Practice final exam solutions


 Austen Davis
 1 years ago
 Views:
Transcription
1 University of Pennsylvni Deprtment of Mthemtics Mth 26 Honors Clculus II Spring Semester 29 Prof. Grssi, T.A. Asher Auel Prctice finl exm solutions 1. Let F : 2 2 be defined by F (x, y (x + y, x y. If F denotes force field, then show tht the work done by this field to move prticle long the curve α : [, b] 2 defined by α(t (f(t, g(t, only depends on the vlues f(, f(b, g(, g(b. Solution. The work done is clculted by the line integrl F dα b b b f(b f( F (α(t α (t dt (f(t + g(t, f(t g(t (f (t, g (t dt ( f(tf (t + g(tf (t + f(tg (t g(tg (t dt ( u du + g(tf(t b b g (tf(t dt + b f(tg (t dt 1 2 (f(b2 f( 2 + f(bg(b f(g( 1 2 (g(b2 g( ((f(b2 + 2f(bg(b g(b 2 (f( 2 + 2f(g( g( 2 g(b using integrtion by prts nd the substitutions u f(t nd v g(t. This nswer might led one to guess tht there s potentil function, i.e. tht F φ, for some φ : 2. Indeed, letting φ(x, y 1 2 (x2 + 2xy y 2 works! b Find the mount of work done when f( 1, f(b 2, g( 3, g(b 4. Solution. The mount of work done is 1 2 (( ( g( v dv 2. A sphere is inscribed in cylinder. The sphere is sliced by two prllel plnes tht re perpendiculr to the xis of the cylinder. Show tht the portions of the sphere nd cylinder lying between these plnes hs the sme surfce re. Solution. We might s well ssume tht the sphere is centered t the origin, hs rdius, nd tht the cylinder (with rdius hs xis long the zxis. Now the sphere hs prmeteriztion r : [, 2π] [ π/2, π/2] 3 given by r(θ, ϕ ( cos(θ cos(ϕ, sin(θ cos(ϕ, sin(ϕ Let s sy tht the plnes re z b nd z c where b c. Then The portion S of the sphere between the two plnes is prmeterized (vi r by the subset [, 2π] 1
2 [sin 1 (b/, sin 1 (c/]. Using the prmeteriztion formul for surfce integrls, we hve 2π sin 1 (c/ 1 det ( (D (θ,ϕ r t D (θ,ϕ r dϕdθ S θ ϕsin 1 (b/ 2π sin 1 (c/ θ ϕsin 1 (b/ 2 cos(ϕ dϕdθ. The re of the portion C of the cylinder between the two plnes is Check for yourself tht these gree. 2π(c b. 3. Let D 2 be the solid dimondshped region in the plne bounded by the points (,, (1, 1, (2,, (1, 1. Compute D (x2 + y 2 using the chnge of coordintes (u, v (x y, x + y. Solution. Define f : 2 by f(x, y x 2 + y 2 nd Φ : 2 2 by Φ(x, y (x y, x + y. To use the chnge of vribles formul (integrl prmeteriztion formul, we wnt to find nice [ subset S] 2 so tht Φ( D, i.e. Φ 1 (D. Note tht Φ is liner mp given by the mtrix 1 1, so the function theoretic inverse Φ is lso liner mp given by the inverse mtrix Since Φ nd Φ 1 re liner mps, they tke prllelogrms to prllelogrms. Then 1 1 Φ 1 (D is the solid region in the plne bounded by the point (, Φ 1 (,, (1, Φ(1, 1, (1, 1 Φ(2,, (, 1 Φ(1, 1, i.e. [, 1] [ 1, ] is the squre in the fourth qudrnt of side length 1 with one corner t the origin. So Φ : D is prmeteriztion of D by nice region in which we cn use strightforwrd iterted Fubini integrl. Now we ll consider the prmeterizion fctors in the integrl. We hve Also (f Φ(x, y (x y 2 + (x + y 2 2(x 2 + y 2. D (x,y Φ since Φ is liner (so equls its own totl derivtive nd so det ( (D (x,y Φ t D (x,y Φ det D (x,y Φ 2. Finll we cn use the prmeteriztion formul f (f Φ det ( (DΦ t DΦ D x 1 x y 1 (x 2 + y 2 dydx (x dx By the wy, if you wnted to compute the originl integrl with n iterted integrl, it would be 1 x x y x (x 2 + y 2 dydx + 2 x1 2 x y 2+x (x 2 + y 2 dydx which you cn check for yourself gives the sme nswer, lbeit fter long ugly clcultion. Try it!
