Solutions to Homework 10


 Marilynn Sharp
 5 years ago
 Views:
Transcription
1 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x [1, 3]. Since f is continuous over R, the function g(y) = f(x, y) is continuous for each x [1, 3]. (Can you prove this?) In particular, g(y) is integrable over [, 4] for each x [1, 3]. Let F (x) = 4 f(x, y) dy. Thus, F (x) = 4 y x dy = 1 x 4 y dy = 6 x. Since F (x) = 6/x is continuous on [1, 3], it is integrable on [1, 3]. Therefore, all of the hypotheses of Fubini s theorem are satisfied. Applying Fubini s theorem, one has that R f dv = f(x, y) dy dx = dx = 6 ln 3. x The above solution gives much more detail than one would normally give. Moreover, one would like an easily applicable test for whether Fubini s theorem can be applied. One such test is given in exercise # of Section 7.: if f(x, y) is continuous on R = [a, b] [c, d] and if f x is continuous on R, then F (x) = d c f(x, y) dy is not only continuous on [a, b], it is differentiable on [a, b] and F (x) = d f c x dy. Another such test is given in Corollary.. The corollary implies that if f is continuous and if F (x) = d c f(x, y) dy and G(y) = b a f(x, y) dx are continuous (on [a, b] and [c, d], respectively), then f is integrable on R and the value of R f dv is equal to the value of each of the iterated integrals. For the function in (b), it is easy to apply either test. For instance, f(x, y) = y/x is continuous on [1, 3] [, 4]. (It is built out of continuous
2 functions using arithmetic operations.) And it is equally easy to see that f x = y/x is continuous on [1, 3] [, 4]. (d) Evaluate R (x + y)z dv where R = [ 1, 1] [1, ] [, 3]. Fubini s theorem applies to the above function since f is continuous and it will be clear from the computation below that the functions which result from computing the iterated integrals are also continuous ( 5 x 15 4 ) dx = 5 (x + y)z dz = (x + y) z 3 = 5 (x + y) 5 (x + y) dy = 5 y (xy + ) 1 = 5 x x dx dx = = 15 4 Section 7., exercise # (b,c,d,e): Interpret the iterated integral as an (double) integral Ω f da for the appropriate region Ω. Sketch Ω and change the order of integration. You may assume that f is continuous. Note: The term double integral is synonymous with iterated integral consisting of two integrals. So, I would not use the phrase double integral as the author has done. I would simply call Ω f da the integral of f over Ω. (b): The result of changing the order of integration is printed in the textbook. (c): 4 1 y f(x, y) dx dy Solution: The region Ω R is the region given by the following inequalities: y x 4 and 1 y. The first inequality describes the closed, bounded region in the plane bounded by the parabola x = y and the line x = 4. The second inequality
3 describes the closed, unbounded region which lies between the lines y = 1 and y =. The set of points which satisfy both inequalities is the intersection of these two regions. Thus, Ω is a closed, bounded region having three sides. It is bounded below by the line y = 1, above (or to the left) by the parabola x = y and on the right by the line x = 4. Since the integrand is assumed to be continuous, Fubini s Theorem applies. (This is not entirely clear. How do we know, for instance, that G(y) = 4 y f(x, y) dx is continuous on [1, ]? We will assume that this is the case for purposes of this problem and the other integrals in this exercise.) Therefore, Ω f da is equal to the value of the iterated integral above. Moreover, it is also equal to the value of the other iterated integral obtained by changing the order of integration. To change the order of integration, we need to describe the region Ω by two inequalities. These inequalities need to take the following form: b(x) y t(x) and l x r, where the names of these functions and constants are suggestive of their geometric significance: the curve y = b(x) is the curve which describes the bottom of the region, y = t(x) describes the top of the region, and l and r describes vertical lines which bound the region on the left and on the right. The following inequalities describe Ω: 1 y x and x 4. This is geometrically clear if one sketches the regions described by each pair of inequalities separately and then considers the intersection of these two regions. One process for deducing the above description is the following. Choose a generic point p in the region Ω. (Generic means that you should not choose a point which lies on the boundary of Ω.) Now let the ycoordinate of the point p vary. By looking at the sketch of Ω it is clear that a generic point p must lie between the line y = 1 and the curve y = x (equivalently, the curve x = y ) as its ycoordinate varies. Thus, 1 y x is a condition on the ycoordinate of points which belong to Ω. Next consider the projection of the region Ω onto the xaxis. This is the interval [, 4]. And
