)du.and be able to compute the integral exactly when necessary. (6)1
(c) Be able to compute the solid volume defined by areas in thex-yplane revolved aboutx- andy-axes,or any other axes given as vertical or horizontal lines. Be able to achieve this by constructing thecorrect inner- and outer-radius functionsr1(u) andr2(u), whereu=xory, depending, and usethe formulaA=Zbaπ r22(u)-r21(u) du.(7,8,10)(d) Be able to sketch a basic 3D picture of the solids of revolution described by areas in thex-y-planerevolved about a given axis.4.(6.3) Volumes by concentric cylinders.(a) Given a radius functionr(u) and a height functionh(u) on [a, b], be able to identify the volume ofthe solid or revolution with the integral. (??)A=Zba2πr(u)h(u)du.(b) Be able to compute the solid volume defined by areas in thex-yplane revolved aboutx- andy-axes,or any other axes given as vertical or horizontal lines. Be able to achieve this by constructing thecorrect radius functionr(u) and height functionh(u), whereu=xory, depending. (9,10)(c) Be able to sketch a basic 3D picture of the solids of revolution described by areas in thex-y-planerevolved about a given axis.5.(6.4) Arc Length.Given a curve in thex-yplane, be able to set up and compute an integralrepresenting the arc length of the curve.(a) Given a parametric curve (x(t), y(t)) fortin [a, b], commit to memory the arc length formulaL=Zbasdxdt2+dydt2dt.(12,13)(b) Given a curve defined by a functiony=f(x) on [a, b], know the arc length formula to beL=Zbas1 +dfdx2dx.(14)6.(6.5) Average value of a functionf(x), and the mean value theorem for integrals.(a) Given a functionf(x) on [a, b], commit to memory average value formulafavg=1b-aZbaf(x)dxand be able to use it in computations. (15)(b) Commit to memory mean value theorem for integrals (MVTI): Iff(x) is continuous on [a, b], thenthere is acin [a, b] such thatf(c) =favg=1b-aZbaf(x)dx.Page 2
(c) Be able to both check the hypotheses of the MVTI and find ac-value when required. (b)(d) Understand the geometric perspective of the MVTI, that the area under the curve off(x) can berepresented by an equivalent-area rectangle with baseb-aand heightf(c) =favg. (c)(e) Letfmaxandfminbe the maxima and minima values off(x) on [a, b]. Know thatfmin≤favg≤fmax.7.(6.6) Applications of integration: work and energy.(a) The general identity:W=Fd, when the force is constant.(b) Given a physical force functionf(u) on [a, b], the energy or workWdone by the force over adistanceb-ais given byW=Zbaf(u)du.(c) Given a description of a physical system involving motion over space [a, b], students should be ableto formulate an appropriate force function that describes the system utilizing common physicallaws – Newton’s laws, gravitation, electrostatics, and so on; and be able to construct the correctintegral that calculates the energy/work done by the system.(d) Specific force relations include, but are not limited to:(i) The Newton’s second law:F=ma(t), wherex(t) is position,a(t) =x00(t) is acceleration,andmis the mass of the object.(ii) Near earth gravitation:F=mg, wheremis the mass of the object that could vary as afunction of time or position – e.g. pulling up leaky buckets, pumping water out of tanks ofvarious shapes. (17,18)(iii) General gravitationF=Gm1m2r2, whereris the linear distance between massesm1andm2,andGa constant. (19)(iv) Electrostatic force between chargesq1andq2:F=kq1q2r2, whereris the linear distance andkis a constant.(v) Hookean springs:F=-kx, wherekis the spring constant. (20)8.(6.6) Applications of integration: hydrostatic pressure(a) Know the definition of pressure as a force per unit of area areaP=FA, and to calculate a force ona given surface and pressure, one needs to multiply the area by the pressure –F=PA.(b) Know that liquids with densityρin a container under the force of near-earth gravity exert pressureon the walls of the container equal toP=ρgd, wheredis the depth below the fill-line – i.e., pressureincreases directly with depth.(c) The wall of a filled tank with length‘(y) as a function of depthyexperiences a total force con-structed from the areas dA=‘(y)dyof segments at each dydepth, each with force dF=ρgy‘(y)dy,The total force is thenF=Rd0ρgy‘(y)dy.(d) Based on the tank description, be able to construct the length function‘(y) as a function of heightyand use it to compute the total force on a tank wall. (21)(e) Naturally, the total force on all walls of the vessel must equal the weight of the liquidgρV.9.(6.6) Applications of integration: center of mass, moments.(a) If given a density of a linear objectλ(x) (mass/length) laid on thex-axis on [a, b](i) The total mass:m=Rbaρ(x)dx.(ii) The center of mass, a position in space ¯x=1mRbaxρ(x)dxrepresenting the balance point ofthe object.Page 3
Chapter 810.(8.1) Sequences(a) Know the definition of a sequence as an infinite-sized ordered list of numbers{an}∞n=1={a1, a2, a3, . . .}.A sequence is indexed by a whole numbernthat indicates the point in the sequence.(b) Sequences can be any list of numbers, but there are several types that are of specific interest: (22)(i) Functions ofn:an=f(n) defined by a specified mathematical operation on an inputn; e.g.,an= 1/n2, oran=e-nsin(n).(ii) Recursively defined sequences:an+1is defined in terms ofan; e.g.,a1= 2 andan+1=0.5(3-an).(iii) Geometric: givena,r, definean=arn.(iv) Alternating sign: (-1)nor (-1)n-1.(c) Be able to recognize when two differently-indexed or represented sequences are equivalent, or not.(23)(d) Know the definition of convergence for sequences:Iflimn→∞an=Lexists, then the sequence isconvergent.(e) Be able to compute limits of sequences when possible; particularly, ifan=f(n), andf(x) is anindeterminate form, you can apply l’Hospital’s rule when needed.(f) Know how to utilize the Monotone Sequence Theorem (MST): If the sequence is monotonic in-creasing/decreasing, and the sequence is bounded above/below, respectively, then the sequence isconvergent.When applying the MST be sure to verify/assert/prove that the hypotheses of thetheorem are met in order to assert convergence.11.(8.2) Series(a) Aninfinite seriessis a sum of a sequence:s=∞Xn=1an.(b) Apartial sumsNis the sum of the firstNterms:sN=NXn=1an.(c) Know the definition of convergence and divergence of a series:sconverges when the sequence ofpartial sums converges, that islimN→∞sN=sis a finite number;sdiverges when it is undefined ordiverges to infinity. (24)(d) Be able to recognize special categories of series and their rules for convergence/divergence:(i) Geometric series:a.rnumbers, thens=∞Xn=0arnconverges when|r|<1, equal toa1-r. (25)(ii)p-series:pany number,s=∞Xn=11npconverges whenp >1 and diverges whenp≥1, by theintegral test (see below).(e) Know the basic series manipulation identities. If∑anand∑bnare convergent andcany number,then (26)(i)∑(an±bn) =∑an±∑bnis convergent.(ii)∑can=c∑an.12.(8.2-4) Series convergence tests fors=∞Xn=han, wherehis some integer.Page 4
(a) Test for divergence: A series is divergent iflimn→∞an6= 0 or undefined. (27)(b) Integral test: Ifan=f(n), andf(x)>0 forxin [h,∞), andf(x) is decreasing, thenR∞hf(x)dxhas the same convergence/divergence property ass. (28)(c) Comparison test (bounding by a known series): Supposean, bn>0, and suppose∑bnis known toconverge, and ifan≤bnthensconverges; conversely, if∑bnis known to diverge, and ifan≥bnthensdiverges. (29)(d) Limit comparison test: suppose∑bnis known to converge or diverge, andan, bn>0. If limn→∞anbn=c >0, wherecis finite, thensshares the same convergence property. (30)(e) Alternating series test: Supposean= (-1)n-1bn, wherebn>0,bn+1≤bn, andlimn→∞bn= 0, thensconverges. (31)(f) Know the definition of absolute convergence:∑anconverges absolutely if∑|an|is convergent.An absolutely convergent series is convergent, but a convergent series does not necessarily convergeabsolutely.(g) The ratio test: Computelimn→∞an+1an=L. The scenarios are as follows (32)(i) IfL <1 the series is absolutely convergent.(ii) IfL >1 or is =∞, then the series is divergent.(iii) IfL= 1 or DNE, then the test is indeterminate.13.(8.2-4) Series estimation methods for convergents=∞Xn=1an.(a) A geometric series does not need to be estimated, the exact value is known:s=a1-rif|r|<1.(b) Students should understand the need for series estimation methods. Anypartial sumsNcan beused as an approximation, but one might want to know how accurate eachsNis – estimationtechniques give information as to how close a given partial sumsNis tos.(c) Know the definition of a remainderRN=s-sN=∞Xn=N+1an. The remainder can be consideredthe error between an approximation and the exact value.(d) The remainder estimate for integral testsan=f(n)>0:Z∞N+1f(x)dx≤RN≤Z∞Nf(x)dx.(33)(e) Alternating series estimation theorem:ifs=∑∞n=1(-1)nbn, wherebnsatisfies the alternatingseries test hypotheses, then|RN| ≤bN+1. (34)14.(8.5-6) Power series(a) Know the definition of a function defined by a power series inx:f(x) =∞Xn=0cn(x-a)n.(b) Be able to use convergence tests, principally the ratio test, to determine for which valuesxdoesthe series converge.(c) Know the definition of the radius and interval of convergence and be able to determine them for agiven series. (35)(d) Using the known representation of a geometric power series bya/(1-r), be able to construct otherpower series by algebraic manipulation, integration, and differentiation. (36,37,38)Page 5
(e) Know under which conditions a functionf(x) represented by a power series∞Xn=0cn(x-a)nisdifferentiable (f0(x) exists) and computable by differentiation of the series term by term∞Xn=1ncn(x-a)n-1.(f) Know under which conditions a functionf(x) represented by a power series∞Xn=0cn(x-a)nis inte-grable (Rf(x)dxexists) and computable by integration of the series term by termC+∞Xn=0cn(x-a)n+1n+ 1.15.(8.7-8) Taylor and Maclaurin series(a) Be able to determine/compute the special power series representation of a function near a pointx=a– termed a Taylor series – asf(x) =∞Xn=0cn(x-a)n,wherecn=f(n)(a)n!,for|x-a|< R.This is known as aTaylor expansion offatx=a. Whena= 0, the Taylor series is termed aMaclaurin series. (39,40)(b) Know the definition of theNth partial sum as the order-NTaylor expansionTN(x) =NXn=0f(n)(a)n!(x-a)n(c) Know how to use the Taylor’s Inequality to determine the bound on the errorRN(x) in a neigh-borhood ofx=a:|x-a| ≤d: If|f(N+1)(x)|< M– the bound on the (N+ 1)th derivative forpoints|x-a| ≤d, then the remainder/error betweenTN(x) andf(x) is within the range (d,41,42)|f(x)-TN(x)|=|RN(x)| ≤M(N+ 1)!|x-a|N+1,for|x-a| ≤d.(d) If givenf(x) and a required error tolerance≥ |RN(x)|on|x-a| ≤d, students should beable to find the correct Taylor polynomial of orderNthat ensures this error tolerance is met;conversely, given anN, students should be able to determine the maximal error bound of|RN(x)|on|x-a| ≤d. (43)Chapter 916.(9.1) 3D coordinate systems(a) Know how to compute distances between points in the Cartesian 3D coordinate system.(b) Know how to represent a set of points forming a sphere as an equation of equal distancerfrom acenter point (x0, y0, z0) to the sphere surface:(x-x0)2+ (y-y0)2+ (z-z0)2=r2.(c) Know how to represent a set of points forming a cylinder as an equation of equal distancerfromthe vertical axis located at (x0, y0):(x-x0)2+ (y-y0)2=r2,wherez∈R.Page 6
(d) If given an equation of a sphere or a cylinder, be able to identify the center point or axis, respec-tively, and the radius. (44)(e) Be able to write out in set notation the points lying in a given surface. (45)17.(9.2) Vectors in 2- and 3D space(a) Know how to compute the algebraic vector computations of vector addition, subtraction, and scalarmultiplication using the algebraic properties on page 643. (46)(b) If given graphical depictions of two or more vectors, be able how to graphically represent vectoraddition, subtraction, and scalar multiplication. (47)(c) Given a vector~u, be able to produce a unit-length vector~Uin the same direction~U=~u|~u|; also befamiliar with the standard unit vectors~i,~j, and~k. (48)(d) Be able to recognize and use various vector notations of angle brackets~u=hu1, u2, u2i, and usingstandard unit vectors~u=u1~i+u2~j,+u2~k.(e) Be able to compute the magnitude of a vector using the formula|~u|=pu21+u22+u23.18.(9.3) Dot product(a) Know how to compute the dot product between two vectors~u·~v=u1v1+u2v2+u3v3.(49)(b) Be able to use the dot product angle-formula:~u·~v=|~u||~v|cos(θ).(50)(c) Know how to identify the dot product number in relation to the angleθbetween the vectors whendepicted graphically.(d) Given a force vector~Fon an object moving on a path given by~d, students should be able tocompute the workWdone on the object asW=~F·~d.(52)(e) Be able to compute the angle between vectors using the dot product-based formulaθ= cos-1~u·~v|~u||~v|.(50)(f) Know that two vectors are orthogonal when~u·~v= 0. (52)(g) Be able to perform vector calculations using the algebraic properties of the dot product shown onpage 651. (53)(h) Be able to find the scalar component of vector~uin the direction of vector~vusing thecompformulacomp~v(~u) =~u·~v|~v|.(i) Be able to graphically depict the quantitycomp~v(~u) and explain how it relates to the vectors. (54)Page 7
