Homework 3 Solutions(Part 2) Due Friday Sept. 8
|
|
- Mark Tate
- 5 years ago
- Views:
Transcription
1 MATH 315 Differential Equations (Fall 017) Homework 3 Solutions(Part ) Due Frida Sept. 8 Part : These will e graded in detail. Be sure to start each of these prolems on a new sheet of paper, summarize the prolem, and explain what ou are doing so that a classmate who is struggling can follow what ou are doing just reading our work (do not refer to an external references, including the text). 1. (10 pts) Solve the following initial value prolem using the Integrating Factor Method hand Check to e sure our answer is correct. (x + 1) + 3x = 6x, (0) = 1 We want to solve the first-order ODE aove using the Integrating Factor Method. Since x is the independent variale, we want to e sure the equation can e written in the form: + p(x) = g(x) first. Naturall, we can isolate dividing oth sides x + 1 to otain the equation: + ( 3x ) = 6x. With p(x) = 3x and g(x) = 6x, we are read to find our Integrating x +1 x +1 x +1 x +1 Factor µ(x) = e p(x)dx. p(x)dx = 3x dx. Appling a u-sustitution with u = x +1 x + 1, du = xdx and therefore p(x)dx = 3 du = 3 ln u = 3 ln u x + 1 since 1 du = ln u + C and for the integrating u factor, the aritrar constant is unnecessar. Since x + 1 > 0 and ln a = ln a for real a > 0, this reduces to ln ((x + 1) 3 ). Therefore, our integrating factor is µ(x) = e ln ((x +1) 3 ) = (x + 1) 3. The next step is to multipl oth sides of our modified equation + ( 3x ) = x +1 integrating factor to simplif our work. 6x x +1 µ(x)( + ( 3x 6x )) = µ(x)( ) = x +1 x +1 (x + 1) 3 + (x + 1) 3 ( 3x ) = x +1 (x + 1) 3 ( 6x ). x +1 This simplifies down to (x + 1) 3 + 3x(x + 1) 1 = 6x(x + 1) 1. the Notice that d dx ((x +1) 3 ) = 3 (x +1) 1 (x +1) = 3 (x +1) 1 x = 3x(x +1) 1. This matches the second term in the equation. Therefore, we have: (x + 1) 3 + ((x + 1) 3 ) = 6x(x + 1) 1. Recalling the product rule, we know that fg + f g = (fg) with f = (x + 1) 3, g =, our equation simplifies down to: ((x + 1) 3 ) = 6x(x + 1) 1. We can now solve for integration. Appling the Fundamental Theorem of Calculus Part I, we know that d (f(x))dx = f(x) + C. dx Therefore: d dx ((x + 1) 3 )dx = 6x(x + 1) 1 dx = ((x + 1) 3 ) = 6x(x + 1) 1 dx and so = 6x(x +1) 1 dx (x +1) 3. To evaluate the integral in the numerator, we need to appl a u-sustitution. With u = x + 1, du = x = 1 du = xdx. This means that: dx
2 6x(x + 1) 1 dx = 6 (u) 1 1 du = 3 u 1 du = 3 3 u 3 + C = (x + 1) 3 + C since u n du = un+1 n+1 + C for real n 1. We can now write down our general solution: (t) = We are almost finished. 6x(x +1) 1 dx (x +1) 3 Appling our initial condition (0) = 1, we have: ((0) +1) 3 +C = 1+C 1 = 1 = C = 3. ((0) +1) 3 Our particular solution is then given : (t) = (x +1) (t) = (x + 1) 3. (16 pts) Consider the initial value prolem (x +1) 3 = (x +1) 3 +C. (x +1) 3 = 1 or equivalentl: = 1 + cos(t), (0) = 0 (a) (6 pts) Solve using Integrating Factor Method hand. Our equation is in the form + p(t) = g(t), meaning we can use the Method of Integrating Factors directl. With p(t) = 1, g(t) = 1 + cos t we are read to compute our integrating factor. Our integrating factor is µ(t) = e p(t)dt = e 1dt = e t+c. Again, we can ignore the aritrar constant and take our integrating factor to e e t. The next step is to multipl oth sides of the equation this function: e t ( ) = e t (1 + cos t) = (e t ) + ( e t ) = e t + e t cos t. Noticing that (e t ) = e t like the previous prolem, we have an expression which reminds us of the product rule: (e t ) + (e t ) = e t + e t cos t = (e t ) = e t + e t cos t. We can now solve for : = (e t +e t cos t)dt e t. To tackle the integral aove, we can first split it up: (e t + e t cos t)dt = e t dt + e t cos tdt. The first integral can e tackled a u-sustitution with u = t = du = dt: e t dt = e u du = e u + C = e t + C. The second integral requires a recursive integration parts(ibp). It is a special case of the following integral: e at cos tdt. To solve an integral like this, we want to integrate parts until we see a similar integral appear. This will require appling IBP twice. The IBP formula is given udv = uv vdu. Setting u = e at, dv = cos t, du = (e at ) dt = ae at dt. v is the antiderivative of cos t = v = 1 sin t, appling the single-variale Chain Rule: d (f(g(t)) = f (g(t)) g (t) = d sin t cos t cos t ( ) = (t) = = cos t. dt dt