3 4. Find the volume of the solid tetrhedrl region bounded in the first octnt by the plne 3x + 4y + 2z 12. Set up the complete iterted integrl. Solution. The solid region we re considering is S {(x, y, z 3 : x, y, z, 3x + 4y + 2z 12}. First note tht the intersection of the plne with the positive x, y, nd zxes is 4, 3, nd 6, respectively. In prticulr, x hs the freedom to go from to 4 in S. Next, we ll consider the freedom of y for ech fixed x. For this, projecting the region to the xyplne (i.e. setting z, we get the eqution 3x + 4y 12. Solving for y in terms of x gives y 3 3 4x. This shows tht for fixed x, y hs the freedom to move from to 3 3 4x. Finlly, we consider the freedom of z for ech fixed x nd y. Solving the eqution of the plne for z in terms of x nd y yields z x 2y. This shows tht for fixed x nd y, z hs the freedom to move from to z 6 3 2x 2y in the region. Thus the iterted integrl is x y 4 x x 2y z 1 dzdydx. 5. Using Green s theorem, evlute the line integrl F dr, where F (x, y (x3 2y 3, x 3 +2y 3 nd D {(x, y 2 : x 2 + y 2 2, x, y }, for some fixed positive rel number. Solution. Since the region D is contined in the xyplne, we cn use version (1 of Green s theorem. In this cse, F dr (x 3 2y 3 dx + (x 3 + 2y 3 dy ( (x 3 + 2y 3 (x3 2y 3 D x y (3x 2 + 6y 2. D Now to compute this integrl, we use the chnge of coordintes (prmeteriztion formul to use circulr coordintes. Indeed, let Φ : 2 2 be defined by Φ(r, θ (r cos(θ, r sin(θ. Then Φ : [, ] [, π/2] D gives prmeteriztion of the region D in terms of nice box. As usul with circulr coordintes, the prmeteriztion fctor is given by det ( (D (r,θ Φ t D (r,θ Φ det D (r,θ Φ cos(θ r sin(θ det r. sin(θ r cos(θ Thus we use the prmeteriztion formul F dr 3 (x 2 + 2y D r r r π/2 θ (r 2 cos 2 (θ + 2r 2 sin 2 (θ r dθdr π/2 r 3 (1 + sin 2 (θ dθ dr θ r 3 ( π 2 + π 4 dr 9π4 16. Do you think this is esier thn computing the originl line integrl? First, there re three pieces to the curve, γ 1 (t ( cos(t, sin(t, t π/2, γ 2 (t (, t, t, γ 3 (t (t,, t,
4 so the line integrl breks into three integrls F dr π/2 + ( 3 cos 3 (t 2 3 sin 3 (t, 3 cos 3 (t sin 3 (t γ 1(t dt ( 2t 3, 2t 3 γ 2(t dt + (t 3, t 3 γ 3(t dt π/2 4 ( cos 3 (t sin(t + 2 sin 4 (t + cos 4 (t + 2 sin 3 (t cos(t dt 2 t 3 dt + Then you will need to do freshmn clculus tricks to compute this. Wht do you think is esier? t 3 dt. 6. Using Stokes Theorem, compute the surfce integrl S F n where F (x3 z, xz 2, 3 nd where S is the solid region {(x, y, z 3 : x 2 + y 2 4z 2, z 1} nd n is some choice of unit norml. NOTE: there s chnge of nottion from the originl sttement of the problem. Solution. By Stokes Theorem we hve F n S S 1 z divf 2z y 2z 4z 2 y 2 x 4z 2 y 2 3x 2 z dxdydz constructing the iterted Fubini integrl just s in problem 4. But this integrl looks hrd, but doble with pproprite trig substitutions. Insted, let s chnge to cylindricl coordintes. Define Φ : 3 3 by φ(r, θ, z (r cos(θ, r sin(θ, z. We see tht the region {(r, θ, z 3 : z 1, θ 2π, r 2z} prmeterizes S vi Φ, i.e. Φ : S. The prmeteriztion fctor is cos(θ r sin(θ det D (r,θ,z Φ det sin(θ r cos(θ 1 r. Finlly, using the prmeteriztion formul, we hve F n divf S S (divf Φ r π 1 θ 2π θ 2π θ 2π θ z 2z cos 2 (θ cos 2 (θ r 1 r 2 cos 2 (θz r drdzdθ z 1 z [ r 4 z 4 cos 2 (θ dθ 2π. ] 2z z 5 dzdθ dzdθ