(j) Be able to find the vector projection of vector~uin the direction of vector~vusing theprojformulaproj~v(~u) =comp~v(~u)~v|~v|=~u·~v|~v|2~v.(55)(k) Be able to graphically depict the vector projection as the vector of lengthcomp~v(~u) in the directionof~v.19.(9.4) Cross product(a) Students should be able to compute cross product between two vectors to be~u×~v=hu2v3-u3v2, u3v1-u1v3, u1v2-u2v1i.(56)(b) Students should be able to utilize all the algebraic properties of the cross product (page 656) tocompute various vector calculations. (57)(c) The magnitude of the cross product relates to the angle between vectors according to the torqueformula in (a)|~u×~v|=|~u||~v|sin(θ),and is equal to the area of the parallelogram formed by the~uand~vsides. (58)(d) Students should be able to infer the properties of the cross product for two special~uand~vrelationships.(i)~u×~vis orthogonal to both~uand~v.(ii) If~uand~vpoint along the same direction—eitherθ= 0 orθ=π–then~u×~v=~0.(e) Students should be able to compute torque: If a vector force~Fis applied to a lever arm~rto effecta twisting force on an axis passing through the origin (the base of~r), the torqueτis the forcevector that points along the axis of twisting, and has magnitude|~τ|=|~F||~r|sin(θ),whereθis the angle between~Fand~r. The direction of~τis chosen with the right hand rule. (59)(f) Students should be able to compute the volumeVof a parallelepiped formed by three vectors~u,~v, and~wusing the triple product formulaV=~w·(~u×~v).(60)20.(9.5) Equations for lines and planes(a) Students should commit to memory the two equation forms for lines and be able to be able toproduce one from the other:(i) Symmetric:x-x0a=y-y0b=z-z0c(ii) Parametric:~r(t) =~vt+~r0,t∈R.(61)(b) Students should commit to memory the three equation forms for planes and be able to be able toproduce any one from the others:(i) Normal form:~n·(~r-~r0) = 0(ii) Scalar:ax+bx+cy=d.Page 8
(iii) Parametric:~r(s, t) =~ut+~vs+~r0,s, t∈R.(62)(c) Students should be able to produce the correct equation for a line or plane if given two, or threepoints respectively, that reside in the line/plane. (62)(d) Given two objects, either points, lines, or plane objects, students should be able to the objectdefined by their intersection (if it exists). (63)(e) Students should be able to find the angle of intersection between two line or plane objects. (64)(f) Students should be able to find the distance between a plane or line and a point in space by findingthe closest-distance point. (66)(g) Given a line or a plane, students should be able to find orthogonal or parallel lines or planes,possibly subject to additional conditions. (67,68)21.(9.6) Multivariate functions and surfaces(a) Be able to determine the domain and range of a multivariable functionz=f(x, y) and express thedomain as a 2D set.(b) Be able to draw or graphically depict the rudiments of the surface generated from a multivariablefunctionz=f(x, y). This can be done several strategies, when applicable.(i) Holding one variable fixed (e.g.,y) and drawing cross sections in the other variable (e.g.,x).(ii) Using special geometric principles if applicable (e.g., distance formulas, etc).(iii) Finding level sets: fixzand draw a contour lines in thex-yplane.(c) Be familiar with certain prototypical surfaces and their corresponding functions:Paraboloids,hyperbolic paraboloids, ellipsoids, cones, ... (see page 679).(69)22.(9.7) Cylindrical and spherical coordinates(a) Students should be able to convert between cartesian coordinates and cylindrical coordinates:x=rcos(θ),y=rsin(θ),tan(θ) =yx,r2=x2+y2,z=z.(70)(b) Students should be able to convert between cartesian coordinates and spherical coordinates:x=ρsin(φ) cos(θ),y=ρsin(φ) sin(θ),z=ρcos(φ)tan(θ) =yx,ρ2=x2+y2+z2, r=ρsin(φ).(70)(c) Be able to identify basic surfaces defined by equations expressed in these coordinate systems, likecones, spheres, etc. (71)Chapter 1023.(10.1) Vector functions(a) Students should be able to graphically represent space curves given by vector functions of the form~r(t) =hx(t), y(t), z(t)i,t∈[a, b].(72)Page 9
(b) Given a description of a basic space curve, students should be able to formulate a vector function~r(t) that represents the curve.(73)(c) Be able to formulate a space curve as a vector function~r(t) if the space curve is defined by a setof equations. (74)24.(10.2) Derivatives and integrals of vector functions.(a) Students should be able to compute derivatives of vector functions~r0(t) =hx0(t), y0(t), z0(t)i,t∈[a, b].(75)(b) Students should be able to graphically depict the direction and magnitude of~r0(t) as tangent tothe space curve with magnitude equal to the speedv(t) =|~r0(t)|.(c) Students should be able to compute derivatives of vector expressions using the differentiation ruleson page 704. (76)(d) Students should be able to compute integrals of vector functionsZba~r(t)dt=hZbax(t)dt,Zbay(t)dt,Zbaz(t)dti.25.(10.3) Arc length and curvature(a) Students should be able to produce and compute the arc lengthLformula of a space curve, definedasL=Zds=Zba|~r0(t)|dt,whereds=|~r0(t)|dt. (77,78)(b) Know how the graphical meaning and how to compute the tangent vector:~T(t) =~r0(t)|~r0(t)|.(c) Students should be able to explain and compute the curvatureκ(t) of a space curve as the magni-tude of rate of change of~Twith respect to arc lengthds:κ=d~Tds. The constructive formulas forcomputing curvature areκ(t) =|~T0(t)||~r0(t)|=|~r0(t)×~r00(t)||~r0(t)|3,whereds=|~r0(t)|dt. (79)(d) Students should know how to compute the, tangent, normal and binormal vectors forming a localcoordinate system (~T(t),~N(t),~B(t)) at every point~r(t) of a space curve (see page 712-713. Studentsshould be able to graphically depict the directions of these vectors where appropriate. (80)(e) Know the definitions of the normal and osculating planes of a space curve, know their normalvectors and how to find equations of the planes. Be able to describe the behavior of these planesas a function of timet.26.(10.4) Motion in space: velocity and acceleration(a) Students should be able to compute the velocity and acceleration vectors given a position vectorfunction~r(t)”~v(t) =~r0(t),~a(t) =~v0(t) =~r00(t).(b) Students should be able to decompose the velocity vector into its direction unit vector~T(t), andmagnitude given by speedv(t) =|~v(t)|:~v(t) =v(t)~T(t)Page 10
(c) Students should be able to decompose the acceleration vector into its tangential component~T(fore/aft direction), and its normal component~N(side-to-side, or steering direction) as expressedas:~a(t) =v0(t)~T(t) +κ(t)v2(t)~N(t),wherev0(t) is the tangential component of accelerationv0(t) =aT(t) =~T(t)·~a(t),andκ(t)v2(t) is the normal component of acceleration:κ(t)v2(t) =aN(t) =~N(t)·~a(t).(81)(d) Given a vector~x=hx1, x2, x3iin the standard coordinate system (~i,~j,~k), be able to represent~xin an alternative coordinate system defined by orthogonal vectors~u,~v,~was~x=a~u+b~v+c~wwhere the triplet (a, b, c) are the coordinates for~xin the new system given bya=~x· · ·~u|~u|2,b=~x· · ·~v|~v|2,a=~x· · ·~w|~w|2.(82)Chapter 1127.(11.1) Functions of several variablesf(x, y) =z.(a) Students should be able to determine the domain and range of multivariate functions.(b) Students should be able to determine the nature of the relationship between points (x, y) andzusing a methodical approach involving the synthesis of multiple approaches:(i) Forming a table of values for a relevant range of example points (x, y, z).(ii) Studying the algebraic relationship between (x, y) andz.(iii) Form graphical depictions of the surface by setting a variable equal to constants and drawingthe resulting traces or level curves of the remaining variables. Note that for each trace formedat each (x, y) point, the traces must intersect for the surface to be defined by a function.(83,84,85)(c) The aforementioned items must also be performed on any 3D coordinate system: Euclidian, cylin-drical, spherical.28.(11.2) Limits and continuity off(x, y) =z.(a) Students should understand that lim(x,y)→(a,b)f(x, y) =Lmeans thatf(x, y) takes onz-valuesthat are exceedingly close toLwhenever (x, y) is near (a, b).This means that any trajectory(x, y)→(a, b) yieldsz→L.(b) Students should be able to craft a clear argument that a given limitdoes not existby taking twolimits (x, y)→(a, b) along two distinct pathsC1andC2and establishing that the limits (if theyexists) are different. If the limits on the two paths are the same, then the no conclusions can bemade about the existence of the limit. (86,87)Page 11
(c) Continuity definition: Students should know the definition of continuity and use it to assert exis-tence of a limit. Definition: If a limit exists at a point lim(x,y)→(a,b)f(x, y) =L, andf(a, b) =L,then the function is continuous at (a, b). Limit existence: If a function is known to be continuous,students should be able to use this fact to assert what the limit is (L).(d) Determining continuity: Just as with single-variable functions, certain categories of functions likepolynomials, trigonometrics, logs, exponentials,, and function compositions, sums, products, anddivisions of these functions, are continuous, if not at singular points (e.g., division by zero pointsor other singularities). That is, if a function is in such a category and non singular at the pointin question, the function is continuous at that point.Students should be able to identify thesefunctions and be able to assert continuity when needed. (88)29.(11.3) Partial derivatives(a) Students should be able to compute partial derivatives using the limit-based definition.∂f∂x(a, b) = limh→0f(a+h, b)-f(a, b)h,∂f∂y(a, b) = limh→0f(a, b+h)-f(a, b)h.(b) Students should be able to compute partial derivatives using standard differentiation rules. (89,90)(c) Students should be familiar with the various partial derivative notations∂f∂x=fx=∂xf=Dxf,∂f∂y=fy=∂yf=Dyf,and second partial derivative notations, for instance∂2f∂x2=fxx∂xxf=Dxxf,∂2f∂x∂y= (fy)x=fyx=∂yxf=Dyxf.(d) Given a differentiable function (see below), students should be able to determine the slope of afunction’s tangent lines in thexorydirections using partial derivatives and be able to use thisinformation to understand the local properties of the surface off.(e) Students should be able to use Clairaut’s theorem to assert the equality of mixed partials: Iffisdefined in a region around (a, b) and the mixed partialsfxyandfyxare both continuous there (i.e.the mixed partials smooth), thenfxy=fyxat (a, b).30.(11.4) Tangent planes and linear approximations(a) Students should be able to compute/find tangent plane of a surface defined byf(x, y) =zat apoint (x0, y0, z0) is defined byz-z0=fx(x0, y0)(x-x0) +fy(x0, y0)(y-y0),where the normal vector to the tangent plane is~n=hfx(x0, y0), fy(x0, y0),-1i.(91)(b) Students should be able to use the tangent plane of function at a point (x0, y0, z0) as a way toapproximatef(x, y) at points (x, y) near (x0, y0):f(x, y) =z≈z0+fx(x0, y0)(x-x0) +fy(x0, y0)(y-y0).(92)(c) Differentiability definition:fis differentiable at a point (a, b) if the errorR(x, y) between a functionand its linear approximation goes to zero as (x, y)→(a, b) in such a manner thatR(x, y)/(x-a)→0 andR(x, y)/(y-b)→0. That is, a tangent plane well approximates the surface in a way thatRgoes to zero faster than linearly (e.g., quadratically). This is a difficult definition put here forreference only.Page 12