3 e at cos tdt = uv vdu = e t sin t ae at sin t dt = e t sin t a( e at sin tdt). Appling the same process with the new integral in parentheses: e at sin tdt = UdV = UV V du Setting U = e at, du = ae at dt, dv = sin t = V = Putting it all together: e at sin tdt = eat cos t ae at cos tdt = eat cos t + a e at cos tdt Simplifing, we have: e at cos tdt = eat sin t a( eat cos t + a e at cos tdt) = eat sin t + ( a e at cos t a e at cos tdt) cos t Continuing: e at cos tdt = eat sin t + a e at cos t a ( e at cos tdt) = (1 + a ) e at cos tdt = eat sin t + a e at cos t This simplifies to: a + ( e at cos tdt) = eat sin t + a e at cos t = e at cos tdt = ( eat sin t a + + a e at cos t) + C = eat (a cos t+ sin t) a + + C. The integral that we were tring to solve is a special case of the aove formula we derived. With a = 1, = 1, we have: e at cos tdt = eat (a cos t+ sin t) + C a + = e t cos tdt = e t ( cos 1 t+1 sin 1 t) + C = e t (sin t cos t) ( 1) +(1) Convenientl, e t cos tdt = e t (sin t cos t) + C. This means that our general solution is: (t) = (e t +e t cos t)dt e t = 1 + sin t cos t + C e t = e t +e t (sin t cos t)+c e t = sin t (1 + cos t) + Ce t Appling the initial condition (0) = 0, we otain: sin 0 (1 + cos 0) + Ce 0 = 0 = 0 + C = 0 = C = 0 +. The particular solution to the ODE must then e: (t) = sin t (1 + cos t) + ( 0 + )e t. () (4 pts) i. For what values of 0 does lim t (t) =? Recall that sin t 1 and cos t 1. This means that 1 sin t 1, 1 cos t 1. This implies that the expression sin t (1 + cos t) is at least 1 (1 + 1) = 3 and at most 1 (1 1) = 1 therefore 3 sin t (1 + cos t) 1, this term is ounded. Since the term ( 0 + )e t appears in the equation, as t this term will dominate: it will either vanish or grow quickl to ± depending on the value of 0. This is what we need to determine our long term ehavior. If 0 + > 0, lim t ( 0 + )e t = + = lim t (t) = + So we need 0 > to guarantee upward growth. ii. For what values of 0 does lim t (t) =? B the same reasoning, if ( 0 + ) < 0, lim t ( 0 + )e t = = lim t (t) =. So we need 0 < to guarantee oundless deca.
4 (c) (6 pts) Graph the direction field and 3 particular solutions on one graph using technolog (Maple is proal easiest). For the three particular solutions, choose 0 so that one solution goes to +, one goes to, and one remains ounded as t. Be sure to lael each curve. B our reasoning from part (), we need 0 > for +, 0 < to guarantee and we can pick 0 = to ensure that ( 0 +)e t = 0 so that (t) = sin t (1+cos t), which we showed must oscillate etween -3 and 1, and thus remains ounded aove and elow. The following Maple code generates a lue graph with 0 = 0 approaching +, a green graph with (0) = with confined oscillations and a purple graph with (0) = 3, approaching. with(detools): prolem1:=diff((t),t)=(1+*cos(t)) + (t); vars:={(t)}; t_values:=t=-6..6; _values:==-5..5; ICs:={[0,0],[0,-], [0,-3]}; DEplot(prolem1,vars,t_values,_values,ICs,stepsize=0.1,linecolor=[purple, green, lue]); 3. (10 pts) Consider the following initial value prolem d dx = + 1, (1) = 3. (a) Solve the initial value prolem using Separation of Variales. The Separation of Variales Technique allows us to solve first-order ODEs which can e written as a product of functions of the dependent and independent variales respectivel. We will assume x to e the independent variale. Setting h() = +1 d, g(x) = 1 we have: = ( +1 )(1) = (h())(g(x)), the equation is dx separale. We can then separate the functions and integrate: +1 = 1 = +1 dx = 1dx = d = 1dx +1 Since x, we know that from studing differentials and linear approximations that this relation carries through the approximation: d = dx.