5 7. Let S 2 {(, } nd define F : S 2 by ( y F (x, y x 2 + y 2, x x 2 + y 2. Compute the line integrl C F dr for ny closed curve C 2 not enclosing the origin. Solution. Since the problem is sking to compute line integrl for pretty much rbitrry curve, the nswer probbly doesn t depend on the curve. So let s just sy we hve ny curve C no enclosing the origin nd let 2 be the region it bounds. Then by Green s theorem version (1 (which we cn pply, since F is nice in the region, C F dr y x 2 + y 2 dx + ( x x x 2 + y 2 y. x x 2 + y 2 dy y x 2 + y 2 8. Find bsis for the vector spce of 3 3 symmetric mtrices. Here symmetric mens A t A. Solution. A bsis is given by e ij for ll 1 i j 3, where e ij is the 3 3 mtrix of ll zeros except for 1 in the ij nd ji spots. If i j then the mtrix e ii just hs 1 in the ii spot. For exmple e 11 1, e In totl, there re 6 such mtrices, so the vector spce of 3 3 symmetric mtrices hs dimension Let V n be the vector spce of polynomils of degree n. Consider the liner trnsformtion T : V 1 V 3 defined by T (p(x x 2 p(x 1. Compute bsis of ker(t nd im(t. Solution. Choose bses 1, x nd 1, x, x 2, x 3 for V 1 nd V 3, respectively. Then T (1 x 2 T (x x 2 (x 1 x 3 x 2 so the mtrix of T with respect to these bses is In prticulr, we see tht the rnk of T is 2, so by the rnknullity theorem, the kernel of T hs dimenion, i.e. ker(t {} nd T is injective. The imge of T is spnned by x 2 nd x 3 x 2, which re linerly independent, so form bsis. A nicer bsis is x 2 nd x 3, which is gotten by tking liner combintions. 1. Let A b c where ±b. Compute the dimensions of ker(a nd im(a. b c
6 Solution. The condition ±b implies, in prticulr, tht nd b re both not zero. So A is never the zero mtrix, so cn never hve rnk. Since A is 2 3 mtrix, dim ker(a 3, dim im(a 2, nd dim ker(a + dim im(a [ 3. Let s [ consider the imge, which is the column spn of A. We b lredy see tht the vectors nd re linerly independent. Indeed, b] ] b det b 2 b 2, which is never zero s long s ±b. In prticulr, the imge of A hs dimension t lest 2, thus is must hve dimension 2. By the rnknullity theorem, the dimension of the kernel is 1. By the wy, bsis for the kernel is given by the (column vector ( c +b, c +b, 1t. 11. Let A be n n n rel mtrix. Wht re the possible vlues of det A if A is symmetric. Solution. Any rel number. Indeed, for ny rel number, the n n digonl mtrix with digonl consisting of one nd n 1 1 s (nd ll the offdigonl entries zero is symmetric nd hs determinnt. b Wht re the possible vlues of det A if A is invertible. Solution. Any nonzero rel number. We know tht n n n invertible mtrix hs nonzero determinnt. Now we ll show tht the determinnt chieves ll nonzero vlues. Given ny nonzero rel number, the digonl mtrix considered bove is invertible nd hs determinnt. 12. For ll (x, y (,, define f(x, y sin(x2 + y 2 x 2 + y 2. Cn f(, be ssigned to mke f continuous t the origin? Solution. It s wellknown from clculus (when you find the derivtive of sin tht nd tht the function sin sin(z ( lim cos( 1, z z g(z { sin(z z for z 1 for z is continuous. Thus we should set f(, 1. To prove tht f, s defined, is continuous, let γ : 2 be continuous curve in the plne, with γ( (,. Note tht r : 2 defined by r(x, y x 2 +y 2 is continuous. Finlly, we now relize f s composition of functions f g r, which re ech continuous everywhere, mking f continuous everywhere, with f(, g(r(, g( Show tht the function f : 2 defined by f(x, y xy, is not differentible t the origin (,. Solution. A good necessry criterion for function f to be differentible t p is tht the difference quotient, f(p + h f(p, h
7 is bounded s h p. In our sitution, unfortuntely our difference quotient is f(u, v f(, uv (u, v u 2 + v, 2 for (u, v (,, which is bounded (by 1 2. So we must use different strtegy. We wnt to sk if liner mp D (, f exists so tht the limit f(h + D (, f(h lim. h h Now long the pth h (, b with b, we hve + D (, f(1, 1 lim lim (,b (, 1 + D (,(1, 1, b which gives limit of either 1+D (, f(1, 1 or 1 D (, f(1, 1 depending on whether the pther is pproching from the first or third qudrnt in the plne. Now both of these vlues cn t be, so the originl limit cn t equl. Thus f is not differentible t the origin.