(d) Students should be a able to use the differentiability theorem—If the partials offexists and arecontinuous at (a, b), thenfis differentiable—to determine differentiability and the usability of atangent plane to make good approximations.31.(11.5) Chain rule(a) Given a differentiable trajectory in 2D~r(t) =hx(t), y(t)i, students should be able to compute thederivative off(x, y) =zwith respect to time on the trajectory using the chain rule in 2D:dzdt=∂f∂x(x(t), y(t))dxdt+∂f∂y(x(t), y(t))dydt=∂z∂xdxdt+∂z∂ydydt.(93)(b) Student should be able to compute derivatives similarly defined for functions of three variablesu=f(x, y, z) defined on 3D trajectories~r(t) =hx(t), y(t), z(t)i:dudt=fxx0+fyy0+fzz0.(94)(c) Students should be able to graphically depict and/or describez(t) and its surrounding local surfacebased on information inz0(t) computed as above in (a). (95)(d) Implicit functions: Given an equationF(x, y) = 0 that defines implicitly a functionf(x) =y(x) =y, students should be able to compute the derivative off(x) =yusing the chain rule:ddxF(x, y(x))= 0 =∂F∂xdxdx+∂F∂ydydx=⇒dydx=-FxFy.32.(11.6) Directional derivatives and gradient vector.(a) Students should be able to compute the directional derivative of a differentiable functionf(x, y) =zat a point (x, y) in any direction~u=ha, bi=hcos(θ),sin(θ)idefined as a unit vector|~u|= 1 givenby the formulaD~uf(x, y) =fx(x, y)a+fy(x, y)b.(96)(b) Students should be able to graphically identifyD~uf(x, y) as the slope of the tangent line on thesurface offin the direction~u.(c) Students should know how to compute the gradient vector∇f=hfx, fyi(d) Students should be able identify direction of travel with the steepest ascent of the surface is definedby the maximalDuf, which will always be in the direction of∇f. Moreover,∇fwill be orthogonalto any level curve tangent vector. (97)(e) Students should be able to express directional derivative by the gradient vector asD~uf(x, y) =∇f·~u.(f) Given a surface defined by the equationF(x, y, z) =k, students should be able to find the normalvector of the tangent plane of the surface at a point (x0, y0, z0) as~n(x0, y0, z0) =∇F(x0, y0, z0),with tangent plane equation given by0 =~n(x0, y0, z0)· hx-x0, y-y0, z-z0i.(98)33.(11.7) Minimum and maximum values of a surfacePage 13
(a) Students should utilize Fermat’s theorem (not his last theorem) to aid in finding local min andmax values: iffhas a local min/max, at (x, y), then∇f(x, y) =~0—the point (x, y) is a criticalpoint if∇f(x, y) =~0. (99)(b) Students should be able to use the second derivative test of a differentiablefto determine if acritical point (x, y) is a local minimum, maximum, saddle, or indeterminate point: DefineD(x,y)=fxxfxyfyxfyy=fxxfyy-(fxy)2.IfD >0 andfxx>0 =⇒local min; ifD >0 andfxx<0 =⇒local max;D <0 =⇒saddle; ifD= 0 =⇒indeterminate.(99,100)(c) Students should be able to determine absolute min and max values on a closed setUby checkingall critical points withinUand its boundary points. (101)34.(11.8) Lagrange multipliers—optimization under constraints.(a) Students should recognize that many optimization problems in science and engineering involvemaximizing/minimizing a given valuez=f(x, y) subject to some given constraint on (x, y) tosatisfy a given equationg(x, y) = 0—this could also be expressed in three or more independentvariables.(b) Solving the problem in (a) involves solving∇f=λ∇g,for an unknown scalarλto find critical point (x, y). (102,103)(c) Students should be able to graphically and mathematically explain the rationale behind (c):g(x, y) = 0 defines a level curve~r(t).The min/maxz(t) critical points will be attained trav-eling along on~r(t):z(t) =f(~r(t)).z(t) will be a critical point whenz0(t) = 0. By the chain rulez0(t) =fxx0+fyy0= 0 =∇f·~r0(t), meaning that the gradient offis orthogonal to the level curveonly at critical points. Note also that 0 =dgdt=∇g·~r0(t), meaning the gradient ofgis orthogonalto the level curve everywhere. Therefore∇gand∇fmust pointing along the same direction atcritical points; or rather, they are scalar multiples of each other, thus verifying the equation in (c)defining the critical points.Chapter 1235.(12.1) Double integrals over rectangular regions.(a) Be able to visualize functionsf(x, y) =zover rectangular regions [a, b]×[c, d].(b) Understand that the volume under the surface of a functionf(x, y) =zover rectangular regionR= [a, b]×[c, d] can be computed by a double Riemann sumV≈NXi=1MXj=1f(x*ij, y*ij)ΔA ,whereRis broken up into many rectangular patchesRij= [xi-1, xi]×[yj-1, yj] with area ΔA=ΔxΔy, wherexi=a+iΔx ,yj=c+jΔy ,Δx=b-aN,Δy=d-cMwith sample points (x*ij, y*ij) inRij.Page 14
(c) Be able to numerically approximate the volume using the double Riemann sum and choosing thesample points by one of the following methods (104)i. Basic midpoint rule: The sample points are in the middle of the rectangleRij, that isx*ij=¯xi=xi+xi-12andy*ij= ¯yj=yj+yj-12.ii. Left-endpoint rule: The sample points are in the lower left corner of the rectangleRij, thatisx*ij=xi-1andy*ij=yj-1.iii. Right-endpoint rule: The sample points are in the upper right corner of the rectangleRij,that isx*ij=xiandy*ij=yj.(d) Be able specify the a Riemann sum as a limit of finite sums from above as the definition of adouble integralV=limN→∞M→∞NXi=1MXj=1f(x*ij, y*ij)ΔA≡ZZRf(x, y)dA .36.(12.2) Iterated integrals.(a) Be able to apply Fubini’s theorem: Iff(x, y) =zis continuous on the rectangleR= [a, b]×[c, d],then the double integral can be computed by either of the following iterated integrals:ZZRf(x, y)dA=ZdcZbaf(x, y)dx dy=ZbaZdcf(x, y)dy dx ,where either iterated integral is computed by first computing the inner integral in one variable,keeping the other variable a fixed parameter, then subsequently computing the outer integralafterwards.(b) Be able to identify, where relevant, when one iterated integral is more advantageous or computa-tionally simpler than the other. (105)(c) When the integrand is a product of two functions in one variable, that isf(x, y) =g(x)h(y), beable to exploit the identity: (116)ZdcZbag(x)h(y)dx dy=Zbag(x)dx! Zdch(y)dy!.37.(12.3) Integration over general regions.(a) Be able to specify type 1 and type 2 regionsDin set notation:Type I:D={(x, y)|a≤x≤b, g1(x)≤y≤g2(x)}Type II:D={(x, y)|c≤y≤d, h1(y)≤x≤h2(y)}(b) Given a set of equations and/or lines defining the boundaries of a regionD, be able find theintersection points for the boundary curves and formulate the regionDas a type I or II regionwhen possible. If the full region is not type I or II, be able to split the region into subsets thatare when possible. (109)(c) Given a type I or II region, be able to express the double integralRRDf(x, y)dAas one of theiterated integrals containing functions of the outer integral variable in limits of integration of theinner integral:Type I:ZbaZg2(x)g1(x)f(x, y)dy dxType II:ZdcZh2(y)h1(y)f(x, y)dx dyBe able to compute these integrals. (106,107)Page 15
(d) If a regionDis both type I and II, be able to assess which iterated integral is most efficient tocompute. (108)(e) Be able to compute the area of a regionD, as (112)A=ZZD1dA .(f) If a regionDcannot be written as a region of type I or II, be able to break it up into regionsD1andD2of type I and/or II, that do not overlap, and knowZZDf(x, y)dA=ZZD1f(x, y)dA+ZZD2f(x, y)dA .38.(12.4) Integration with polar coordinates.(a) Be able to convert 2D regions expressed in Cartesian coordinates (x, y) into polar coordinate (r, θ)through the relations: (110)x=rcos(θ),y=rsin(θ),r2=x2+y2.(b) Commit to memory the area element in polar coordinates:dA=r dr dθ.(c) Be acquainted with the derivation of the area element formula, and understand the essentialprinciple that the area ΔAcovered due to an angle change Δθand radial change Δris in directproportion to the distancerof the area from the origin. That is, if the area in question is furtherfrom the origin, the area in question will be larger than if it were closer to the origin: ΔA∝r.(d) Be able to compute a double integral over a regionD={(r, θ)|a≤r≤b, α≤θ≤β}bytransforming it into an iterated integral in polar coordinates:ZZDf(x, y)dA=ZβαZbaf(rcos(θ), rsin(θ))r dr dθ .Be able to do the same when the bounds ofr, respectivelyθ, depend on the other variable. (111,112)39.(12.5) Applications of double integrals.(a) Mass from density: Ifρ(x, y) is the density (mass/area) of a laminar object with spatial extentgiven by a 2D regionD, know the massmis (113)m=ZZDρ(x, y)dA .(b) Center of mass: Given the laminar object defined in (a), be able to compute the center of mass,or balance point, (¯x,¯y) of the object from the formulas¯x=1mZZDxρ(x, y)dAand¯y=1mZZDyρ(x, y)dA .(c) Probability: Let (X, Y) be a pair of random events that could be observed to take on values ina given regionR, where the likelihood of the events is given byp(x, y) is a probability densityfunction. Know the properties of the probability density: (115)p(x, y)≥0andZZRp(x, y)dA= 1.Page 16
(d) LetDbe a subset of events ofR. Know how to interpret a verbal/written description ofDandformulate it in set notation:D={(x, y)|. . .}.(e) Be able to compute the probabilityP((x, y)∈D) of the eventsDas computed by the doubleintegral: (115,116)P((x, y)∈D) =ZZDp(x, y)dA .Page 17
Example Questions1. (6.1) Consider the curvesf(x) =√x, andg(x) =x2.(a) Determine the correct integral that expresses the area enclosed between the curves.Solution:A=Z10(√x-x2)dx(b) Compute the integral in (a).Solution:A=23x3/2-13x310=13.2. (6.1) Find the area enclosed by the given curves. Decide whether integration with respect toxoryismore pracitcal.(a)y= 2x-x2andy=xSolution:First the intersection points between the curves must be found.2x-x2=x,x-x2=x(1-x) = 0,x= 0,1.Note also that 2x-x2> xover the interval (0,1). So the area is given byA=Z10(2x-x2)-xdx=12-13=16.(b)y= cos(x) andy=x2-π24Solution:The intersection points are (-π2,0) and (π2,0), and cos(x)> x2-π24in between. Sothe area isA=Zπ2-π2cos(x)-x2-π24dx=sin(x)-13x3+π24xπ2-π2=2-23π38+ 2π24π2= 2 +π36(c)y2-1 =xand-y2+ 1 =xSolution:The intersection points are (0,-1) and (0,1). So the area isA=Z1-1((-y2+ 1)-(y2-1))dy=-23y3+ 2y1-1=-43+ 4 =83(d)y2=-xandy=x2Page 18
Solution:The intersection points are (0,0) and (-1,1). So the area isA=Z0-1(√-x-x2)dx=-23(-x)32-13x30-1=13Alternatively integrating overygivesA=Z10(-y2+√y)dy=-13y3+23y3210=133. (6.1) Consider the region bound byy=x2,y= (x-2)2, andy= 0.(a) Determine the integral that expresses the area.Solution:First, the intersection points between the curves must be found.x2= 0x= 0,and(x-2)2=x24x= 4x= 1,and(x-2)2= 0x= 2.So,x= 0 tox= 2 is the interval. Note also that (x-2)2>0, andx2>0 over (0,2). So thearea is given byA=Z10x2dx+Z21(x-2)2dx(b) Compute the integral.Solution:A=12+12= 1.4. (6.1) Find the area of the region enclosed by4x2+y2= 4,y= 2x-2,y=-2x-2(a) Draw the region.Page 19
-2-112-2-112xy(b) Set up and evaluate the integral with respect toxfor the area. (Hint: Split up the area into twopieces.)Solution:Split up the area atx= 0. The upper curve isy= 2√1-x2and the lower curve isy=-2x-2 respectivelyy= 2x-2. SoA=Z0-12p1-x2-(-2x-2)dx+Z102p1-x2-(2x-2)dx=2Z1-1p1-x2dx+Z0-1(2x+ 2)-Z10(2x-2) dxThe first integral describes a half-circle of radius 1. ThusA= 212π12+x2+ 2x0-1-x2-2x10=π+ 1 + 1 = 2 +π(c) Set up and evaluate the integral with respect toyfor the area. (Hint: Split up the area into twopieces.)Solution:Split up the area aty= 0. For the first part, the upper curve isx=12p4-y2andthe lower curve isx=-12p5-y2. For the second part, the upper curve isx=12y+ 1 and thelower curve isx=-12y-1. SoA=Z2012p4-y2+12p4-y2dy+Z0-212y+ 1--12y-1dy=Z20p4-y2dy+Z0-2y+ 2dyThe first integral describes a quarter-circle of radius 2. ThusA=14π22+12y2+ 2y0-2=π+ 2Page 20
5. (6.1) Consider the parametric curve given byx=t3,y=1t2-4,0≤t≤1Find the area under the parametric curve.Solution:A=Zx(1)x(0)ydx=Z10ydxdtdt=Z103t2t2-4dt=Z103 +3t2-4dtUsing partial fraction3t2-4=At-2+Bt+ 23 =At+ 2A+Bt-2B3 = 2A-2B0 =A+BA=-B3 = 4AA=34B=-34This givesA=Z103 +341t-2-1t+ 2dt=3t+34(ln|t-2| -ln|t+ 2|)10=1 +34(ln(1)-ln(3))-34(ln(2)-ln(3)) = 1-34ln(2)6. (6.2) Use the disk/washer method to find the volume of the following solids.(a) (a) The solid of revolution formed by revolving the area between the curvesy= sin(x) andy= 0on [0, π] about thex-axis.Solution:The volume has a radius defined byr(x) = sin(x), so the volume is given byA=Zπ0πsin2(x)dx=π2Zπ0(1-cos(2x)) dx=π22(b) The solid obtained by rotating the region bound byy=xex,x= 1 andy= 0 about thex-axis.Solution:V=πZ10(xex)2dx=πZ10x2e2xdx=πx212e2x10-Z102x12e2xdx!=π12e2-x12e2x10-Z1012e2xdx!!=π12e2-12e2+14e2x10!=π4(e2-1)using integration of parts twice.(c) The solid obtained by rotating the region bound byx=y1+y,y= 1 andx= 0 about they-axis.Page 21