5 To solve the integral on the left-hand side, we can use a u-sustitution with u = + 1, du = = d = 1 du, therefore: d d = 1 1du = 1 ln u + C = 1 ln +1 u C Finishing up our solution: d = 1dx = 1 ln = 1dx = x + C after comining constants. We can solve for. Notice that + 1 > 0 for all real-valued, which means we can drop the asolute value: ln + 1 = x + C = + 1 = Ce x = (x) = ± Ce x 1. Notice that we have two families of functions rather than one due to the ±. The real-valued function = x has a nonnegative range. This means that the sign indicated the initial condition tells us which solution to choose. Our initial condition is (1) = 3. This means that when x = 1, (x) = 3. Since the square root function does not change sign for real inputs, we can assume that the negative output of the solution (x) at one point implies a negative value for an other point in its range. Therefore, (x) = Ce x 1. We can now solve for C: 3 = Ce 1 1 = 9 = Ce 1 = C = 10. e The particular solution is given : 10 (x) = e e x 1 or equivalentl: (x) = 10e x 1. () For what values of x, does the solution exist? (This means the solution is well-defined and is well-defined). Appling the Chain Rule, the derivative is given : = d ( 10e dx x 1) = 1 (10ex 1) 1 (10e x 1) = 1 (10ex 1) 1 (10(x ) e x 0) = 1 (10ex 1) 1 (10()e x 0) = (10e x 1) 1 (10e x ) Finall, = 10e x 1 = = 10ex. 10e x 1 The derivative does not exist when 10e x 1 0 and that is enough to guarantee discontinuit in our solution. The solution exists when: 10e x 1 > 0 e x > e x > 1 e ln x > 1 ( ln e ln 10) This means that ln 10 x > (c) Graph the solution and denote on the graph the interval of x-values otained from part () where the solution exists. Explain graphicall wh it does not exist eond that interval. The following Maple code generates a direction field for the ODE and a graph of the ln 10 solution in lue. The lue graph nearl touches the x-axis at x = 1 : a warning is given maple claiming that the solution cannot e evaluated eond this point.
6 with(detools): prolem1:=diff((x),x)=(((x))^ + 1)/((x)); vars:={(x)}; x_values:=x=-6..6; _values:==-5..5; ICs:={[1,-3]}; DEplot(prolem1,vars,x_values,_values,ICs,stepsize=0.1,linecolor=[purple]); Correspondingl, the graph is not visile for x-values to the left of this point. That is ecause near this point, the derivative tends to infinit, as shown the vertical tendenc of the graph near the x-intercept. ln 10 Furthermore, since the graph is complex-valued for x < 1, it the graph will not e visile in this region, as we can see. Producing this graph using other software such as Desmos, our output looks like the following: The point (1 ln 10, 0) is emphasized.
OBJECTIVES Use the area under a graph to find total cost. Use rectangles to approximate the area under a graph.
4.1 The Area under a Graph OBJECTIVES Use the area under a graph to find total cost. Use rectangles to approximate the area under a graph. 4.1 The Area Under a Graph Riemann Sums (continued): In the following
More informationSolutions to Exam 2, Math 10560
Solutions to Exam, Math 6. Which of the following expressions gives the partial fraction decomposition of the function x + x + f(x = (x (x (x +? Solution: Notice that (x is not an irreducile factor. If
More informationGeneralized Geometric Series, The Ratio Comparison Test and Raabe s Test
Generalized Geometric Series The Ratio Comparison Test and Raae s Test William M. Faucette Decemer 2003 The goal of this paper is to examine the convergence of a type of infinite series in which the summands
More informationIntegration by Parts
Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u
More informationMath 106 Answers to Exam 3a Fall 2015
Math 6 Answers to Exam 3a Fall 5.. Consider the curve given parametrically by x(t) = cos(t), y(t) = (t 3 ) 3, for t from π to π. (a) (6 points) Find all the points (x, y) where the graph has either a vertical
More informationAnnouncements. Topics: Homework:
Announcements Topics: - sections 7.3 (the definite integral +area), 7.4 (FTC), 7.5 (additional techniques of integration) * Read these sections and study solved examples in your textbook! Homework: - review
More informationIntegration by Parts. MAT 126, Week 2, Thursday class. Xuntao Hu
MAT 126, Week 2, Thursday class Xuntao Hu Recall that the substitution rule is a combination of the FTC and the chain rule. We can also combine the FTC and the product rule: d dx [f (x)g(x)] = f (x)g (x)
More informationMath 1310 Lab 10. (Sections )
Math 131 Lab 1. (Sections 5.1-5.3) Name/Unid: Lab section: 1. (Properties of the integral) Use the properties of the integral in section 5.2 for answering the following question. (a) Knowing that 2 2f(x)
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More information1 Introduction; Integration by Parts
1 Introduction; Integration by Parts September 11-1 Traditionally Calculus I covers Differential Calculus and Calculus II covers Integral Calculus. You have already seen the Riemann integral and certain
More informationMATH 131P: PRACTICE FINAL SOLUTIONS DECEMBER 12, 2012
MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, This is a closed ook, closed notes, no calculators/computers exam. There are 6 prolems. Write your solutions to Prolems -3 in lue ook #, and your solutions to
More informationDEFINITE INTEGRALS & NUMERIC INTEGRATION
DEFINITE INTEGRALS & NUMERIC INTEGRATION Calculus answers two very important questions. The first, how to find the instantaneous rate of change, we answered with our study of the derivative. We are now
More informationMath Refresher Course
Math Refresher Course Columbia University Department of Political Science Fall 2007 Day 2 Prepared by Jessamyn Blau 6 Calculus CONT D 6.9 Antiderivatives and Integration Integration is the reverse of differentiation.
More information1. Evaluate the integrals. a. (9 pts) x e x/2 dx. Solution: Using integration by parts, let u = x du = dx and dv = e x/2 dx v = 2e x/2.