4 so, x 4 is a condition on the xcoordinates of points in Ω. Thus, every point p in Ω must satisfy these two inequalities. Sketching the region described by these two inequalities results in the same region Ω. A second process for deducing the above description is to (carefully) solve the inequalities. The goal of such a solution is a pair of inequalities of the form b(x) y t(x) and l x r. Care needs to be taken when applying operations to inequalities. For example, multiplying by a negative number reverses the inequality. Applying an increasing function to both sides of an inequality preserves the inequality. Applying a decreasing function to both sides reverses an inequality. It the present case, the original inequalities 1 y and y x 4 imply that (by applying the increasing function of taking the square root) y x and that (by rearranging the original inequalities) 1 y x and that x 4. Finally, since 1 y, 1 y x and so both 1 x 4 and 1 y x hold true. (d) 1 1 y f(x, y) dx dy The region Ω described by the inequalities 1 y 1 and x 1 y is right half of the closed unit disk, i.e. the region inside the disk x + y 1 and inside the halfspace x. This can be deduced geometrically or by solving the system of inequalities: squaring implies that x 1 y, and the condition 1 y 1 places no further restriction since x + y 1 implies that y 1 and so 1 y 1. Choosing a generic point p, one sees that as the ycoordinate varies, the point is bounded below by the semicircle y = 1 x and above by the semicircle y = 1 x. Projecting the region Ω onto the xaxis results in the interval [, 1]. Therefore, x 1. A quick sketch verifies that Ω is indeed described by the inequalities 1 x y 1 x and x 1. These inequalities can also be obtained algebraically. Either way, the resulting integral with the order of integration reversed is the following: 1 x 1 x f(x, y) dy dx
5 and above by the semicircle y = 1 x, and so y 1 x. Projecting onto (e) x x f(x, y) dy dx Solution: Algebraically manipulating the inequalities, one sees that x y x and x 1 implies that x y and y x 1 and x y x 1. Therefore, y 1 and y x y. (I do not recommend this method. The preferred method is to sketch the region and let a generic point vary its position.) Answer: y y f(x, y) dx dy. (f) The solution appears in the textbook. Section 7., exercise # 3 (a,b): Solution: (a) Change the order of integration and evaluate the resulting integral: x (x + y) dy dx The region y x, x 1 is a triangular region having vertices (, ), (1, 1), and (, 1). Choosing a point in this region an varying its x coordinate shows that the minimum value of the x coordinate is determined by the line y = x and that the maximum value of x is equal to x = 1. So, y x 1, y 1. x (x + y) dx dy The above integral is straightforward to compute. Answer: 1/. The original integral is easier to evaluate; so, in practice, there would be no reason to change the order of integration.
6 (b) Change the order of integration and evaluate the resulting integral: 1 y 1 y y dx dy The integral describes a double integral over the region enclosed by the semicircle y = 1 x and the line y =. 1 x 1 y dy dx The inner integral evaluates to (1/)( 1 x ) = (1/)(1 x ). Answer: /3. The original integral is not difficult to compute; they are perhaps of equal difficulty. This integral is also easy to compute using polar coordinates. Section 7., exercise # 4: Does the integral f(x, y) dy dx exist, where f(x, y) = 1 if y Q and f(x, y) = x if y / Q? Solution: No, this integral does not exist. The meaning of the above is that the expression is an iterated integral. But the first integral is not defined: f(x, y) dy. Suppose that x [, 1] is chosen and held constant; to emphasize this, let c = x. The function f(c, y) = g(y) = 1 if y Q and f(c, y) = g(y) = c. But the function g(y) is no integrable unless c = 1/. This is because the upper sum of g(y) with respect to any partition of [, 1] is always equal to the larger of 1 and c (since every subinterval must contain both rational and irrational points) and the lower sum of g(y) with respect to any partition is always equal to the smaller of 1 and c for the same reason. Therefore, the upper and lower sums approach the same value as the partition is refined if and only if c = 1. Thus, the function F (x) = f(x, y) dy is undefined at every value of x except when x = 1/. Section 7., exercise # 8 (b,c): Evaluate the integrals.