Solution:V=πZ10y1 +y2dy=πZ10y2(1 +y)2dy=πZ21(u-1)2u2du=πZ211-2u+1u2du=πu-2 ln|u| -1u21=π32-2 ln(2)using substitutionu= 1 +y.(d) The solid obtained by rotating the region bound byy=x3,x= 0 andy= 0 aboutx= 1.Solution:x= 1 is parallel to they-axis, thus one has to integrate overy. Solving the functionforx:x=3√y. ThusV=πZ10(1-3√y)2dy=πZ101-23√y+y23dy=πy-234y43+35y5310=π1-32+35=110π(e) The solid obtained by rotating the region bound byy=-cos(x),y= 0,x=-π2andx=π2abouty=-1.Solution:y=-1 is parallel to thex-axis, thus one has to integrate overx.V=πZπ2-π2(-1 + cos(x))2dx=πZπ2-π2(1-2 cos(x) + cos(x)2)dx=πx-2 sin(x) +1212sin(2x) +xπ2-π2=ππ-2·2 +12π= 2π(π-2)using the trigonometric identity cos(x)2=12(cos(2x) + 1).(f) The solid of revolution formed by revolving the area enclosed byy=xe-x,x= 0,y= 0 on [0,∞)about thex-axis.Solution:The radiusr(x) =xe-x, soA=Z∞0π(xe-x)2dx=Z∞0πx2e-2xdx.Use integration by parts, twice.Notev0=e-2x, sov=-12e-2x, andu=x2, sou0= 2x;therefore,A=π-12e-2xx2∞0+12Z∞02xe-2xdx.The limits of integration in the first term both vanish when evaluated. Likewise, the remainingintegral can be computed by integration by parts again:A=πZ∞0xe-2xdx=π-12e-2xx∞0+12Z∞0e-2xdx=π14-e-2x∞0=π4Page 22
7. (6.2) Use the washer method to determine the correct integral, but do not compute, that expresses thevolume of the following solids.(a) The solid of revolution formed by revolving the area enclosed byy=√x,x= 1, on [0,1] andy= 0about they-axis.Solution:The integral will sum thin washers stacked along they-axis. At eachy, the washerhas an inner radius defined by thex-distance from thex= 0 point to√x=y; hencex=y2.Sor1(y) =y2. Also, the outer radius is simplyr2(y) =yso the volume is given byA=Z10π(1-y4)dy.(b) The solid of revolution formed by revolving the area enclosed byy=x,y=-x+ 4, andy= 0about they= 2 axis.Solution:y=xintersects thex-axis (y= 0) atx= 0, andy=-x+ 4 intersects at 4. So,the domain of area cross section isx∈[0,4], from whichyvaries across 0 and up to 2, sothe domain of integration will bey∈[0,2]. The radius of the inner diameter of the washer isr1(y) =y+ 2 The outer radius isr2(y) = 6-y=x. Therefore, the integral isA=Z20π((6-y)2-(y+ 2)2)dy.8. (6.2) Use the washer method to find the volume of the following solids or revolution.(a) The solid obtained by rotating the region bound byy= sin(x),y= 1 + sin(x),x= 0 andx=πabout thex-axis.Solution:V=πZπ0(1 + sin(x))2-sin(x)2dx=πZπ0(1 + 2 sin(x)) dx=π[x-2 cos(x)]π0=π(π+ 4)(b) The solid obtained by rotating the region bound byy=x3-1 andx= (y+ 1)2about they-axis.Solution:First find the intersections of the two curves. Substituting the first equation intothe second givesx=x6. Thusx= 0 orx= 1. So the intersections are (0,-1) and (1,0). Forrotation about they-axis, one has to integrate overy. So solving the first equation forxgivesx=3√y+ 1. SoV=πZ0-13py+ 12-((y+ 1)2)2dy=πZ10u23-u4dy=π35u53-15u510=25πusing substitutionu=y+ 1.9. (6.3) Use the shell method to find the volume of the following solids.Page 23
(a) The solid obtained by rotating the region bound byy=x4-x2,y= 0 andx= 1 about they-axis.Solution:V= 2πZ10xx4-x2dx= 2πZ10-1 +44-x2dxUsing partial fraction44-x2=A2-x+B2 +x4 = 2A+Ax+ 2B-Bx2A+ 2B= 4A-B= 0ThusA=B= 1 andV=2πZ10-1 +12-x+12 +xdx= 2π[-x-ln|2-x|+ ln|2 +x|]10=2π(1-ln(1) + ln(3) + ln(2)-ln(2)) = 2π(1 + ln(3))(b) The solid obtained by rotating the region bound byy=x2+ 1 andy= 3-x2aboutx= 2.Solution:For finding the intersection points, set the functions equalx2+ 1 = 3-x22x2= 2x=±1Thus the intersection points are (-1,2) and (1,2). SoV=2πZ1-1(2-x)(3-x2-(x2+ 1))dx= 4πZ1-1(4-x-2x2+x3)dx=4π4x-12x2-23x3+14x41-1= 4π8-43=803π(c) The solid obtained by rotating the region bound byy=xandy=x2abouty=-1.Solution:The intersection of the two curves are (0,0) and (1,1). Sincey=-1 is parallel tothex-axis, one has to integrate overy. Solving the equations forxgivesx=yandx=√y. SoV=2πZ10(y+ 1) (√y-y) dy= 2πZ10y32-y2+√y-ydy=2π25y52-13y3+23y32-12y210=715π10. (6.3) Consider the finite region bounded by the graphs ofy= (x+ 1)2andx=y2-1.(a) Sketch the the two curves. Label each curve and indicate the points of intersection. Shade in theenclosed region.-2-112-112yroutrinxyPage 24
-2-112-2-112xhrxySolution:The intersection points are (-1,0) and (0,1).(b) Find the volume of the solid of revolution formed by rotating the region aroundx= 1 using thewasher method. Label the outer radius and the inner radius in the first graph above.Solution:To use the washer method, solve forx:x=√y-1x=y2-1V=Z10π((y2-1)-1)2-((√y-1)-1)2dy=πZ10(y4-4y2+ 4-(y-4√y+ 4))dy=πZ10(y4-4y2-y+ 4√y)dy=π15y5-43y3-12y2+83y3210=π630-4030-1530+8030=3130π(c) Find the volume of the solid of revolution formed by rotating the region around thex= 1 usingthe shell method. Label the height and radius in the second graph above.Solution:To use the shell method, solve fory:y= (x+ 1)2y=√x+ 1V=Z0-12π(1-x)(√x+ 1-(x+ 1)2)dxUse substitutionu=x+ 1V=Z102π(2-u)(√u-u2)du= 2πZ10u3-2u2-u32+ 2√udu=2π14u4-23u3-25u52+43u3210=π1530-4030-2430+8030=3130πPage 25
11. (6.3) Consider the region bound byy=x3y=√x-11-11xyMatch the given integral with the solid, whose volume it computes, and the method that was used.Explain.A)Z10π(3√y+ 1)2-(y2+ 1)2dyB)Z10π(√x+ 1)2-(x3+ 1)2dxC)Z102π(x+ 1)(√x-x3)dxD)Z102π(y+ 1)(3√y-y2)dy1) the solid obtained by rotating the region aboutx=-1using thewasher method2) the solid obtained by rotating the region aboutx=-1using theshell method3) the solid obtained by rotating the region abouty=-1using thewasher method4) the solid obtained by rotating the region abouty=-1using theshell methodSolution:A) and B) are of the formπr2, so they are obtained by the washer method, and C) andD) are of the form 2πhr, so they are obtained by the shell method. A) corresponds to 1), since theintegral is integrated overy, so B) corresponds to 3).C) corresponds to 2), since the integral isintegrated overx, so D) corresponds to 4).12. (6.4) Consider the curve defined parametrically byy=e-tcos(t),x=e-tsin(t) fromt= 0 tot= 2π.Calculate the arc length.Page 26
Solution:dxdt=-e-tsin(t) +e-tcos(t) =e-t(cos(t)-sin(t))dydt=-e-t(cos(t) + sin(t))L=Z2π0sdxdt2+dydt2dt=Z2π0p(e-t)2((cos(t)-sin(t))2+ (cos(t) + sin(t))2)dt=Z2π0e-tp2 cos(t)2+ 2 sin(t)2dt=Z2π0e-t√2dt=h-e-t√2i2π0=√2(1-e2π)13. (6.4) Set-up (don’t evaluate) the integrals for the arc length for the following curves.(a)x=t2andy= cos(t) for 0≤t≤πSolution:L=Zπ0sdxdy2+dydt2dt=Zπ0p4t2+ sin(t)2dt(b)x(t) = ln(t+ 1) cos(t) andy(t) = ln(t+ 1) sin(t) for 0≤t≤2πSolution:The derivatives:x0(t) =-ln(t+ 1) sin(t) +cos(t)t+ 1,y0(t) = ln(t+ 1) cos(t) +sin(t)t+ 1.So,L=Z2π0s-ln(t+ 1) sin(t) +cos(t)t+ 12+ln(t+ 1) cos(t) +sin(t)t+ 12dt.14. (6.4) Set-up (don’t evaluate) the integrals for the arc length for the following curves.(a)x=y2-1 for-1≤y≤1Solution:L=Z1-1s1 +dxdy2dy=Z1-1p1 + 4y2dy(b)y=xexfor 0≤x≤1Page 27
Solution:L=Z10s1 +dydx2dx=Z10q1 + (ex+xex)2dx15. (6.5) Find the average value of the following functions over the given interval.(a)f(x) = 4-x2over [-2,2]Solution:fave=12-(-2)Z2-2f(x)dx=144x-13x32-2=1416-163=83(b)f(x) =xsin(x) over [0,2π]Solution:fave=12πZ2π0xsin(x)dx=12π[x(-cos(x))]2π0-Z2π0-cos(x)dx=12π(-2π+ 0) =-1using integration by parts.16. (6.5) Consider the functionf(x) =x3-x+ 1 on [-1,1].(a) Find the average value.Solution:fave=12Z1-1f(x)dx= 1(b) Find allc, such thatf(c) =faveSolution:1 =fave=f(x) =c3-c+ 1c3-c= 0c=-1,0,1(c) Sketch the graph offand a rectangle with a base [-1,1], whose area is the same as the area undergraph off.Page 28
-2-112-2-112xy17. (6.6) Compute the work required to lift a load 50kg up a 20-meter building using a cable. The cableweights 1kg/meter, and is spooled up as it is pulled up with the load.Solution:The massmof the system as a function ofy= 0 toy= 20 ism(y) = 50+20-y= 70-y.The force isF=gm(y). The work isW=Z200g(70-y)dy=g70×20-g2202.18. A conical tank as the pointy end aty= 0 and heighty= 2 meters. The top of the cone has an upperradiusr= 1.The tank is filled with water.Set up the integral that calculates the work required topump the water out over the top of the tank, but do not calculate. Water has mass density 1000 kg/m3.Solution:The radius of the tank as a function ofyisr(y) =12y. This makesr(0) = 0 andr(2) = 1,as required. The volume of water atyfor a thindysegment isdV=πr2(y)dy, so the mass at eachdysegment isdm= 1000πr2(y)dy. The force of gravity isF=mass×gand the work to move itfromyto 2, isdW=g(2-y)1000πr2(y)dyThe total work is found by summing alldWwork fromy= 0 toy= 2.W=Z20g(2-y)1000π14y2dy.19. (6.6) Compute the work required to launch a 1200 kg satellite vertically into the orbit of 20,000 km.Assume the mass of the earth isM= 5.97×1024kg, the radius of the earth isR= 6371 km and thegravitational constantG= 6.67×10-11 N m2kg2.Solution:The satellite has to be lifted fromR= 6371 km to 20,000 km and the gravitational forcePage 29
isF(r) =GmMr2, wherem= 1200 kg is the mass of the satellite. So the work required isW=Z20,000 km6371 kmGmMr2dr=-GmMr20,000 km6371 km=-GmM120,000 km-16371 km≈5.11×10-13J20. Calculate the work to compress a (Hookean) spring that exerts a forceF(x) =-2xwhen the spring’spositionxis compressed away from it’s starting neutral pointx= 0 up tox= 3.Solution:W=-Z302xdx= [-x2]30=-9.A positive work is done by the compressor, the spring does negative work.Either sign will beaccepted.21. Calculate the total force on one of the sidewalls of a two-meter cube tank of water. Note, the pressurePat every depthdisP=ρgd, whereρ= 1000 kg/m3. Set up the integral for the force, but do notcalculate.Solution:The length of a single sidewall is‘= 2 meters, so the force on eachdysegment at depthd= 2-yisdF=ρg(2-y)2dy.Recall that 2dyis the area, and the pressure at that depth isρg(2-y). The total force is the integral of the force at each depth fromy= 0 toy= 2:Ftotal=ZdF=Z20ρg(2-y)2dy.22. Write out the describe sequenceanindexed bynin set notation:{??}?n=?. Indicate start and endingindexes (if there is no end, indicate∞).(a)ancounts up by one, starting from 1 to infinity.(b)ancounts up by two, starting at 6 to infinity.(c)anis all positive odd numbers starting at one to infinity.(d)anhas a positive one at the startn= 1, and alternates sign for allnincreasing.Solution:(a){n}∞n=1(b){2n}∞n=3(c){2n-1}∞n=1(d){(-1)n+1}∞n=1Page 30
23. Match the equivalent sequences forn= 0,1,2,3,4,5,6. . ., or state that there is no match. Note that(-1)12=i.(a)an= sin(nπ/2)(b)bn= cos(nπ/2)(c)cn=((-1)n/2,n= 0, evens0,else(d)dn=((-1)(2n+1)/2,n=odds0,else(e)en=((-1)[1+(n+1)/2],n=odds0,elseSolution:bn=cn,an=en, anddnis unmatched.24. (8.2) The given series are telescoping series.Find the partial sums and decide whether the series isconvergent or divergent. If it is convergent, find its sum.(a)∞Xk=12k(k+ 2)Solution:Write the summand as a difference2k(k+ 2)=Ak+Bk+ 22 =Ak+ 2A+BkA= 1B=-1So thenth partial sum issn=nXk=11k-1k+ 2=nXk=11k-nXk=11k+ 2=nXk=11k-n+2Xk=31k=0n= 01-13n= 11 +12-13-14n= 21 +12-1n+1-1n+2n >2Now the limit of the partial sums exists and it is∞Xk=12k(k+ 2)= limn→∞sn= limn→∞1 +12-1n+ 1-1n+ 2= 1 +12=32(b)∞Xn=2lnnn+ 1Solution:Write each log ratio term as a difference of logs:lnnn+ 1= ln(n)-ln(n+ 1)Page 31
So, the partial sum toNcan be expressed assN=NXn=2lnnn+ 1= [ln(2)-ln(3)] + [ln(3)-ln(4)] +. . .+ [ln(N)-ln(N+ 1)].= ln(2) + [-ln(3)] + ln(3)] + [-ln(4) + ln(4)] +. . .-ln(N+ 1)= ln(2)-ln(N+ 1).limN→∞sN=limN→∞[ln(2)-ln(N+ 1)] =-∞,and so the series is divergent.(c)∞Xk=2lnk2-1k2+ 2kSolution:Write the summand as a differencelnk2-1k2+ 2k= ln(k2-1)-ln(k2+ 2k) = ln(k2-1)-ln((k+ 1)2-1)So thenth partial sum forn≥2 issn=nXk=2lnk2-1k2+ 2k=nXk=2ln(k2-1) +nXk=2ln((k+ 1)2-1)=nXk=2ln(k2-1)-n+1Xk=3ln(k2-1) = ln(2)-ln((n+ 1)2-1)Nowlimn→∞sn= limn→∞(ln(2)-ln((n+ 1)2-1))=∞and so the series is divergent.25. (8.2) Decide whether the given geometric series is convergent or divergent. If it is convergent, find itssum.(a)∞Xk=032k+12-kSolution:∞Xk=032k+12-k=∞Xk=0392kSo the common ratior=92>1 and the series is divergent.(b)∞Xk=2(-2)k33k-1Solution:∞Xk=2(-2)k33k-1=∞Xk=23-227=3(-227)21--227=49·272927=49·29=4261Page 32