MATH 8 Test -SOLUTIONS Spring 4. Evaluate the integrals. a. (9 pts) e / Solution: Using integration y parts, let u = du = and dv = e / v = e /. Then e / = e / e / e / = e / + e / = e / 4e / + c MATH 8
More informationLecture 5: Integrals and Applications
Lecture 5: Integrals and Applications Lejla Batina Institute for Computing and Information Sciences Digital Security Version: spring 2012 Lejla Batina Version: spring 2012 Wiskunde 1 1 / 21 Outline The
More informationMath 180 Written Homework Assignment #10 Due Tuesday, December 2nd at the beginning of your discussion class.
Math 18 Written Homework Assignment #1 Due Tuesday, December 2nd at the beginning of your discussion class. Directions. You are welcome to work on the following problems with other MATH 18 students, but
More informationMATH 1207 R02 MIDTERM EXAM 2 SOLUTION
MATH 7 R MIDTERM EXAM SOLUTION FALL 6 - MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () (5 pts) Find
More information" $ CALCULUS 2 WORKSHEET #21. t, y = t + 1. are A) x = 0, y = 0 B) x = 0 only C) x = 1, y = 0 D) x = 1 only E) x= 0, y = 1
CALCULUS 2 WORKSHEET #2. The asymptotes of the graph of the parametric equations x = t t, y = t + are A) x = 0, y = 0 B) x = 0 only C) x =, y = 0 D) x = only E) x= 0, y = 2. What are the coordinates of
More informationMath RE - Calculus II Antiderivatives and the Indefinite Integral Page 1 of 5
Math 201-203-RE - Calculus II Antiderivatives and the Indefinite Integral Page 1 of 5 What is the Antiderivative? In a derivative problem, a function f(x) is given and you find the derivative f (x) using
More informationAnnouncements. Topics: Homework:
Announcements Topics: - sections 7.4 (FTC), 7.5 (additional techniques of integration), 7.6 (applications of integration) * Read these sections and study solved examples in your textbook! Homework: - review
More informationCalculus II Practice Test Problems for Chapter 7 Page 1 of 6
Calculus II Practice Test Problems for Chapter 7 Page of 6 This is a set of practice test problems for Chapter 7. This is in no way an inclusive set of problems there can be other types of problems on
More information1 Antiderivatives graphically and numerically
Math B - Calculus by Hughes-Hallett, et al. Chapter 6 - Constructing antiderivatives Prepared by Jason Gaddis Antiderivatives graphically and numerically Definition.. The antiderivative of a function f
More informationAnnouncements. Topics: Homework:
Announcements Topics: - sections 7.5 (additional techniques of integration), 7.6 (applications of integration), * Read these sections and study solved examples in your textbook! Homework: - review lecture
More informationMath 3B: Lecture 11. Noah White. October 25, 2017
Math 3B: Lecture 11 Noah White October 25, 2017 Introduction Midterm 1 Introduction Midterm 1 Average is 73%. This is higher than I expected which is good. Introduction Midterm 1 Average is 73%. This is
More informationLecture 4: Integrals and applications
Lecture 4: Integrals and applications Lejla Batina Institute for Computing and Information Sciences Digital Security Version: autumn 2013 Lejla Batina Version: autumn 2013 Calculus en Kansrekenen 1 / 18
More informationf(x) g(x) = [f (x)g(x) dx + f(x)g (x)dx
Chapter 7 is concerned with all the integrals that can t be evaluated with simple antidifferentiation. Chart of Integrals on Page 463 7.1 Integration by Parts Like with the Chain Rule substitutions with
More informationERASMUS UNIVERSITY ROTTERDAM Information concerning the Entrance examination Mathematics level 2 for International Business Administration (IBA)
ERASMUS UNIVERSITY ROTTERDAM Information concerning the Entrance examination Mathematics level 2 for International Business Administration (IBA) General information Availale time: 2.5 hours (150 minutes).