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationThings to Memorize: A Partial List. January 27, 2017
Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors  Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved
More informationMTH 5102 Linear Algebra Practice Exam 1  Solutions Feb. 9, 2016
Nme (Lst nme, First nme): MTH 502 Liner Algebr Prctice Exm  Solutions Feb 9, 206 Exm Instructions: You hve hour & 0 minutes to complete the exm There re totl of 6 problems You must show your work Prtil
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More information(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence
Problem ( points) Find the vector eqution of the line tht joins points on the two lines L : r ( + t) i t j ( + t) k L : r t i + (t ) j ( + t) k nd is perpendiculr to both those lines. Find the set of ll
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re openended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnkoutnnswer problems! (There re plenty of those in the
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationFINAL REVIEW. 1. Vector Fields, Work, and Flux Suggested Problems:
FINAL EVIEW 1. Vector Fields, Work, nd Flux uggested Problems: { 14.1 7, 13, 16 14.2 17, 25, 27, 29, 36, 45 We dene vector eld F (x, y) to be vector vlued function tht mps ech point in the plne to two
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4
More informationMath 520 Final Exam Topic Outline Sections 1 3 (Xiao/Dumas/Liaw) Spring 2008
Mth 520 Finl Exm Topic Outline Sections 1 3 (Xio/Dums/Liw) Spring 2008 The finl exm will be held on Tuesdy, My 13, 25pm in 117 McMilln Wht will be covered The finl exm will cover the mteril from ll of
More informationAnonymous Math 361: Homework 5. x i = 1 (1 u i )
Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More informationChapter 14. Matrix Representations of Linear Transformations
Chpter 4 Mtrix Representtions of Liner Trnsformtions When considering the Het Stte Evolution, we found tht we could describe this process using multipliction by mtrix. This ws nice becuse computers cn
More information4402 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
4402 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More information14.4. Lengths of curves and surfaces of revolution. Introduction. Prerequisites. Learning Outcomes
Lengths of curves nd surfces of revolution 4.4 Introduction Integrtion cn be used to find the length of curve nd the re of the surfce generted when curve is rotted round n xis. In this section we stte
More informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More informationMATH 260 Final Exam April 30, 2013
MATH 60 Finl Exm April 30, 03 Let Mpn,Rq e the spce of nyn mtrices with rel entries () We know tht (with the opertions of mtrix ddition nd sclr multipliction), M pn, Rq is vector spce Wht is the dimension
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationMATH 13 FINAL STUDY GUIDE, WINTER 2012
MATH 13 FINAL TUY GUI, WINTR 2012 This is ment to be quick reference guide for the topics you might wnt to know for the finl. It probbly isn t comprehensive, but should cover most of wht we studied in
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationSpace Curves. Recall the parametric equations of a curve in xyplane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xyplne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More informationMath 426: Probability Final Exam Practice
Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationJackson 2.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jckson.7 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: Consider potentil problem in the hlfspce defined by, with Dirichlet boundry conditions on the plne
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationSOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014
SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationMath 32B Discussion Session Session 7 Notes August 28, 2018
Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos(  1 2 ) = rcsin( 1 2 ) = rcsin(  1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationdt. However, we might also be curious about dy
Section 0. The Clculus of Prmetric Curves Even though curve defined prmetricly my not be function, we cn still consider concepts such s rtes of chnge. However, the concepts will need specil tretment. For
More informationQuadratic Forms. Quadratic Forms
Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions
Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationLecture 3: Curves in Calculus. Table of contents
Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationWeek 10: Line Integrals
Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationOrthogonal Polynomials
Mth 4401 Gussin Qudrture Pge 1 Orthogonl Polynomils Orthogonl polynomils rise from series solutions to differentil equtions, lthough they cn be rrived t in vriety of different mnners. Orthogonl polynomils
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationHW3, Math 307. CSUF. Spring 2007.