since the common ratio isr=-227,|r|<1 and the series is convergent.26. (8.2) Decide whether the following series are divergent or convergent.(a)∞Xk=11k+12kSolution:∞Xk=11kis divergent, since it is a harmonic series, and∑∞k=112kis convergent, since itis a geometric series with common ration12. So∞Xk=11k+12khas to be divergent.(b)∞Xk=11k2+12kSolution:∞Xk=11k2is convergent, since it is ap-series withp >1, and∑∞k=112kis convergent,since it is a geometric series with common ration12. So∞Xk=11k2+12k=∞Xk=11k2+∞Xk=112kis convergent.27. (8.2) Use the Test for Divergence on the following series. Decide whether it shows the series is divergentor it is inconclusive.(a)∞Xk=0kk2+ 1Solution:limk→∞kk2+ 1= limk→∞1k+1k= 0So the test is inconclusive.(b)∞Xk=1kcos(k)Solution:The limit limk→∞kcos(k) does not exist, so the series is divergent.(c)∞Xk=1cos1kSolution:limk→∞cos(1k= 1So the series is divergent.Page 33
28. (8.3) Use the integral test to decide whether the following series are divergent or convergent. Make sureto check, that it is possible to use the integral test.(a)∞Xk=0kk2+ 1Solution:The functionxx2+1is positive and continuous. It is decreasing, sinceddxxx2+ 1=(x2+ 1)-x(2x)(x2+ 1)2=1-x2(x2+ 1)2<0forx >1Z∞0xx2+ 1dx=Z∞11u12du=12[ln|u|]∞1=∞using substitutionu=x2+ 1, du= 2xdx. Since the integral diverges, the series is divergent.(b)∞Xk=21k(ln(k))2Solution:The function1xln(x)2is positive and continuous. It is decreasing, since the denomi-nator is increasing.Z∞21xln(x)2dx=Z∞ln(2)1u2du=-1u∞ln(2)=1ln(2)using substitutionu= ln(x), du=1xdx. Since the integral converges, the series is convergent.(c)∞Xk=1k2Solution:It is not possible to use the integral test, since the sequence{k2}∞k=1is increasingand thus any function, that takes these values at the integers, can’t be decreasing.29. (8.3) Use the comparison test to compare the given series to the indicated series. Decide(i) whether the indicated series is convergent or divergent(ii) whether the comparison test shows the given series is convergent, divergent or if the comparisontest is inconclusive(a) Compare∞Xk=11k+ 1to∞Xk=11k.Solution:(i)∞Xk=11kis the harmonic series, so it is divergent.(ii) Now1k+ 1≤1kSo the comparison test is inconclusive.Page 34
(b) Compare∞Xk=03k5k+ 3to∞Xk=035kSolution:(i)∞Xk=035kis a geometric series with common rationr=35<1, so the series is convergent.(ii) Now3k5k+ 3≤3k5k=35kSo by the comparison test,∞Xk=03k5k+ 3is convergent.(c) Compare∞Xk=03k+12k-4to∞Xk=032kSolution:(i)∞Xk=032kis a geometric series with common rationr=32>1, so the series is divergent.(ii) Now3k+12k-4≥33k2k≥3k2k=32kSo by the comparison test,∞Xk=03k+12k-4is divergent.30. (8.3) Use the limit comparison test to decide whether the given series are convergent or divergent. Useeither a geometric series or ap-series to compare the series to.(a)∞Xk=1√k2+ 1k3-√2kSolution:To find a series to compare this series to, take the highest powers of numerator anddenominator, that is∞Xk=1kk3=∞Xk=11k2. This is ap-series withp= 2>1, so it is convergent.Now using the limit comparison testlimk→∞√k2+1k3-√2k1k2= limk→∞k2√k2+ 1k3-√2k= limk→∞1k31k3k2√k2+ 1k3-√2k= limk→∞q1 +1k21-q21k5= 1So∞Xk=1√k2+ 1k3-√2kis also convergent.(b)∞Xk=032k-kPage 35
Solution:Since 2kgrowths faster thank, compare this series to∞Xk=012k. This is a geometricseries withr=12<1, so it is convergent. Now using the limit comparison testlimk→∞32k-k12k= limk→∞32k2k-k= limk→∞31-k2k= 3sincelimk→∞k2k= 0. So∞Xk=032k-kis convergent.31. (8.4) Use the alternating series test on the following series.Decide whether is shows the series isconvergent or if the test is inconclusive.(a)∞Xk=1(-1)nnn2+ 1Solution:To see the sequencebk=kk2+1is decreasing, consider the functionf(x) =xx2+1. Ifthis function is decreasing, then the sequence is decreasing. Nowf0(x) =x2+ 1-x(2x)(x2+ 1)2=1-x2(x2+ 1)2<0forx >1So the function and especially the sequence is decreasing. Alsolimk→∞bk= limk→∞kk2+ 1= 0So by the alternating series test the given series is convergent.(b)∞Xk=1k(-1)k(k+ 2)Solution:Consider the sequencebk=kk+2. It islimk→∞kk+ 2= 1So this series does not satisfy the conditions of the alternating series test and the test isinconclusive.32. (8.4) Use the ratio test on the following series. Decide whether it shows the series is convergent, divergentor if the test is inconclusive.(a)∞Xk=1k!3kSolution:limk→∞(k+1)!3k+1k!3k= limk→∞(k+ 1)!k!3k3k+1= limk→∞k+ 13=∞So the series is divergent.Page 36
(b)∞Xk=2k2k4-√kSolution:limk→∞(k+1)2(k+1)4-√k+1k2k4-√k= limk→∞(k+ 1)2k2k4-√k(k+ 1)4-√k+ 1!= 1since in each fraction the highest power in numerator and denominator are the same. So theroot test is inconclusive.(c)∞Xk=1(-1)kk2k+k2Solution:limk→∞(-1)k+1(k+1)2k+1+(k+1)2(-1)kk2k+k2= limk→∞k+ 1k2k+k22k+1+ (k+ 1)2= limk→∞k+ 1k1 +k22k2 +(k+1)22k!=12Sincelimk→∞k22k= 0 = limk→∞(k+ 1)22k. So the series is convergent.33. Consider the convergingp-seriess=∑∞n=11n2.(a) Find the remainder of thes9partial sum. Express using summation notation.Solution:R9=s-s9=∞Xn=101n2.(b) Using the integral remainder estimate, estimate how closes9is tos.Solution:Z∞91x2dx= 1/9,Z∞101x2dx= 1/10.So110≤ |R9| ≤19(c) Iss9within 0.1 ofs?Solution:no,s9is more than a tenth fromsbecause the error/remainder is larger than 1/10.34. Consider the converging alternating seriess=∑∞n=0(-1)ne-√n.(a) Find the remainder of thes15partial sum. Express using summation notation.Solution:R15=s-s15=∞Xn=16(-1)ne-√n.Page 37
(b) Using the alternating series remainder estimate, estimate how closes15is tos.Solution:|R15|=|a16|=e-√16=e-4≈0.0183.(c) Iss15within 1/16th ofs?Solution:Yes, 1/16 = 2-4= 0.0625> e-4= 0.0183.35. (8.5) Find the interval of convergence and the radius of convergence for the following power series.(a)∞Xk=0(x+ 3)k2kSolution:This is a geometric series with common rationr=x+32. This converges forx+32<1and diverges forx+32≥1. So it converges if and only if|x+ 3|<2, that is-2< x+ 3<2-5< x <-1So the radius of convergence is 2 and the interval of convergence is (-5,-1).(b)∞Xk=1(2x+ 1)k32k-1kSolution:Use the ratio testlimk→∞(2x+1)k+132(k+1)-1(k+1)(2x+1)k32k-1k= limk→∞|2x+ 1|132kk+ 1=|2x+ 1|9So the test gives, that the series converges for|2x+ 1|<9, diverges for|2x+ 1|>9 and it isinconclusive for|2x+ 1|= 9. So now testx= 4 andx=-5 seperately.x= 4 :∞Xk=19k32k-1k=∞Xk=13kThis series is a multiple of the harmonic series, so it diverges.x=-5 :∞Xk=1(-9)k32x-1k=∞Xk=13(-1)kkThis series is a multiple of the alternating harmonic series, so it converges. So the interval ofconvergence for the power serie is [-5,4) and the radius of convergence is92.(c)∞Xk=0k(x-2)kk!Solution:Use the ration testlimk→∞(k+1)(x-2)k+1(k+1)!k(x-2)kk!= limk→∞k+ 1kk!(k+ 1)!|x-2|= limk→∞k+ 1k1k+ 1|x-2|= 0Page 38
So the series converges for allx. That is the interval of convergence is (-∞,∞) and the radiusof convergence isR=∞.36. (8.6) Use11-x=∞Xk=0xkto find a power series for the following functions.(a)33-xSolution:33-x=11-x3=∞Xk=0x3k=∞Xk=013kxk(b)x22 +xSolution:x22 +x=x2211-(-x2)=x22∞Xk=0-x2k=∞Xk=2(-1)k2k-1xk(c)1 +x1-xSolution:1 +x1-x= (1 +x)∞Xk=0xk=∞Xk=0xk+∞Xk=0xk+1Note that∑∞k=0xk+1=∑∞k=1xk, and∑∞k=0xk= 1 +∑∞k=1xk, so1 +x1-x= 1 + 2∞Xk=1xk.37. (8.6) Use11-x=∞Xk=0xkto find a power series in (x-a) for the following functions.(a)12 +xwitha=-1Solution:12 +x=11-(-(x+ 1))=∞Xk=0(-(x+ 1))k=∞Xk=0(-1)k(x+ 1)k(b)11 +xwitha= 1Solution:11 +x=12-(-(x-1))=1211--(x-1)2=12∞Xk=0-(x-1)2k=∞Xk=0(-1)k2k+1(x-1)kPage 39
(c)4x2+ 4xwitha=-2Solution:4x2+ 4x=4(x+ 2)2-4=-11-(x+2)24=-∞Xk=0(x+ 2)24k=∞Xk=0-142k(x+ 2)2k38. (8.6) Consider the power series11-x=∞Xk=0xkwith radius of convergence 1.Use differentiation andintegration to find power series for the following functions. What is the radius of convergence?(a)1(1-x)3Solution:It isddx11-x=1(1-x)2d2dx211-x=2(1-x)3This gives1(1-x)3=12d2dx2∞Xk=0xk=12∞Xk=2k(k-1)xk-2=∞Xk=0(k+ 2)(k+ 1)2xkDifferentiated series have the same radius of convergence as the original series, so the radius ofconvergence isR= 1.(b)1(1 +x)2Solution:It isddx11 +x=-1(1 +x)2This gives1(1 +x)2=-ddx11 +x=-ddx11-(-x)=-ddx∞Xk=0(-x)k=-∞Xk=1(-1)kkxk-1=∞Xk=0(-1)k(k-1)xkThe radius of convergence isR= 1.(c) ln(1 +x2)Solution:It isln(1 +x) =Z11 +xdx=Z∞Xk=0(-1)kxkdx=C+∞Xk=0(-1)kk+ 1xk+1Since ln(1 + 0) = 0,C= 0 and it isln(1 +x2) =∞Xk=0(-1)kk+ 1x2k+2=∞Xk=1(-1)k+1kx2kPage 40
The radius of convergence isR= 1.39. (8.7) Find the Taylor series of the given functions about the indicated points. Use Sigma notation toexpress it, if possible.(a)f(x) =e2xaboutx= 0Solution:f0(x) = 2e2xf00(x) = 22e2xf(k)(x) = 2ke2xThis givesf(k)(0) = 2kand the Taylor series isT(x) =∞Xk=02kk!xk(b)f(x) = sin(x) aboutx=π4(Hint: Write the Taylor series using two or more sums.)Solution:f0(x) = cos(x)f00(x) =-sin(x)f(3)(x) =-cos(x)f(4)(x) = sin(x)This givesf(k)π4=√22k= 4l√22k= 4l+ 1-√22k= 4l+ 2-√22k= 4l+ 3and the Taylor series isT(x) =√22∞Xl=0x4l(4l)!+∞Xl=0x4l+1(4l+ 1)!-∞Xl=0x4l+2(4l+ 2)!-∞Xl=0x4l+3(4l+ 3)!!=√22∞Xk=0(-1)kx2k(2k)!+∞Xk=0(-1)kx2k+1(2k+ 1)!!(c)f(x) =x5+ 1 aboutx=-1Solution:f0(x) = 5x4f00(x) = 20x3f(3)(x) = 60x2...This givesT(x) =0·(x+ 1)0+51(x+ 1)1+-202(x+ 1)2+606(x+ 1)3+-12024(x+ 1)4+120120(x+ 1)5=5(x+ 1)-10(x+ 1)2+ 10(x+ 1)3-5(x+ 1)4+ (x+ 1)5Page 41
40. (8.7) Remember the hyperbolic functionssinh(x) =ex-e-x2cosh(x) =ex+e-x2(a) Find the first three derivatives of sinh(x), that isddxsinh(x),d2dx2sinh(x) andd3dx3sinh(x).Solution:ddxsinh(x) = cosh(x)d2dx2sinh(x) = sinh(x)d3dx3sinh(x) = cosh(x)(b) What is the Maclaurin series of sinh(x)? Use the sum notation to write it down.Solution:It is sinh(0) = 0 and cosh(0) = 1, so the Maclaurin series of sinh(x) is0 + 1x+02!x2+13!x3+· · ·=∞Xk=0x2k+1(2k+ 1)!(c) What is the radius of convergence of the Maclaurin series?Solution:Use the ratio testlimk→∞x2(k+1)+1(2(k+1)+1)!x2k+1(2k+1)!= limk→∞x2(2k+ 3)(2k+ 2)= 0So the radius of convergence isR=∞and the Maclaurin series converges everywhere.(d) Show, that the Maclaurin series is equal to sinh(x).Solution:It isdndxnsinh(x) =(sinh(x)nevencosh(x)noddNow it is for|x| ≤d|sinh(x)| ≤ex2≤ex≤ed|cosh(x)| ≤ex≤edand sodndxnsinh(x)≤edfor allxwith|x| ≤d. SetM=ed. So Taylor’s Inequality gives|Rn(x)| ≤ed(n+ 1)!|x|n+1andlimn→∞|Rn(x)|=edlimn→∞|x|n+1(n+ 1)!= 0Since the limit of the remainder converges to 0, the Taylor polynomials converge to sinh(x).That is the Maclaurin series is equal to sinh(x).Page 42
41. (8.7) Use Taylor’s inequality to show that the given function and their Taylor series are equal on theopen interval of convergence. That is, show that|Rn(x)|goes to zero asn→ ∞by bounding|Rn(x)|by an expression involvingn.(a)f(x) = cos(x) with Taylor series∞Xk=0x2k(2k)!, which converges for allxSolution:The derivatives of cos(x) are up to a sign either sin(x) or cos(x). Both are boundedabove by 1, so|f(n+1)(x)| ≤1 for allx. So setM= 1 and Taylor’s inequality gives|Rn(x)| ≤1(n+ 1)!|x|n+1n→∞----→0So cos(x) is equal to its Taylor series.(b)f(x) = (x-1)5with Taylor seriesx5-5x4+ 10x3-10x2+ 5x-1 which converges for allxSolution:The higher derivatives are zero, that isf(n+1)(x) = 0 forn≥5. So 0 is an upperbound and Taylor’s inequality gives|Rn(x)| ≤0n→∞----→0So (x-1)5equals its Taylor series.42. Considerf(x) =√x. Use the linearT1(x) and quadraticT2(x) truncated Taylor series, expanded abouta= 1 to approximate√2. Which approximation does better?Solution:The, zeroth, first, and second derivatives aref(1) = 1,f0(1) =12,f00(1) =-14So, the linear approximation isT1(x) = 1 +12(x-1),and the quadratic isT1(x) = 1 +12(x-1)-18(x-1)2.The approximations areT1(2) = 1 + 1/2 = 3/2 = 1.5≈√2 = 1.4142. . . ,andT2(2) = 1 + 1/2-1/8 = 11/8 = 1.3758≈√2 = 1.4142. . .The errors:|T1(2)-p(2)|= 0.0858, and|T2(2)-p(2)|= 0.0392, so the quadratic approximationdoes better.43. (8.7) Letf(x) =e-xanda= 1.Use Taylor’s inequality to find the smallestNso that the TaylorpolynomialTN(x) is within a 0.1 error bound off(x) on the interval (0,2).Page 43
Solution:The derivatives aref(n)(x) = (-1)ne-x. For eachn, the bound for the absolute value ofthe derivatives on the interval (0,2) is found on its left edge:x= 0. That is,M=e0= 1>|f(n)(x)|forx∈(0,2). The Taylor inequality states|f(x)-TN(x)| ≤M(N+ 1)!|x-1|N+1,for|x-1|<1 (i.e.,x∈(0,2)). Therefore, the boundis=1(N+ 1)!.The goal is to make the bound less than 0.1 = 1/10, so (N+ 1)!>10, the smallestNfor which thisis satisfied isN= 3:= 1/4! = 1/24.44. (9.1) Decide whether the given equation describes a sphere. If yes, find its radius and center.(a)x2+y2+z2+ 2z-6x+ 3 = 0Solution:It is0 =x2+y2+z2+ 2z-6x+ 3 = (x-3)2+y2+ (z+ 1)2-9-1 + 3 = 0(x-3)2+y2+ (z+ 1)2= 7So the equation describes a sphere with radius√7 and center (3,0,-1).(b)x2+ 2y2= 4x-2 + 4y-z2Solution:It is0 =x2+ 2y2+z2-4x-4y+ 2 = (x-2)2+ 2(y-1)2+z2-4-2 + 2(x-2)2+ 2(y-1)2+z2= 4This does not describe a sphere, since (y-2)2is weight more than (x-2)2andz2.Thisdescribes an ellipsoid.(c)x2+y2+z2+ 4z+ 2y+ 12 = 0Solution:It is0 =x2+ (y+ 1)2+ (z+ 2)2-1-4 + 12x2+ (y+ 1)2+ (z+ 2)2=-7This does not describe a sphere, since not tuple (x, y, z) satisfies this equation.45. (9.1) Write the set of all the points, that lie in the described object.(a) A cylinder about thex-axis with radius 2 and height fromx= 0 tox= 3.Solution:{(x, y, z)|y2+z2= 4 and 0≤x≤3}Page 44