More informationExploring Substitution
I. Introduction Exploring Substitution Math Fall 08 Lab We use the Fundamental Theorem of Calculus, Part to evaluate a definite integral. If f is continuous on [a, b] b and F is any antiderivative of f
More informationSection 5.6. Integration By Parts. MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10
Section 5.6 Integration By Parts MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10 Integration By Parts Manipulating the Product Rule d dx (f (x) g(x)) = f (x) g (x) + f (x) g(x)
More informationHomework Solutions: , plus Substitutions
Homework Solutions: 2.-2.2, plus Substitutions Section 2. I have not included any drawings/direction fields. We can see them using Maple or by hand, so we ll be focusing on getting the analytic solutions
More informationExam 3 review for Math 1190
Exam 3 review for Math 9 Be sure to be familiar with the following : Extreme Value Theorem Optimization The antiderivative u-substitution as a method for finding antiderivatives Reimann sums (e.g. L 6
More informationINTEGRATION: THE FUNDAMENTAL THEOREM OF CALCULUS MR. VELAZQUEZ AP CALCULUS
INTEGRATION: THE FUNDAMENTAL THEOREM OF CALCULUS MR. VELAZQUEZ AP CALCULUS RECALL: ANTIDERIVATIVES When we last spoke of integration, we examined a physics problem where we saw that the area under the
More informationMath 152 Take Home Test 1
Math 5 Take Home Test Due Monday 5 th October (5 points) The following test will be done at home in order to ensure that it is a fair and representative reflection of your own ability in mathematics I
More informationFunction Practice. 1. (a) attempt to form composite (M1) (c) METHOD 1 valid approach. e.g. g 1 (5), 2, f (5) f (2) = 3 A1 N2 2
1. (a) attempt to form composite e.g. ( ) 3 g 7 x, 7 x + (g f)(x) = 10 x N (b) g 1 (x) = x 3 N1 1 (c) METHOD 1 valid approach e.g. g 1 (5),, f (5) f () = 3 N METHOD attempt to form composite of f and g
More informationIntegrated Calculus II Exam 1 Solutions 2/6/4
Integrated Calculus II Exam Solutions /6/ Question Determine the following integrals: te t dt. We integrate by parts: u = t, du = dt, dv = e t dt, v = dv = e t dt = e t, te t dt = udv = uv vdu = te t (
More informationGrade: The remainder of this page has been left blank for your workings. VERSION D. Midterm D: Page 1 of 12
First Name: Student-No: Last Name: Section: Grade: The remainder of this page has been left blank for your workings. Midterm D: Page of 2 Indefinite Integrals. 9 marks Each part is worth marks. Please
More information1 Solution to Homework 4
Solution to Homework Section. 5. The characteristic equation is r r + = (r )(r ) = 0 r = or r =. y(t) = c e t + c e t y = c e t + c e t. y(0) =, y (0) = c + c =, c + c = c =, c =. To find the maximum value
More informationCALCULUS II MATH Dr. Hyunju Ban
CALCULUS II MATH 2414 Dr. Hyunju Ban Introduction Syllabus Chapter 5.1 5.4 Chapters To Be Covered: Chap 5: Logarithmic, Exponential, and Other Transcendental Functions (2 week) Chap 7: Applications of
More information2.12: Derivatives of Exp/Log (cont d) and 2.15: Antiderivatives and Initial Value Problems
2.12: Derivatives of Exp/Log (cont d) and 2.15: Antiderivatives and Initial Value Problems Mathematics 3 Lecture 14 Dartmouth College February 03, 2010 Derivatives of the Exponential and Logarithmic Functions
More informationMath 106: Review for Exam II - SOLUTIONS
Math 6: Review for Exam II - SOLUTIONS INTEGRATION TIPS Substitution: usually let u a function that s inside another function, especially if du (possibly off by a multiplying constant) is also present
More informationShort Solutions to Review Material for Test #2 MATH 3200
Short Solutions to Review Material for Test # MATH 300 Kawai # Newtonian mechanics. Air resistance. a A projectile is launched vertically. Its height is y t, and y 0 = 0 and v 0 = v 0 > 0. The acceleration
More informationMath 142, Final Exam, Fall 2006, Solutions
Math 4, Final Exam, Fall 6, Solutions There are problems. Each problem is worth points. SHOW your wor. Mae your wor be coherent and clear. Write in complete sentences whenever this is possible. CIRCLE
More informationLecture 19: Solving linear ODEs + separable techniques for nonlinear ODE s
Lecture 19: Solving linear ODEs + separable techniques for nonlinear ODE s Geoffrey Cowles Department of Fisheries Oceanography School for Marine Science and Technology University of Massachusetts-Dartmouth
More informationMath 162: Calculus IIA
Math 62: Calculus IIA Final Exam ANSWERS December 9, 26 Part A. (5 points) Evaluate the integral x 4 x 2 dx Substitute x 2 cos θ: x 8 cos dx θ ( 2 sin θ) dθ 4 x 2 2 sin θ 8 cos θ dθ 8 cos 2 θ cos θ dθ
More informationdy = f( x) dx = F ( x)+c = f ( x) dy = f( x) dx
Antiderivatives and The Integral Antiderivatives Objective: Use indefinite integral notation for antiderivatives. Use basic integration rules to find antiderivatives. Another important question in calculus
More informationFinal Problem Set. 2. Use the information in #1 to show a solution to the differential equation ), where k and L are constants and e c L be
Final Problem Set Name A. Show the steps for each of the following problems. 1. Show 1 1 1 y y L y y(1 ) L.. Use the information in #1 to show a solution to the differential equation dy y ky(1 ), where
More informationTIPS FOR WRITING PROOFS IN HOMEWORK ASSIGNMENTS. 1. Simple rules