HW, Mth 7. CSUF. Spring 7. Nsser M. Abbsi Spring 7 Compiled on November 5, 8 t 8:8m public Contents Section.6, problem Section.6, problem Section.6, problem 5 Section.6, problem 7 6 5 Section.6, problem
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationStudent Handbook for MATH 3300
Student Hndbook for MATH 3300 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.5 0 0.5 0.5 0 0.5 If people do not believe tht mthemtics is simple, it is only becuse they do not relize how complicted life is. John Louis
More informationMATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The halfangle formula cos 2 θ = 1 2
MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xyplne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points
More information12 TRANSFORMING BIVARIATE DENSITY FUNCTIONS
1 TRANSFORMING BIVARIATE DENSITY FUNCTIONS Hving seen how to trnsform the probbility density functions ssocited with single rndom vrible, the next logicl step is to see how to trnsform bivrite probbility
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationFinal Exam Solutions, MAC 3474 Calculus 3 Honors, Fall 2018
Finl xm olutions, MA 3474 lculus 3 Honors, Fll 28. Find the re of the prt of the sddle surfce z xy/ tht lies inside the cylinder x 2 + y 2 2 in the first positive) octnt; is positive constnt. olution:
More informationChapter 3. Vector Spaces
3.4 Liner Trnsformtions 1 Chpter 3. Vector Spces 3.4 Liner Trnsformtions Note. We hve lredy studied liner trnsformtions from R n into R m. Now we look t liner trnsformtions from one generl vector spce
More informationName Solutions to Test 3 November 8, 2017
Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier
More informationSpring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17
Spring 07 Exm problem MARK BOX points HAND IN PART 0 555=x5 0 NAME: Solutions 3 0 0 PIN: 7 % 00 INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationHung problem # 3 April 10, 2011 () [4 pts.] The electric field points rdilly inwrd [1 pt.]. Since the chrge distribution is cylindriclly symmetric, we pick cylinder of rdius r for our Gussin surfce S.
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
More information1 Linear Least Squares
Lest Squres Pge 1 1 Liner Lest Squres I will try to be consistent in nottion, with n being the number of dt points, nd m < n being the number of prmeters in model function. We re interested in solving
More informationMath Advanced Calculus II
Mth 452  Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationProblem Set 3 Solutions
Msschusetts Institute of Technology Deprtment of Physics Physics 8.07 Fll 2005 Problem Set 3 Solutions Problem 1: Cylindricl Cpcitor Griffiths Problems 2.39: Let the totl chrge per unit length on the inner
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More informationDifferential Geometry: Conformal Maps
Differentil Geometry: Conforml Mps Liner Trnsformtions Definition: We sy tht liner trnsformtion M:R n R n preserves ngles if M(v) 0 for ll v 0 nd: Mv, Mw v, w Mv Mw v w for ll v nd w in R n. Liner Trnsformtions
More informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationA. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c
More informationSample Problems for the Final of Math 121, Fall, 2005
Smple Problems for the Finl of Mth, Fll, 5 The following is collection of vrious types of smple problems covering sections.8,.,.5, nd.8 6.5 of the text which constitute only prt of the common Mth Finl.
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationChapter One: Calculus Revisited
Chpter One: Clculus Revisited 1 Clculus of Single Vrible Question in your mind: How do you understnd the essentil concepts nd theorems in Clculus? Two bsic concepts in Clculus re differentition nd integrtion
More informationElementary Linear Algebra
Elementry Liner Algebr Anton & Rorres, 1 th Edition Lecture Set 5 Chpter 4: Prt II Generl Vector Spces 163 คณ ตศาสตร ว ศวกรรม 3 สาขาว ชาว ศวกรรมคอมพ วเตอร ป การศ กษา 1/2555 163 คณตศาสตรวศวกรรม 3 สาขาวชาวศวกรรมคอมพวเตอร
More information1 Probability Density Functions
Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationMath 113 Exam 2 Practice
Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationTABLE OF CONTENTS 3 CHAPTER 1
TABLE OF CONTENTS 3 CHAPTER 1 Set Lnguge & Nottion 3 CHAPTER 2 Functions 3 CHAPTER 3 Qudrtic Functions 4 CHAPTER 4 Indices & Surds 4 CHAPTER 5 Fctors of Polynomils 4 CHAPTER 6 Simultneous Equtions 4 CHAPTER
More information