(b) A sphere of radius 3 with center (0,3,1).Solution:{(x, y, z)|x2+ (y-3)2+ (z-1)2= 9}(c) A disk with center (3,1,-5) of radius 2 parallel to theyz-plane.Solution:{x, y, z|x= 3 and (y-1)2+ (x+ 5)2≤4}46. (9.2) Consider~u=<3,-2,12>~v=<-2,5,0>~w= 3~i-3~kEvaluate the following expressions.(a)~u+~vSolution:~u+~v=<1,3,12>(b) 3~wSolution:3~w= 9~i-9~k(c) 2~v-~wSolution:2~v-~w=<-7,10,3>=-7~i+ 10~j+ 3~k(d)|~w|Solution:|~w|=√9 + 0 + 9 = 3√247. (9.2) Consider~uand~vin the figure. Draw on the figure the following vector calculations(a)~u+~v(b)~u-~v(c)~v-~uuvPage 45
Solution:uvv-uu-vu+v48. (9.2) For the given vectors, find the unit vector with the same direction as the given vector. Express itin component formandusing the standard basis.(a)~u=<2,-3,5>Solution:|~u|=√4 + 9 + 25 =√38So the unit vector is~u|~u|=1√38<2,-3,5>=<2√38,-3√38,5√38>=2√38~i-3√38~j+5√38~k(b)~v=-~j+ 4~kSolution:|~v|=√1 + 16 =√17So the unit vector is~v|~v|=1√17-~j+ 4~k=h0,-1√17,4√17i(c)~wthe vector fromP(0,12,-3) toQ(3,8,1)Solution:~w=<3,-4,4>|~w|=√9 + 16 + 16 =√41So the unit vector is~w|~w|=1√41<3,-4,4>=<3√41,-4√41,4√41>=3√41~i-4√41~j+4√41~k49. (9.3) Compute the dot product~u·~vfor the given vectors~uand~v.(a)~u=<3,7,-1>and~v=<2,0,3>Solution:~u·~v= 6 + 0-3 = 3(b)~u=-2~i+ 4~k+~jand~v=<-1,2,1>Page 46
Solution:~u·~v= (-2)(-1) + 1·2 + 4·1 = 8(c)~uthe vector fromP(1,3,2) toQ(2,3,-2),~v= 5~i-3~kSolution:~u=<1,0,-4>~u·~v= 5 + 12 = 1750. (9.3) Compute the angle between the given vectors~uand~v.(a)~u=<3,-2,1>and~v=<2,0,-1>Solution:cos(θ) =~u·~v|~u||~v|=6-1√9 + 4 + 1√4 + 1=5√14√5θ= arccos5√14√5≈0.93 = 53.3◦(b)~uthe vector fromP(3,-1,4) toQ(-1,3,0) and~vthe vector fromPtoR(-2,9,-1)Solution:~PQ=<-4,4,-4>= 4<-1,1,-1>~PR=<-5,10,-5>= 5<-1,2,-1>Since the length does not have an effect on the angle, one can use multiples to calculate theanglecos(θ) =<-1,1,-1>·<-1,2,-1>|<-1,1,-1>||<-1,2,-1>|=1 + 2 + 1√3√6=43√2θ= arccos43√2≈0.34 = 19.47◦51. (9.3) A sled is pulled along a horizontal path by a rope. A force of 30 N is applied to the rope at anangle of 30◦from the horizontal. How much work is done, when the sled moved 20 m?Solution:The force~Fhas a magnitude of|~F|= 30 N and has an angle of 30◦from the horizontal,so~F= 30 Ncos(30◦)~i+ sin(30◦)~j= 30 N√32~i+12~j!The sled moves along the horizontal, so the distance vector is~d= 20 m~i. So the work isW=~F·~d= 30 N√3220 m =√3300 J≈519.6 J52. (9.3) Consider vectorsh3,2,-1iandh1,2, ci, wherecis a constant. Find a value forcthat makes thetwo vectors orthogonal.Page 47
Solution:h3,2,-1i · h1,2, ci= 3 + 4-c= 0 =⇒c= 7.53. (9.3) Consider the vectors~u=h3,1i,~v=h1,2iand~w=h-1,2i. Find numbersaandbin order representthe vector~win terms of~uand~v:~w=a~u+b~v.Hint: take dot products of the above expression with~uand~vto solve foraandb. Verify your answerusing direct computation.Solution:~w=a~u+b~v ~w·~u=-1~w·~v= 3~u·~v= 5|~u|2= 10|~v|2= 5~w·~u= (a~u+b~v)·~u=-1 =a10 +b5~w·~v= (a~u+b~v)·~v= 3 =a5 +b5=⇒ -4 = 5a=⇒a=-4/5=⇒3 =-4 + 5b=⇒b= 7/5.-4/5~u+ 7/5~v=-4/5h3,1i+ 7/5h1,2ih-12/5 + 7/5,-4/5 + 14/5i=h-5/5,10/5i=h-1,2i=~w54. (9.3) Consider the vectors~uand~vin the figure.Draw on the figure the componentscomp~v(~u) andcomp~u(~v)uvSolution:uvcompu(v)compv(u)Page 48
55. (9.3) Consider the orthogonal vectors~u=h1,1i,~v=h-1,1i.Verify by direct computation that thevector~w=h-1,2ican be expressed as~w=proj~u(~w) +proj~v(~w).Note this representation of~wis only possible when~uand~v.Solution:~w·~u= 1~w·~v= 3|~u|2= 2|~v|2= 2proj~u(~w) =12h1,1iproj~v(~w) =32h-1,1iproj~u(~w) +proj~v(~w) =h-1,2i=~w.56. (9.4) Calculate the cross product of the given vector~uand~v.(a)~u=<3,-1,4>and~v=<2,-1,6>Solution:<3,-1,4>×<2,-1,6>=<-2,-10,-1>(b)~u= 3~i+ 2~kand~v= 2~j-~kSolution:~u×~v=-4~i+ 3~j+ 7~k57. (9.4) Compute the following(a)~i×~j(b)~i×~i(c)~j×~k(d)~i×~k(e)~i×(~j×~i)(f)~k×(~i×~j)Solution:(a)~k(b)~0(c)~iPage 49
(d)-~j(e)~j(f)~058. (9.4) Consider the triangle formed by the pointsP(1,0,-3),Q(2,-1,5) andR(0,3,1).(a) Find the length of the sides of the triangle.Solution:|PQ|=√1 + 1 + 64 =√66|QR|=√4 + 16 + 16 =√36|PR|=√1 + 9 + 16 =√26(b) Find the angle atQ.Solution:Ifθis the angle atQ, thencos(θ) =~QP·~QR|~QP||~QR|=<-1,1,-8>·<-2,4,-4>√66√36=406√66=203√66θ= arccos203√66≈0.608≈34.85◦(c) Find the area of the triangle.Solution:The area of the triangle is half of the area of the parallelogram spanned by~QPand~QR. SoA=12|~QP×~QR|=12|<1·4-(-8)(-4),(-8)(-2)-(-4)(-1),(-1)4-1(-2)>|=12|<-28,12.2>|=|<-14,6,1>|=√196 + 36 + 1 =√23359. (9.4) To tighten a bolt a force of 12 N is applied to a wrench at a distance of 15 cm. The angle betweenthe wrench and the applied force is 30◦. What is the magnitude of the torque?Solution:|~τ|= 15 cm·12 N·sin(30◦) = 90 cN m = 0.9 N m60. (9.4) Find the volume of the parallelepiped spanned by~a=<3,-1,3>,~b=<1,0,4>and~c=<0,-3,1>.Page 50
Solution:V=|~a·(~b×~c)|=|<3,-1,3>·<12,-1,-3>|=|36 + 1-9|= 28The volume of the parallelopiped is 28.61. (9.5) Find the vector equation and the parametric equations for the given lines.(a) The line throughP(3,-1,12) parallel to the vector<1,0,3>.Solution:The vector equation is~r=<3,-1,12>+t <1,0,3>and the parametric equations arex= 3 +ty=-1z= 12 + 3t(b) The line throughP(4,-1,0) andQ(3,2,1).Solution:A possible direction vector is<3-4,2-(-1),1-0>=<-1,3,1>, so the vectorequation is~r=<4,-1,0>+t <-1,3,1>alternatively:~r=<3,2,1>+t <-1,3,1>and the parametric equations arex= 4-ty=-1 + 3tz=t62. (9.5) Find the vector equation and the scalar equation for the given planes.(a) The plane containingP(3,1,6),Q(-2,1,4) andR(3,1,-1).Solution:The vectors~PQand~PRlie on the plane, so the normal vector is~n=~PQ×~PR=<-5,0,-2>×<0,0,-7>=<0,-35,0>= (-35)<0,1,0>So the vector equation is<0,1,0>·(~r-<3,1,6>) = 0and the scalar equation isy-1 = 0(b) The plane containingP(6,2,-2) andQ(1,-2,0), that is parallel to the vector<-1,4,2>.Solution:The vector~PQlies in the plane, so the normal vector is perpendicular to it. Thenormal vector also has to be perpendicular to<-1,4,2>, so~n=~PQ×<-1,4,2>=<-5,-4,2>×<-1,4,2>=<-8-8,-2 + 10,-20-4>=<0,8,-24>= 8<0,1,-3>Page 51
So the vector equation is<0,1,-3>·(~r-<6,2,-2>) = 0and the scalar equation isy-2 + (-3)(z+ 2) = 0y-3z= 8(c) The plane perpendicular to<2,-2,5>containingP(0,3,-2).Solution:The vector equation is<2,-2,5>·(~r-<0,3,-2>) = 0and the scalar equation is2x+ (-2)(y-3) + 5(z+ 2) = 02x-2y+ 5z=-463. (9.5) Find the intersection of the given objects.(a) The line~r=h3,5,-4i+th2,-1,0iand the linex= 3 +s,y=-2-4s,z= 2 + 3s.Solution:To find the intersection, set the lines equal, that gives three equations3 + 2t= 3 +s5-t=-2-4s-4 = 2 + 3sThe third equation givess=-2, plugged into the first two3 + 2t= 3-25-t=-2 + 82t=-2-t= 1Both are solved byt=-1. So the intersection point is (1,6,-4).(b) The line~r=h0,4,-1i+th3,-7,7iand the plane 3x-5y+ 2z= 10Solution:To parametric equations of the line arex= 3ty= 4-7tz=-1 + 7tPlugging these into the scalar equation of the plane gives3(3t)-5(4-7t) + 2(-1 + 7t) = 109t-20 + 25t-2 + 14t= 1048t= 32t=3248=23So the point of intersection is (2,4-143,-1 +143) = (2,-23,113).(c) the plane<4,0,-2>·(~r-<1,0,5>) = 0 and the plane-2x+ 6y-z= 9Solution:The planes are not parallel, since the normal vectors<4,0,-2>and<-2,6,-1>are not parallel. So the intersection is a line. The direction of the line is orthogonal to bothnormal vectors, so<4,0,-2>×<-2,6,-1>=<12,8,2>= 2<6,4,1>Page 52
is the direction vector of the line of intersection. It remains to find one point, that lies on bothplanes. Consider the scalar equation of each plane4x-2z=-6-2x+ 6y-z= 9Settingx= 0, givesz= 3 andy= 2. So (0,2,3) lies on both planes and the line of intersectionis~r=<0,2,3>+t <6,4,1>64. (9.5) Find the angle in which the given planes intersect.(a)<1,-3,0>·(~r-<1,0,4>) = 0 and<0,-2,4>·(~r-<5,-2,0>) = 0Solution:The planes are clearly not parallel, so the intersect in a line. The angle in whichthey intersect is the same as the angle between the normal vectors. Socos(θ) =<1,-3,0>·<0,-2,4>|<1,-3,0>||<0,-2,4>|=6√102√5=35√2θ= arccos35√2≈1.13 = 64.9◦So the planes intersect at an angle of 64.9◦.(b) 2x-7y+ 3z= 12 and<1,5,2>·(~r-<5,-6,2>) = 0Solution:The planes are clearly not parallel, so the intersect in a line. The angle in whichthey intersect is the same as the angle between the normal vectors. Socos(θ) =<2,-7,3>·<1,5,2>|<2,-7,3>||<1,5,2>|=2-35 + 6√4 + 49 + 9√1 + 25 + 4=-27√62√30θ= arccos-27√62√30≈2.25 = 128.8◦So the planes intersect at an angle of 51.2◦(the angle of intersection between two planes canalways be choosen≤90◦).65. (9.5) Derive the equation to find the point~xin a plane~n·(~r-~p) = 0 that is closest to a given point inspace~yis given by the formula in terms of the plane normal vector~nand plane reference point~p. Alsofind the distance formula between the points.Solution:The minimum-distance point~xis found when~x-~y=A~nfor someAscalar. So, we needto findA. That is, the vectors are collinear, but may differ by a constant. Solving for~x:~x=~y+A~n~xneeds to also be in the plane, so0 =~n·(~x-~p) =~n·(~y+A~n-~p),so~n·~y+A||~n||2-~n·~p= 0Page 53
Solving forA:A=~n·~p-~n·~y||~n||2,which yields the formula~x=~y-~n(~y-~p)·~n||~n||2and the distance equation is||~x-~y||=|(~y-~p)·~n|||~n||66. (9.5) Find the distance of the given objects.(a) The planeh3,0,-4i ·(~r- h1,3,7i) = 0 and the point (2,6,3)Solution:A point~xin the plane that is closest to a given point in space~yis given by theformula in terms of the plane normal vector~nand plane reference point~p:~x=~y-~n(~y-~p)·~n||~n||2and the distance equation is||~x-~y||=|(~y-~p)·~n|||~n||To use the formula for the distance between a point and a plane, one needs the scalar equationof the plane. For this plane the scalar equation is3x-4z+ 25 = 0So the distance isd=|3·2-4·3 + 25|√32+ 42=195(b) The plane 5x-8y+ 2z= 2 and the line~r=h2,-4,3i+th4,2,-2iSolution:To see plane and the line are parallel, check if the normal vector of the plane andthe direction vector of the line are orthogonal<5,-8,2>·<4,2,-2>= 20-16-4 = 0So they are orthogonal. Now the distance of the line and the plane is the same as the distancefrom any point on the line and the plane. A point on the line is (2,-4,3). The distance of theplane to this point isd=|5·2-8·(-4) + 2·3-2|√25 + 64 + 4=46√91This is also the distance of the line and the plane.(c) the plane 2x-6y+ 2z=-11 and the plane<-1,3,-1>·(~r-<-2,3,5>) = 0Page 54
Solution:First check, if the planes are parallel.The normal vectors<2,-6,2>and<-1,3,-1>are clearly multiples of each other. So the planes are parallel. Now the distancebetween the two planes is the same as the distance of any point on one plane to the other plane.So it is enough to find the distance of (-2,3,5), a point on the second plane, to the first plane.The distance between these isd=|2(-2)-6·3 + 2·5 + 11|√4 + 36 + 4=12√11This is the distance between the two planes.(d) the line~r=<-3,0,7>+t <3,-2,1>and the linex= 3t-2,y=-t+ 2,z= 11Solution:First construct a plane, that contains the first line and is parallel to the second line.The normal vector~nof this plane has to be orthogonal to the direction vector of both planes,which are<3,-2,1>and<3,-1,0>, so take~n=<3,-2,1>×<3,-1,0>=<1,3,3>Then the plane is<1,3,3>·(~r-<-3,0,7>) = 0x+ 3y+ 3z+ 18 = 0Now the distance between the two lines is the distance of this plane to a point on the secondline. Take (-2,2,11) as a point on the second line, then the distance isd=|1(-2) + 3(2) + 3(11) + 18|√1 + 9 + 9=55√19This is the distance between the two lines.67. (9.5) Decide whether the given lines are the same, parallel, intersect or neither.(a)L1:~r=<2,4,-1>+t <3,4,1>andL2:y= 2,x-23=z-1-1Solution:The direction vector ofL2, which is given by symmetric equations, is<3,0,-1>.So the lines are neither the same nor parallel. To check, if they intersect, look at the parametricequations ofL1x= 2 + 3ty= 4 + 4tz=-1 +tNow forL2, it isy= 2, sot=12. This gives the point (72,2,-12). Now check, if this point is onL2:72-23=32=-12-1-1So the lines intersect.(b)L1:~r=<3,-1,2>+t <3,-6,9>andL2:~r=<0,5,-7>+t <-1,2,-3>Solution:The lines are the same or parallel, since the direction vectors are multiplies of eachother<3,-6,9>= (-3)<-1,2,-3>. To check, if they are equal, just check, if a point, thatlies onL1, also lies inL2. TakeP(3,-1,2). Now solve the equations3 =-t-1 = 5 + 2t2 =-7-3tPage 55