TIPS FOR WRITING PROOFS IN HOMEWORK ASSIGNMENTS MARK SKANDERA 1 Simple rules I require my students to follow the rules below when submitting problem sets While other instructors may be more lenient than
More informationMath 104 Calculus 8.8 Improper Integrals. Math Yu
Math 04 Calculus 8.8 Improper Integrals Math 04 - Yu Improper Integrals Goal: To evaluate integrals of func?ons over infinite intervals or with an infinite discon?nuity. Method: We replace the bad endpoints
More informationUNIT 3 INTEGRATION 3.0 INTRODUCTION 3.1 OBJECTIVES. Structure
Calculus UNIT 3 INTEGRATION Structure 3.0 Introduction 3.1 Objectives 3.2 Basic Integration Rules 3.3 Integration by Substitution 3.4 Integration of Rational Functions 3.5 Integration by Parts 3.6 Answers
More informationMATH 162. Midterm 2 ANSWERS November 18, 2005
MATH 62 Midterm 2 ANSWERS November 8, 2005. (0 points) Does the following integral converge or diverge? To get full credit, you must justify your answer. 3x 2 x 3 + 4x 2 + 2x + 4 dx You may not be able
More informationReal option valuation for reserve capacity
Real option valuation for reserve capacity MORIARTY, JM; Palczewski, J doi:10.1016/j.ejor.2016.07.003 For additional information aout this pulication click this link. http://qmro.qmul.ac.uk/xmlui/handle/123456789/13838
More informationLecture 2: Separable Ordinary Differential Equations
Lecture : Separable Ordinar Differential Equations Dr. Michael Doughert Januar 8, 00 Some Terminolog: ODE s, PDE s, IVP s The differential equations we have looked at so far are called ordinar differential
More information1.5 First Order PDEs and Method of Characteristics
1.5. FIRST ORDER PDES AND METHOD OF CHARACTERISTICS 35 1.5 First Order PDEs and Method of Characteristics We finish this introductory chapter by discussing the solutions of some first order PDEs, more
More informationHW2 Solutions. MATH 20D Fall 2013 Prof: Sun Hui TA: Zezhou Zhang (David) October 14, Checklist: Section 2.6: 1, 3, 6, 8, 10, 15, [20, 22]
HW2 Solutions MATH 20D Fall 2013 Prof: Sun Hui TA: Zezhou Zhang (David) October 14, 2013 Checklist: Section 2.6: 1, 3, 6, 8, 10, 15, [20, 22] Section 3.1: 1, 2, 3, 9, 16, 18, 20, 23 Section 3.2: 1, 2,
More informationFirst Order Differential Equations
Chapter First Order Differential Equations Contents. The Method of Quadrature.......... 68. Separable Equations............... 74.3 Linear Equations I................ 85.4 Linear Equations II...............
More informationy = x 3 and y = 2x 2 x. 2x 2 x = x 3 x 3 2x 2 + x = 0 x(x 2 2x + 1) = 0 x(x 1) 2 = 0 x = 0 and x = (x 3 (2x 2 x)) dx
Millersville University Name Answer Key Mathematics Department MATH 2, Calculus II, Final Examination May 4, 2, 8:AM-:AM Please answer the following questions. Your answers will be evaluated on their correctness,
More informationwhose domain D is a set of n-tuples in is defined. The range of f is the set of all values f x1,..., x n
Grade (MCV4UE) - AP Calculus Etended Page o A unction o n-variales is a real-valued unction... n whose domain D is a set o n-tuples... n in which... n is deined. The range o is the set o all values...
More informationGoal: Approximate the area under a curve using the Rectangular Approximation Method (RAM) RECTANGULAR APPROXIMATION METHODS
AP Calculus 5. Areas and Distances Goal: Approximate the area under a curve using the Rectangular Approximation Method (RAM) Exercise : Calculate the area between the x-axis and the graph of y = 3 2x.
More informationMCS 115 Exam 2 Solutions Apr 26, 2018
MCS 11 Exam Solutions Apr 6, 018 1 (10 pts) Suppose you have an infinitely large arrel and a pile of infinitely many ping-pong alls, laeled with the positive integers 1,,3,, much like in the ping-pong
More informationEXAM 3 MAT 167 Calculus I Spring is a composite function of two functions y = e u and u = 4 x + x 2. By the. dy dx = dy du = e u x + 2x.
EXAM MAT 67 Calculus I Spring 20 Name: Section: I Each answer must include either supporting work or an explanation of your reasoning. These elements are considered to be the main part of each answer and
More information(e) 2 (f) 2. (c) + (d). Limits at Infinity. 2.5) 9-14,25-34,41-43,46-47,56-57, (c) (d) 2
Math 150A. Final Review Answers, Spring 2018. Limits. 2.2) 7-10, 21-24, 28-1, 6-8, 4-44. 1. Find the values, or state they do not exist. (a) (b) 1 (c) DNE (d) 1 (e) 2 (f) 2 (g) 2 (h) 4 2. lim f(x) = 2,
More informationLIMITS AT INFINITY MR. VELAZQUEZ AP CALCULUS
LIMITS AT INFINITY MR. VELAZQUEZ AP CALCULUS RECALL: VERTICAL ASYMPTOTES Remember that for a rational function, vertical asymptotes occur at values of x = a which have infinite its (either positive or
More informationMATH 18.01, FALL PROBLEM SET #5 SOLUTIONS (PART II)
MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II (Oct ; Antiderivatives; + + 3 7 points Recall that in pset 3A, you showed that (d/dx tanh x x Here, tanh (x denotes the inverse to the hyperbolic tangent
More informationFinal exam (practice) UCLA: Math 31B, Spring 2017
Instructor: Noah White Date: Final exam (practice) UCLA: Math 3B, Spring 207 This exam has 8 questions, for a total of 80 points. Please print your working and answers neatly. Write your solutions in the
More informationGraphs and polynomials
5_6_56_MQVMM - _t Page Frida, Novemer 8, 5 :5 AM MQ Maths Methods / Final Pages / 8//5 Graphs and polnomials VCEcoverage Areas of stud Units & Functions and graphs Algera In this chapter A The inomial
More informationAssignment 16 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers!