t=-3 solves all these equations simultaneously, so the lines are the same.(c)L1:x= 2 +t, y=-3 + 2t, z=-5tandL2:~r=<1,0,2>+t <-2,-4,10>Solution:A direction vector of the first line is<1,2,-5>. Since<-2,-4,10>= (-2)<1,2,-5>, the lines are parallel or the same. Now check, if the pointP(2,-3,0), which lies onL1, also lies onL2. That is solve the equations2 = 1-2t-3 =-4t0 = 2 + 10tThe point does not lie onL2, since the first equation is solved byt=12and the second byt=34. So the lines are parallel.68. (9.5) Decide whether the given planes are the same, parallel or intersect.(a) Plane 1:<1,0,-3>·(~r-<2,0,3>) = 0 and plane 2: 2x-3y+ 2z+ 4 = 0Solution:The normal vector of the first plane is<1,0,-3>and the normal vector of thesecond plane is<2,-3,2>. These are not parallel, so the planes are neither parallel nor thesame. So they intersect.(b) Plane 1:-4x+ 3y-z= 12 and plane 2:<12,-9,3>·(~r-<-2,1,-1>) = 0Solution:The normal vector of the first plane is<-4,3,-1>and the normal vector of thesecond plane is<12,-9,3>.These vectors are multiplies of each other:<12,-9,3>=(-3)<-4,3,-1>. So the planes are parallel or the same.<-2,1,-1>is a point in thesecond plane, plugging it into the equation of the first plane-4(-2) + 3·1-(-1) = 8 + 3 + 1 = 12So the point also lies in the first plane. So the planes are the same.(c) Plane 1: 8x-4y= 2z-2 and plane 2:-4x+ 2y+z+ 14 = 0Solution:The normal vector of the first plane is<8,-4,-2>and the normal vector ofthe second plane is<-4,2,1>. These are multiples of each other:<8,-4,-2>= (-2)<-4,2,1>. So the planes are parallel or the same. A point on the first plane is (0,0,1). Pluggingthis into the second equation-4·0 + 2·0 + 1 + 14 = 156= 0so the point does not lie in the plane. So the planes are parallel.69. (9.6) Consider the surface 4x2+y2+ 4z2= 16.(a) Draw the traces inx=kfor three different values fork.Page 56
-4-12-4-12yzSolution:The outer trace is fork= 0, the inner fork=±1. If-2> kork >2, the equationhas not solution.(b) Draw the traces iny=kfor three different values fork.-4-12-4-12xzSolution:The outer trace is fork= 0, the inner fork=±1. If-4> kork >4, the equationhas not solution.(c) Draw the traces inz=kfor three different values fork.Page 57
-4-12-4-12xySolution:If-2> kork >2, the equation has not solution.(d) Describe and or sketch the shape of the surface in three dimensions.Solution:The shape is an ellipsoid (an egg), with the long axis on they-axis.70. (9.7) Each point is given in either rectangular, cylindrical or spherical coordinate. Fill in the missingcoordinates.Rectengular(x, y, z)Cylindrical(r, θ, z)Spherical(ρ, θ, φ)(2,2,0)(2√2,π4,0)(2√2,π4,π2)(0,0,3)(0,0,3)(3,0,0)(1,-1,1)(√2,7π4,1)(√3,7π4,arccos1√3≈0.96≈54.7◦)71. (9.7) Consider surface the cone with the tip at the origin and who’s axis goes along the negativez-axisdown to a base with radius radius 1 and height 2 from the tip.(a) Find the equation describing the cone in rectangular coordinates.Solution:x2+y2= (-z2)2,0≤z≤2(b) Find the equation describing the cone in cylindrical coordinates.Solution:r=-z2(c) Find the equation describing the cone in spherical coordinates.Solution:0≤ρ≤√1 + 4φ=5π6θis free to vary.Page 58
72. (10.1) Sketch the curve of the given vector equation.(a)~r(t) =hsin(t),3 cos(t),2i.(b)~r(t) =ht,-1, t2i.(c)~r(t) =hcos(t),sin(t),cos(t)i.Solution:(a) An ellipse in the planez= 2, withy-diameter 3 andx-diameter 1.(b) An upward u-shaped parabola lying in they=-1 plane wherez=x2.(c) An ellipse lying on a slanted plane withy-diameter 1 andx-diameter 1. The normal vector ofthe plane can be computed as~n=~r0×~r00detijk-sin(t)cos(t)-sin(t)-cos(t)-sin(t)-cos(t)=h-1,0,1i.73. (10.1) Specify a vector function that draws the following described paths(a) A circle in thex-yplane (z= 0).(b) Any spiral emanating from the origin and expanding outward that lies in they-zplane(c) A helix that begins in thex-yplane at point (0,-1,0) att= 0 and spirals upward, drawing a circlewith radius 1 when observed looking down along the z-axis onto thex-yplane.Solution:These are examples. Other formulas could work as well.(a)~r(t) =hcos(t),sin(t),0i(b)~r(t) =h0, tcos(t), tsin(t)i(c)~r(t) =hsin(t),-cos(t), ti74. (10.1) Harder: Find the vector function, that represents the curve of intersection of the given surfaces.(a)x2+z2=y2andx+ 2y= 1Solution:Because of the first equation, setx=ycos(t)z=ysin(t)Plugging this into the second one givesycos(t) + 2y= 1y=12 + cos(t)So the vector function for the intersection is~r(t) =<cos(t)2 + cos(t),12 + cos(t),sin(t)2 + cos(t)>0≤t≤2πPage 59
(b)x2-y2+z= 1 withz≥0 and 2y-z= 1Solution:Solving the second forzgivesz= 2y-1, plugging this into the first equationx2-y2+ 2y= 2x2-(y-1)2= 1y≥12So it isy=√x2-1 + 1. Sett=x, so the vector function of the intersection is~r(t) =< t,pt2-1 + 1,2pt2-1 + 1>75. (10.2) Find the tangent line for the given vector functions at the given points.(a)~r(t) =< t2,2t, t3>at (1,2,1)Solution:The tangent vector is~r0(t) =<2t,2,3t2>~r0(1) =<2,2,3>Since thet-value at (1,2,1) ist= 1. So the tangent line isx= 1 + 2ty= 2 + 2tz= 1 + 3t(b)~r(t) =1t~i-ln(t)~j+√t~kat (1,0,1)Solution:The tangent vector is~r0(t) =-1t2~i-1t~j+12√t~k~r0(1) =-~i-~j+12~kSince thet-value at (1,0,1) ist= 1. So the tangent line isx= 1-ty=-tz= 1 +12t76. (10.2) Given two vector function~u(t),~v(t) and~w(t) with~u(1) =<3,-1,3>~u0(1) =<-2,0,6>~v(1) =<0,2,1>~v0(1) =<9,1,3>~w(1) =<1,0,3>~w0(1) =<2,4,1>Evaluate the following expressions.(a)ddt(~u(t)·~v(t)) att= 1Solution:ddt(~u(t)·~v(t))t=1= (~u0(t)·~v(t) +~u(t)·~v0(t))t=1= 6 + (18-1 + 9) = 32(b)ddt(~v(t)×~u(t)) att= 1Page 60
Solution:ddt(~v(t)×~u(t))t=1=~v0(1)×~u(1) +~v(1)×~u(1) =<18,-10,-8>(c)ddt(~w(t)·(~v(t)×~u(t))) att= 1Solution:ddt(~w(t)·(~v(t)×~u(t)))t=1=~w0(1)·(~v(1)×~u(1)) +~w(1)×ddt(~u(t)·~v(t))t=1=<2,4,1>·<7,3,-6>+<1,0,3>·<18,-10,-8>=14 + 12-6 + 18-24 = 1477. (10.3) Find the arc length of the given curves.(a)~r(t) =<1-2t2, t3,-t3>for 0≤t≤√2Solution:~r0(t) =<-4t,3t2,-3t2>|~r0(t)|=p16t2+ 9t4+ 9t4=tp16 + 18t2L=Z√20tp16 + 18t2dt=Z3616136√udu=13623u3/23616=172(63-43)=199with substitutionu= 16 + 18t2, du= 36tdt.(b)~r(t) =et~i-√2t~j+e-t~kfor-1≤t≤1Solution:~r0(t) =et~i-√2~j-e-t~k|~r0(t)|=pe2t-2 +e-2t=q(et-e-t)2=et-e-tL=Z1-1et-e-tdt= 2Z10(et-e-t)dt= 2et+e-t10= 2e+1e-278. (10.3) Find the parametrization with respect to arc length measured from the point wheret= 0.(a)~r(t) = (1 + 3t)~i-t~j+ (2-t)~kSolution:First find the arc length function~r0(t) = 3~i-~j-~k|~r0(t)|=√11s(t) =Zt0|~r0(u)|du=Zt0√11du=√11tSo the reparametrization is~r(t(s)) = (1 +3s√11)~i-s√11+ (2-s√11)~kPage 61
(b)~r(t) =<1, etsin(t),-etcos(t)>Solution:First find the arc length function~r0(t) =<0, etsin(t) +etcos(t),-etcos(t) +etsin(t)>|~r0(t)|=qe2t(sin(t) + cos(t))2+e2t(-cos(t) + sin(t))2=et√2s(t) =Zt0eu√2du=h√2euit0=√2(et-1)Solving this fortgivess√2=et-1s√2+ 1 =ett= lns√2+ 1So the reparametrization is~r(t(s)) =<1,s√2+ 1sinlns√2+ 1,-s√2+ 1coslns√2+ 1>79. (10.3) Find the curvature of the given curves and the specified points.(a)~r(t) =< e2t,3-t2, t >at (1,3,0)Solution:~r0(t) =<2e2t,-2t,1>~r00(t) =<4e2t,-2,0>~r0(t)×~r00(t) =<2,4e2t,-4e2t+ 8te2t>|~r0(t)|=p4e4t+ 4t2+ 1|~r0(t)×~r00(t)|=p4 + 16e4t+ 16e4t(1-4t+ 4t2)κ(t) =|~r0(t)×~r00(t)||~r0(t)|3=p4 + 16e4t+ 16e4t(1-4t+ 4t2)(√4e4t+ 4t2+ 1)3Thet-value at (1,3,0) ist= 0. So the curvature isκ(0) =√4 + 16 + 16(√4 + 1)3=65√5(b)~r=tsin(t)~i+ 4t~j+tcos(t)~kat(π2,2π,0)Solution:~r0(t) = (tcos(t) + sin(t))~i+ 4~j+ (-tsin(t) + cos(t))~k~r00(t) = (-tsin(t) + 2 cos(t))~i+ (-tcos(t)-2 sin(t))~k~r0(t)×~r00(t) = 4(-tcos(t)-2 sin(t))~i+ (t2+ 2)~j+ (-4)(-tsin(t) + 2 cos(t))~k|~r0(t)|=pt2+ 1 + 16|~r0(t)×~r00(t)|=p16t2+ 64 +t4+ 2t2+ 4κ(t) =√t4+ 17t2+ 68(√t2+ 17)3Page 62
Thet-value at(π2,2π,0)ist=π2. So the curvature isκπ2=q(π2)4+ 17(π2)2+ 68q(π2)2+ 17380. (10.3) Find the unit tangent vector, the unit normal vector and the binormal vector of the given curves.(a)~r(t) =t~i-t2~kSolution:~r0(t) =~i-2t~k|~r0(t)|=p1 + 4t2~T(t) =1√1 + 4t2~i-2t~k~T0(t) =-8t2(1 + 4t2)3/2~i-2t~k+1√1 + 4t2-2~k=-1(1 + 4t2)3/24t~i+ 2~k|~T0(t)|=1(1 + 4t2)3/2p16t2+ 4~N(t) =-1√1 + 4t22t~i+~k~B(t) =~j(b)~r(t) =<cos(3t),4t,sin(3t)>Solution:~r0(t) =<-3 sin(3t),4,3 cos(3t)>|~r0(t)|=√9 + 16 = 5~T(t) =<-35sin(3t),45,35cos(3t)>~T0(t) =<-95cos(3t),0,-95sin(3t)>|~T0(t)|=95~N(t) =<-cos(3t),0,-sin(3t)>~B(t) =~T(t)×~N(t) =<-45sin(3t),35,45cos(3t)>81. (10.4) Consider the vector function~r(t) =hsin(t),cos(t), ti. Find the decomposition of the accelerationinto the osculating plane components:~a=aT(t)~T(t) +aN(t)~N(t)Page 63
Solution:~r0(t) =hcos(t),-sin(t),1i~a(t) =~r00(t) =h-sin(t),-cos(t),0i~T(t) =1√2hcos(t),-sin(t),1iaT(t) =~T(t)·~a(t) = 0.~N(t) =h-sin(t),-cos(t),0iaN(t) =~N(t)·~a(t) =-1~a(t) =aN~N(t).82. (10.4) Consider~x=h1,2,3iand an orthogonal coordinate system~u=h1,1,-1i,~v=h0,1,1i,~w=h2,-1,1i. Find the triplet (a, b, c) such that~x=a~u+b~v+c~w.Solution:~x·~u=-1~x·~v= 5~x·~w= 3|~u|2= 3|~v|2= 2|~w|2= 6(a, b, c) =-13,52,12.83. On anx-yplane, draw precise level curves of the functionf(x, y) =e-x2-y2=zat the valuesf(x, y) =k,wherek=e-1, e-1/4, e-1/16. Describe the shape of the surface.Solution:e-x2-y2=e-1=⇒x2+y2= 1, which is a circle of radius 1 centered at the origin.The remaining two level curves are circles of radius 2 and 4. The surface looks like a mountain withsummit located at the origin.84. Draw precise traces and level curves of the functionf(x, y) =x2-4y2=zat the values near the origin(a)x= 0.(b)y= 0(c)f(x, y) = 1(d)f(x, y) =-1Page 64
and then describe the shape of the surface near the origin.Solution:The surface is shown in the figure. The traces and level curves that are embedded in thesurface are found as follows:(a)z=-4y2dark blue downward parabola.(b)z=x2dark red upward parabola.(c)x=±p1 + 4y2light blue hyperbola pairs.(d)y=±12√1 +x2magenta hyperbola pairs.The surface resembles a mountain pass.85. (11.1) Draw a contour map (i.e., level curves) for the functionf(x, y) =xy2Page 65
-4-224-4-224xy86. Compute the two limit paths for the function|x|-|y||x|+|y|=z.(a) The pathy= 0 andx→0+to the origin.(b) The pathx= 0 andy→0+to the origin.and determine if lim(x,y)→(0,0)|x|-|y||x|+|y|exists.Solution:(a) limy=0,x→0+|x|-0|x|+0= 1.(b) limx=0,y→0+0-|y|0+|y|=-1.The limits on the two paths do not agree, so the limit as (x, y)→(0,0) cannot exist.87. (11.2) Show that the following limits do not exist.(a)lim(x,y)→(0,0)sin(x)2+yx2+y2Solution:Setf(x, y) =sin(x)2+yx2+y2. Along thex-axis it isf(x,0) =sin(x)2x2=sin(x)x2x→0---→1Along they-axis it isf(0, y) =yy2=1ywhich diverges. So the limit does not exist.(b)lim(x,y)→(0,0)xyx2+y3Solution:Setf(x, y) =xyx2+y3. Along thex-axis it isf(x,0) = 0x→0---→0Page 66
Along the diagonalx=yit isf(x, x) =x2x2+x3=11 +xx→0---→1Since these limits are different, thelmits does not exist.88. Determine if the following limits exist or not. If so, compute them; if not, explain why.(a) lim(x,y)→(0,0)ln(x+y)2+x+y(b) lim(x,y)→(0,0)x2+2y+32+x+y(c) lim(x,y)→(1,3)x+y6x-2y(d) lim(x,y)→(1,3)sin(πy)x2-y+2x2-2y+3Solution:(a) DNE because the numerator is not defined at the origin.(b) The function is rational with no singularity at the origin, so the function is continuous and thelimit existsL= 3/2.(c) DNE because the denominator is zero at the origin while the numerator is not zero.(d) The function is continuous because it is comprised of continuous component functions.Thelimit isL=-1/(1-6 + 3) = 1/2.89. Compute the partial derivatives(a)∂∂x[xcos(xey)](b)∂∂y[xcos(xey)](c)x2exyxy(d)x2exyxxSolution:Wolfram alpha should do it.90. (11.3) Given the functionf(x, y) =xy3+ sin(xy)-ex2Find the following partial derivatives.(a)fx(x, y)Solution:fx(x, y) =y3+ycos(xy)-2xex2(b)∂f∂y(x, y)Page 67