Assignment 6 Solution Please do not copy and paste my answer. You will get similar questions but with different numbers! Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f
More informationMA261-A Calculus III 2006 Fall Homework 7 Solutions Due 10/20/2006 8:00AM
MA26-A Calculus III 2006 Fall Homework 7 Solutions Due 0/20/2006 8:00AM 3 #4 Find the rst partial derivatives of the function f (; ) 5 + 3 3 2 + 3 4 f (; ) 5 4 + 9 2 2 + 3 4 f (; ) 6 3 + 2 3 3 #6 Find
More informationPrelim 1 Solutions V2 Math 1120
Feb., Prelim Solutions V Math Please show your reasoning and all your work. This is a 9 minute exam. Calculators are not needed or permitted. Good luck! Problem ) ( Points) Calculate the following: x a)
More informationAP Calculus Chapter 9: Infinite Series
AP Calculus Chapter 9: Infinite Series 9. Sequences a, a 2, a 3, a 4, a 5,... Sequence: A function whose domain is the set of positive integers n = 2 3 4 a n = a a 2 a 3 a 4 terms of the sequence Begin
More informationHigher Order Linear Equations
C H A P T E R 4 Higher Order Linear Equations 4.1 1. The differential equation is in standard form. Its coefficients, as well as the function g(t) = t, are continuous everywhere. Hence solutions are valid
More informationChapter 2: First Order DE 2.4 Linear vs. Nonlinear DEs
Chapter 2: First Order DE 2.4 Linear vs. Nonlinear DEs First Order DE 2.4 Linear vs. Nonlinear DE We recall the general form of the First Oreder DEs (FODE): dy = f(t, y) (1) dt where f(t, y) is a function
More informationMath 121 Winter 2010 Review Sheet
Math 121 Winter 2010 Review Sheet March 14, 2010 This review sheet contains a number of problems covering the material that we went over after the third midterm exam. These problems (in conjunction with
More informationSection: I. u 4 du. (9x + 1) + C, 3
EXAM 3 MAT 168 Calculus II Fall 18 Name: Section: I All answers must include either supporting work or an eplanation of your reasoning. MPORTANT: These elements are considered main part of the answer and
More informationQuestions from Larson Chapter 4 Topics. 5. Evaluate
Math. Questions from Larson Chapter 4 Topics I. Antiderivatives. Evaluate the following integrals. (a) x dx (4x 7) dx (x )(x + x ) dx x. A projectile is launched vertically with an initial velocity of
More informationQMI Lesson 19: Integration by Substitution, Definite Integral, and Area Under Curve
QMI Lesson 19: Integration by Substitution, Definite Integral, and Area Under Curve C C Moxley Samford University Brock School of Business Substitution Rule The following rules arise from the chain rule
More informationBusiness and Life Calculus
Business and Life Calculus George Voutsadakis Mathematics and Computer Science Lake Superior State University LSSU Math 2 George Voutsadakis (LSSU) Calculus For Business and Life Sciences Fall 203 / 55
More informationFriday 09/15/2017 Midterm I 50 minutes
Fa 17: MATH 2924 040 Differential and Integral Calculus II Noel Brady Friday 09/15/2017 Midterm I 50 minutes Name: Student ID: Instructions. 1. Attempt all questions. 2. Do not write on back of exam sheets.
More informationPolynomial Degree and Finite Differences
CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson, you Learn the terminology associated with polynomials Use the finite differences method to determine the degree of a polynomial
More information7.3 Hyperbolic Functions Hyperbolic functions are similar to trigonometric functions, and have the following
Math 2-08 Rahman Week3 7.3 Hyperbolic Functions Hyperbolic functions are similar to trigonometric functions, and have the following definitions: sinh x = 2 (ex e x ) cosh x = 2 (ex + e x ) tanh x = sinh
More informationTHE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK. Summer Examination 2009.
OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK Summer Examination 2009 First Engineering MA008 Calculus and Linear Algebra
More informationM343 Homework 3 Enrique Areyan May 17, 2013
M343 Homework 3 Enrique Areyan May 17, 013 Section.6 3. Consider the equation: (3x xy + )dx + (6y x + 3)dy = 0. Let M(x, y) = 3x xy + and N(x, y) = 6y x + 3. Since: y = x = N We can conclude that this
More informationERASMUS UNIVERSITY ROTTERDAM
Information concerning Colloquium doctum Mathematics level 2 for International Business Administration (IBA) and International Bachelor Economics & Business Economics (IBEB) General information ERASMUS
More informationf x,y da ln 1 y ln 1 x dx. To evaluate this integral, we use integration by parts with dv dx
MATH (Calculus III) - Quiz 4 Solutions March, 5 S. F. Ellermeer Name Instructions. This is a take home quiz. It is due to be handed in to me on Frida, March at class time. You ma work on this quiz alone
More informationMath 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3
Math 201 Solutions to Assignment 1 1. Solve the initial value problem: x 2 dx + 2y = 0, y(0) = 2. x 2 dx + 2y = 0, y(0) = 2 2y = x 2 dx y 2 = 1 3 x3 + C y = C 1 3 x3 Notice that y is not defined for some
More informationMATH 225: Foundations of Higher Matheamatics. Dr. Morton. 3.4: Proof by Cases
MATH 225: Foundations of Higher Matheamatics Dr. Morton 3.4: Proof y Cases Chapter 3 handout page 12 prolem 21: Prove that for all real values of y, the following inequality holds: 7 2y + 2 2y 5 7. You
More informationMATH 32A: MIDTERM 2 REVIEW. sin 2 u du z(t) = sin 2 t + cos 2 2
MATH 3A: MIDTERM REVIEW JOE HUGHES 1. Curvature 1. Consider the curve r(t) = x(t), y(t), z(t), where x(t) = t Find the curvature κ(t). 0 cos(u) sin(u) du y(t) = Solution: The formula for curvature is t
More informationMATH 56A SPRING 2008 STOCHASTIC PROCESSES 197
MATH 56A SPRING 8 STOCHASTIC PROCESSES 197 9.3. Itô s formula. First I stated the theorem. Then I did a simple example to make sure we understand what it says. Then I proved it. The key point is Lévy s
More informationChapter 2 Notes, Kohler & Johnson 2e
Contents 2 First Order Differential Equations 2 2.1 First Order Equations - Existence and Uniqueness Theorems......... 2 2.2 Linear First Order Differential Equations.................... 5 2.2.1 First
More informationAPPLICATIONS OF DIFFERENTIATION
4 APPLICATIONS OF DIFFERENTIATION APPLICATIONS OF DIFFERENTIATION 4.9 Antiderivatives In this section, we will learn about: Antiderivatives and how they are useful in solving certain scientific problems.
More informationMath 112 Section 10 Lecture notes, 1/7/04
Math 11 Section 10 Lecture notes, 1/7/04 Section 7. Integration by parts To integrate the product of two functions, integration by parts is used when simpler methods such as substitution or simplifying
More informationReview all the activities leading to Midterm 3. Review all the problems in the previous online homework sets (8+9+10).
MA109, Activity 34: Review (Sections 3.6+3.7+4.1+4.2+4.3) Date: Objective: Additional Assignments: To prepare for Midterm 3, make sure that you can solve the types of problems listed in Activities 33 and
More informationHOMEWORK 3 MA1132: ADVANCED CALCULUS, HILARY 2017
HOMEWORK MA112: ADVANCED CALCULUS, HILARY 2017 (1) A particle moves along a curve in R with position function given by r(t) = (e t, t 2 + 1, t). Find the velocity v(t), the acceleration a(t), the speed
More informationMath 116 Second Midterm November 14, 2012
Math 6 Second Midterm November 4, Name: EXAM SOLUTIONS Instructor: Section:. Do not open this exam until you are told to do so.. This exam has pages including this cover. There are 8 problems. Note that
More informationReview of Lecture 5. F = GMm r 2. = m dv dt Expressed in terms of altitude x = r R, we have. mv dv dx = GMm. (R + x) 2. Max altitude. 2GM v 2 0 R.
Review of Lecture 5 Models could involve just one or two equations (e.g. orbit calculation), or hundreds of equations (as in climate modeling). To model a vertical cannon shot: F = GMm r 2 = m dv dt Expressed
More informationNO CALCULATORS. NO BOOKS. NO NOTES. TURN OFF YOUR CELL PHONES AND PUT THEM AWAY.
FINAL EXAM-MATH 3 FALL TERM, R. Blute & A. Novruzi Name(Print LEGIBLY) I.D. Number Instructions- This final examination consists of multiple choice questions worth 3 points each. Your answers to the multiple
More information20D - Homework Assignment 1
0D - Homework Assignment Brian Bowers (TA for Hui Sun) MATH 0D Homework Assignment October 7, 0. #,,,4,6 Solve the given differential equation. () y = x /y () y = x /y( + x ) () y + y sin x = 0 (4) y =
More information4.1 Analysis of functions I: Increase, decrease and concavity
4.1 Analysis of functions I: Increase, decrease and concavity Definition Let f be defined on an interval and let x 1 and x 2 denote points in that interval. a) f is said to be increasing on the interval
More information1 Review of di erential calculus
Review of di erential calculus This chapter presents the main elements of di erential calculus needed in probability theory. Often, students taking a course on probability theory have problems with concepts
More informationChapter 6. Techniques of Integration. 6.1 Differential notation
Chapter 6 Techniques of Integration In this chapter, we expand our repertoire for antiderivatives beyond the elementary functions discussed so far. A review of the table of elementary antiderivatives (found
More informationLecture Notes in Mathematics. Arkansas Tech University Department of Mathematics
Lecture Notes in Mathematics Arkansas Tech University Department of Mathematics Introductory Notes in Ordinary Differential Equations for Physical Sciences and Engineering Marcel B. Finan c All Rights
More information