Solution:∂f∂y(x, y) = 3xy2+xcos(xy)(c)fxy(x, y)Solution:fxy(x, y) = 3y2+ cos(xy)-xysin(xy)(d)∂2f∂x2Solution:∂2f∂x2= 6xy-x2sin(xy)91. Find the tangent plane equation to the functionf(x, y) =z= 5-x2-y2at the point (1,2).Solution:f(1,2) = 1 =z0,fx(1,2) =-2,fy(1,2) =-4.Soz-1 =-2(x-1)-4(y-2).92. Considerz=p11 +x2+y2. Use a linear approximation at (x, y) = (1,2) to estimate thez-value when(x, y) = (2,3).Solution:Notef(1,2) = 4, andfx=x√11+x2+y2, andfy=y√11+x2+y2, thereforefx(1,2) =14, andfy(1,2) =12. The linear approximationf(2,3) =z≈4 +14(x-1) +12(y-2).At (x, y) = (2,3), the approximatez-value isz=√24≈4 +14+12= 4 +34= 4.75Note that the true value is√24 = 4.8990....93. Consider the functionf(x, y) =z=ysin(x), evaluated on the trajectory~r(t) =ht, t2ifromt= 0 toπ/2.Computedzdtatt= 1.Solution:fx=ycos(x),fy= sin(x),x0(t) = 1,y0(t) = 2t. So,dzdt=π/2 cos(π/2)1 + sin(π/2)2π/2 =π.Page 68
94. (11.5) Considerz=xy+ex2x(r, s, t) =rs+ty(r, s, t) =r2sFind the following expressions.(a)∂z∂rSolution:∂z∂r=∂z∂x∂x∂r+∂z∂y∂y∂r=y+ 2xex2s+x2rsPlugging inx(r, s, t) andy(r, s, t), we get:∂z∂r=r2s+ 2(rs+t)e(rs+t)2s+ (rs+t)2rs.(b)∂2z∂r2(do this if covered in class)Solution:∂z∂r2=∂∂ry+ 2xex2s+x2rs=(2 + 4x2)ex2s+ 2rss+s2rs+x2s.Then substitutex(s, t) andy(s, t) into the above.95. Consider the functionz=x2+ 4y2, evaluated on the trajectory~r(t) =h12cos(t),12sin(t)i.(a) Describe the trajectory graphically overt= 0 to 2π.(b) Computedzdt. The double angle formula may be useful: 2 sin(t) cos(t) = sin(2t).(c) At what timestdoesdzdt= 0. Which are max and min values?(d) Using the above, describe the motion ofz(t) in 3D.Solution:(a) It’s a circular path with radius 1/2 .(b) Using the chain rule:dz/dt=-cos(t) sin(t) + 4 sin(t) cos(t) = 3 sin(t) cos(t) =32sin(2t) (usedouble-angle formula)(c) sin(2t) = 0 whent= 0, π/2, π,3π/2,2π, that alternate from min, max, min, max, min, respec-tively.(d) At what times isdzdtlargest in absolute value?(e) It’s like a up-down-up-down roller coaster.96. (11.6) Consider the functionf(x, y) =x3y-√xy.(a) Find the directional derivative at (2,2) in direction ofh-2,1i.Solution:~∇f(x, y) =h3x2y-y√xy, x3-x√xyi~∇f(2,2) =h23,7iPage 69
The unit vector in the direction ofh-2,1iis15h-2,1i. So the directional derivative ish23,7i ·15h-2,1i=15(-46 + 7) =-395(b) Find the directional derivative at (1,4) in the direction of the vector given by the angleθ=π3.Solution:~∇f(1,4) =h10,12iSo the directional derivative ish10,12i · hcosπ3,sinπ3i=h10,12· h12,√32i= 5 +√3497. (11.6) Find the direction, in which the tangent line off(x, y) =xex+yat (1,-1) is the steepest.Solution:~∇f(x, y) =h(x+ 1)ex+y, xex+yi~∇f(1,-1) =h2,1iSo the tangent line is the steepest in the direction ofh2,1i.98. (11.6) Find the tangent plane to the surface 2x2-y2+ 4z2= 8 at (2,2,1).Solution:This surface is a level surface withF(x, y, z) = 2x2-y2+ 4z2= 8So the normal vector of the tangent plane is~∇F(x, y, z) =h4x,-2y,8zi~∇F(2,2,1) =h8,-4,8i= 4h2,-1,2iSo the tangent plane is2(x-2)-(y-2) + 2(z-1) = 02x-y+ 2z= 499. (11.7) Consider the functionf(x, y) =x-yx2y2+3.(a) Find all the critical points off(x, y).Solution:fx(x, y) =x2y2+ 3-(x-y)2xy2(x2y2+ 3)2fy(x, y) =-(x2y2+ 3)-(x-y)x22y(x2y2+ 3)2Setting each equal to zero gives the equationsx2y2+ 3-2x2y2+ 2xy3= 0-x2y2-3-2x3y+ 2x2y2= 0Page 70
3-x2y2+ 2xy3= 03-x2y2+ 2x3y= 0This givesxy3=x3yand0 =xy3-x3y=xy(y2-x2) =xy(y-x)(y+x)Now clearlyx= 0 ory= 0 does not solve the equations above. Now ify=x, both equationsbecome3 +x4= 0These equations have no solution. Ify=-x, both equations become0 = 3-3x4=-3(x4-1)=-3(x2+ 1)(x+ 1) (x-1)So the critical points are (-1,1) and (1,-1).(b) Decide whether the critical points are local minima, local maxima or saddle points.Solution:To use the second derivative test, find the second derivativesfxx(x, y) =(-2xy2+ 2y3) (x2y2+ 3)-(3-x2y2+ 2xy3)2xy2(x2y2+ 3)3fyy(x, y) =(x22y-2x3) (x2y2+ 3)-(-3 +x2y2-2x3y)x22y(x2y2+ 3)3fxy(x, y) =(-x22y+ 6xy2) (x2y2+ 3)-(3-x2y2+ 2xy3)x22y(x2y2+ 3)3fxx(1,-1) =(-4)(4)-043=-14fxx(-1,1) =(4)(4)-043=14fyy(1,-1) =(-4)(4)-043=-14fyy(-1,1) =(4)(4)-043=14fxy(1,-1) =(8)(4)-043= 1fxy(-1,1) =(-8)(4)-043=-1So it isD(1,-1) =116-1<0D(-1,1) =116-1<0Thus both points are saddle points.100. Find and characterized the critical points as local min, max, or saddles, off(x, y) = 3xy-x3-y3using the second derivative test.Solution:First find the critical pointsfx(x, y) = 3y-3x2fy(x, y) = 3x-3y2Setting both zero givesy=x2x=y2Where these two parabolas intersect at (0,0) and (1,1) defining the critical points.The secondderivative testfxx=-6x,fyy=-6y,fxy=fyx= 3,Page 71
defining theDnumberD(x, y) = 36xy-9.D(0,0) =-9<0, implying the origin critical point is a saddle point.D(1,1) = 36-9>0, andfxx(1,1) =-6<0 implying (1,1) critical point is local maxima.The graph of the surface and the two critical points are shown in the figure: red dot is the saddle,with diagonal traces (black lines) showing one concave up curve and another concave down curve.The green dot is the local max, with both diagonal traces being concave down. Note, the functionis unbounded as (x, y) gets increasingly far from the origin, but we are only concerned with localcritical points.101. (11.7) Find the absolute maximum and the absolute minimum off(x, y) =xy2-x2y+xon the closedrectangle with edges (0,0), (0,4), (3,4) and (3,0).Solution:First find the critical pointsfx(x, y) =y2-2xy+ 1fy(x, y) = 2xy-x2Setting both zero gives0 =y2-2xy+ 10 = 2xy-x2=x(2y-x)From the second equation it isx= 0 orx= 2y. Ifx= 0, then the second equation givesy2+ 1 = 0.This has no solution. Ifx= 2y, the second equation is0 =y2-4y2+ 1 =-3y2+ 1 =-3(y-1√3)(y+1√3)Page 72
So the critical points are2√3,1√3and-2√3,-1√3. The function values at these points aref2√3,1√3=2-4 + 63√3=43√3f-2√3,-1√3=-2 + 4-63√3=-43√3Now check the boundary for extrema.•(0,0) – (0,4):f(0, y) = 0•(0,4) – (3,4):f(x,4) = 16x-4x2+x=-4x2+ 17x, which has critical point for 8x= 17,that is atx=178. So the possible extrema aref(0,4) = 0f(3,4) = 15f(178,4) =17216•(3,4) – (3,0):f(3, y) = 3y2-9y+ 3, which has critical point for 6y= 9, that isy=32. So thepossible extrema aref(3,4) = 15f(3,0) = 3f(3,32) =-154•(3,0) – (0,0):f(x,0) =x, so the possible extrema aref(3,0) = 3f(0,0) = 0So the absolute maximum is17216and the absolute minimum-154.102. (11.8) Find the maximum ofz= 6x2+y, wheny2+x2= 4.Solution:Using Lagrange multipliers gives the equations12x=λ2x1 =λ2yx2+y2= 4The first equation gives0 = 6x-λx=x(6-λ)Sox= 0 orλ= 6. Ifx= 0, theny= 2 from the third equation. Ifλ= 6, theny=112from thesecond equation and the third equation givesx=√57512. So the extrema are at (0,2) and (112,√57512).The first gives the minimum and the second the maximum.103. (11.8) Find the minimum ofz-value ofz=x2-y2, subject to the constraintx+ 2y=-1.Solution:Using Lagrange multipliers∇f=<2x,-2y >,∇g=<1,2>gives the equations2x=λ,-2y= 2λPage 73
The first equation givesx=λ/2.and the second gives-y=λsox=-y/2meaning the liney=-2xfor any lambda-value solves the Lagrange equation. The liney=-2xsatisfies/intersects the constraint equationx+ 2y=-1 whenx-4x=-1.So,x= 1/3 andy=-2/3 is the critical point andλ= 2/3. This point is a minimumz= 1/9-4/9 =-1/3.A figure of the saddle surfacez=x2-y2overlaid by constraint equation linex+ 2y=-1 (blackline) shows that at (1/3,-2/3) the minimum pointz=-1/3 is attained (green dot). For reference,the saddle location at the origin (red dot) is also shown.104. (12.1) Given the functionf(x, y) =x2+yover the rectangleR= [1,3]×[2,5]. Divide the rectangleinto smaller rectangle, by dividing it into eight pieces inx-direction and six pieces iny-direction. Setup the Riemann sum using the midpoint rule to estimate the volume of the function overR. Do notevaluate the sum.Solution:N= 8,Δx=3-18=14,xk= 1 +k4M= 6,Δy=5-26=12,yl= 2 +l2Page 74
The sample points using the midpoint rule arex*kl=xk+xk-12= 1 +k4-18andy*kl=yl+yl-12= 2 +l2-14.So the Riemann sum isV≈8Xk=16Xl=1f(x*kl, y*kl)ΔxΔy=8Xk=16Xl=11 +k4-182+ 2 +l2-14!·18.105. (12.2) Find the volume of the solid that lies above the squareR= [0,2]×[0,4] and below the surfacedefined byz=yexy.Solution:To integrate this function it is better to first integrate with respect tox. That givesV=ZZRyexydA=Z40Z20yexydx dy=Z40[exy]2x=0dy=Z40(e2y-1)dy=12e2y-y40=12e8-8-12106. (12.3) EvaluateRRDxy dA, whereDis the region bounded byx=y2andy=-x+ 2.Solution:To describe the region, first sketch it1234-2-112xyPage 75
To find the intersection points, set the equations, after solving both forx, equaly2=x=-y+ 2=⇒0 =y2+y-2 = (y-1)(y+ 2).Then the intersection points are (1,1) and (4,-2), soD={(x, y)| -2≤y≤1, y2≤x≤ -y+ 2}.Then the integral givesZZDxy dA=Z1-2Z-y+2y2xy dx dy=Z1-212x2y-y+2x=y2dy=12Z1-2y((-y+ 2)2-y4)dy=12Z1-2(-y5+y3-4y2+ 4y)dy=12-16y6+14y4-43y3+ 2y21-2=12-16+14-43+ 2--323+ 4 +323+ 8=-458107. (12.3) Integrate the functionf(x, y) =xyover the setD={(x, y)|0≤x≤1, y≥x, y≤4}. Choosethe integration order so that can you computeRRDf(x, y)dAin only a single iterated integral.Solution:ZZDf(x, y)dA=Z10Z4xxy dy dx=Z1012xy24y=xdx=12Z10(16x-x3)dx=128x2-14x410=128-14=318108. Integrate the functionf(x, y) =xyover the setD={(x, y)|0≤x≤1, y≥x, y≤4}.Choose theintegration order so that can you compute the integralRRDfdAin only a single iterated integral. Donot compute the integral.Solution:Z10Z4xxdydx109. Suppose a laminar object with boundariesy=xandy=√xhas a density functionρ(x, y) =y(1-x2/2). Find the mass of the object.Page 76
Solution:Z10Z√xxy(1-x2/2)dydx=17240110. (12.4) Write the setD={(x, y)|x≥0, y≤0, x2+y2≤4}in polar coordinates.Solution:This is a quarter of a circle in the fourth quadrant, soD=n(r, θ)r≤2,-π2≤θ≤0oor equivalentlyD=(r, θ)r≤2,3π2≤θ≤2π.111. (12.4) Using polar coordinates evaluateRRDf(x, y)dA, whereDis a disk of radius 2 andf(r, θ) =r2cos(2θ).Solution:The regionDcan be described asD={(r, θ)|r≤2}.ZZDf(x, y)dA=Z2π0Z20r2cos(2θ)·r dr dθ=244Z2π0cos(2θ)dθ=24412sin(2θ)2π0= 0112. (12.4) Find the area of the flower-shaped regionD={(r, θ)|r≤sin(8θ) + 4}.MATH 1320-006, Spring 2019Instructor: Janina LetzFinalExpected Learning Outcomes Checklist-4-224-4-224xySolution:Use polar coordinates:dA=r dr d✓A=ZZD1dA=Z2⇡0Zsin(8✓)+40r dr d✓=Z2⇡012r2sin(8✓)+4r=0d✓=12Z2⇡016 + 8 sin(8✓) + (sin(8✓))2d✓=Z2⇡08 + 4 sin(8✓) +14-14cos(16✓)d✓=334⇡sinceZ2⇡0cos(✓)d✓= 0andZ2⇡0sin(✓)d✓= 0.8. (12.5) Suppose a laminar object with boundariesy=xandy=pxhas a density function⇢(x, y) =y⇣1-x22⌘. Find the mass of the object.Page 77
Solution:Use polar coordinates:dA=r dr dθA=ZZD1dA=Z2π0Zsin(8θ)+40r dr dθ=Z2π012r2sin(8θ)+4r=0dθ=12Z2π016 + 8 sin(8θ) + (sin(8θ))2dθ=Z2π08 + 4 sin(8θ) +14-14cos(16θ)dθ=334πsinceZ2π0cos(θ)dθ= 0andZ2π0sin(θ)dθ= 0.113. (12.5) Suppose a laminar object with boundariesy=xandy=√xhas a density functionρ(x, y) =y1-x22. Find the mass of the object.Solution:The intersections of the boundaries are at (0,0) and (1,1). So the region on which theobject lies can be described asD=(x, y) 0≤x≤1, x≤y≤√x.m=Z10Z√xxy1-x22dy dx=Z1012y21-x22√xy=xdx=12Z10x-12x3-x2-12x4dx=1212x2-18x4-13x3+110x510=1212-18-13+110=17240114. Some objects can have the center of mass outside itself, e.g., a boomerang. Let’s consider an idealizedboomerang with constant density and the shape shown in the figure. Find its center of mass.115. (12.5) Letp(x, y) be the probability density function defined for the random variablesXandYin theunit squareD= [0,1]×[0,1], given byp(x, y) = 4y(1-x).Page 78
(a) Check that this function is a probability density function.Solution:ZZD4y(1-x)dA=4Z10Z10y(1-x)dy dx=4Z1012y2(1-x)1y=0dx= 2Z10(1-x)dx=2x-12x210= 1So this function is probability density function.(b) Set up the iterated integral for the probability that (x, y)∈A, whereAcontains all points (x, y) inDfor whichxis greater thany. Do not compute the integral.Solution:The regionAcan be described asA={(x, y)|0≤x≤1,0≤y≤x}.Then the probability isP((x, y)∈A) =ZZAp(x, y)dA=Z10Zx04y(1-x)dy dx .116. (12.5) Suppose a factory must perform two operations on items they are producing. The time (in hours)it takes for operation 1 isXand for operation 2 it isY. The joint probability distribution isp(x, y) =(4xye-x2-y2x≥0, y≥00otherwise.What is the probability that operation 1 takes more than 2 hours, AND operation 2 takes more than 2hours?Solution:The region that describes this occurrence isD={(x, y)|x≥2, y≥2}.Then the regionDcannot be integrated in only one iterated integral.Therefore it is easier tointegrate the region outside ofD, call itR=D={(x, y)|x≤2, y≤2}and subtract the result fromPage 79
1, to get the probabilityP((x, y)∈D) = 1-P((x, y)∈D)1-ZZRp(x, y)dA=Z20Z204xye-x2-y2dy dx1-Z202ye-y2dyZ202xe-x2dx=Z∞22xe-x2dxZ∞22ye-y2dy=h-e-x2i∞2·h-e-y2i∞2=e-4·e-4=e-8.The integralR202ye-y2dy= 1-e-4, soP((x, y)∈D) = 1-(1-e-4)2Page 80