7 Solution: (b) 3 x e y4 dy dx Since there is no apparent antiderivative with respect to y, change the order of integration. The integral defines a double integral over a region bounded below by y = 3 x, above by y = 1, on the left by x =, and the upper and lower curves meet in a single point, namely (1, 1), on the rightside. Thus, y 3 e y4 dx dy The inner integral evaluates to y 3 e y4. And this can be integrated with respect to y by using the substitution u = y 4, du = 4y 3 dy. Answer: (e 1)/4. (c) y e y/x dx dy The integral above describes a double integral over the region bounded below by the line y =, above by the parabola y = x, on the right by the line x = 1, and the upper and lower curves meet in a single point, namely (, ) on the left. The integrand e y/x is undefined at (, ), but is otherwise continuous on this region. Since a single point has dimensional volume zero, the integrand is integrable. There is no obvious way to evaluate the innermost integral, so change the order of integration: x e y/x dy dx. The inner integral evaluates to xe y/x y = xex x. Using integration by parts xe x dx = xe x e x + C. And so, (xe x x) dx = (xe x e x (1/)x 1 ) = 1
8 Section 7., exercise # 1 (b,c,e): Sketch the region of integration and change the order of integration so that the innermost integral is with respect to y. Assume that the integrand is continuous. Solution: (b) x y f dz dy dx The region lies in the octant x, y, z. It is bounded above by the plane z = y and below by the xyplane. One side of the region is a triangle in the yzplane and the other side is part of the parabolic cylinder y = 1 x. Choosing a point in this region and varying its ycoordinate reveals that the minimum ycoordinate is determined by the plane z = y and that the maximum ycoordinate is determined by the cylinder y = 1 x. This can also be seen from the inequalities: z y and y (1 x ) imply that z y (1 x ). Next project the region onto the xzplane. This region determines the bounds of the outer two integrals. Since the intersection of the plane z = y and the cylinder y = 1 x is the arc whose points have the form (1 t, t, t), where t 1, this arc projects to the parabola in the xzplane given by the equation z = 1 x. Thus, z 1 x. So, the new integral is Another answer is x x z 1 z x z f dy dz dx. f dy dx dz. (c) 1 x 1 1 x x +y f dz dy dx The first inequality, x + y z 1, say, geometrically, that the region is bounded below by the cone z = x + y and above by the plane z = 1. The second two inequalities say, geometrically, that the projection of the region onto the xyplane is the unit disk centered at the origin. So, the region is the region bounded above by the plane z = 1, below by the plane
9 z = 1. (The projection of this region is the unit disk, i.e. the second two inequalities impose no additional restrictions.) If a point in this region is chosen and its ycoordinate is varied, then the minimum ycoordinate is determined by the cone and, likewise, the maximum ycoordinate is determined by the cone. Solve z = x + y for y: y = z x, and so z x y z x. The projection of the region onto the xzplane is a triangle bounded by the lines z = 1, x = z, and x = z. So, this needs to be divided into two regions if you wish to integrate next with respect to z. Answer: x z x z x f dy dz dx + x z x z x f dy dz dx If instead we integrate with respect to x before z, then it is much simpler: z z x z z x f dy dx dz (e) x x+y f dz dy dx The region of integration is a pyramid with vertex at the origin and with a rectangular base whose vertices are (1,, ), (1,, 1), (, 1, ), and (, 1, 1). To see this, sketch the plane z = x + y. For example, the points (,, ), (1,, 1), and (, 1, 1) lie on this plane. The region lies between the xyplane (since z) and the aforementioned plane. The outer two integrals describe the region which results by projecting onto the xyplane. This projection is a triangle bounded by the line y = 1 x and by the line x = and y =. If a point is chosen in the solid region above and its ycoordinate is varied, then the largest yvalue is determined by the plane y = 1 x; but the smallest yvalue is determined either by the plane z = x + y or by the plane y =. Two integrals are required. If z x, then z x y 1 x. If z x, then y 1 x. The projection onto the xzplane is a triangle in both cases. In the first case, it is a triangle bounded by the lines x = z, x =, and z = 1. In the second case it is a triangle bounded by the lines
10 x = z, x = 1, and z =. Another solution is x x z x z x z x f dy dz dx + f dy dx dz + x x x z f dy dz dx f dy dx dz Section 7.3, exercise # 3: Determine the area of the cardioid r = 1 + cos θ. Solution: Use da = r dr dθ. From the figure (refer to the textbook) it is clear that if a point is chosen inside the cardioid and if its rcoordinate is varied, then r 1 + cos θ. There are points in the region for all values of θ, and so θ π. π π +cos θ (1 + cos θ) dθ = π r dr dθ = (1 + cos θ + cos θ) dθ, where in the second integral I have doubled the integral and integrated over [, π] rather than [, π]. To integrate cos θ use the halfangle formula cos θ = (1/)(1 + cos θ) or use integration by parts (which gives a rather nice and easy to remember formula: cos θ dθ = 1 (x + cos x sin x) + C. Let s use the halfangle formula: π (1 + cos θ cos θ) dθ = π + π = 3π. There is no need to integrate the terms involving cosine: these values are clearly equal to zero over [, π] (think about the graph of y = cos θ). Section 7.3, exercise # 1: Determine the volume of the region bounded above by z = y and below by z = x + y.
11 Solution: The region is bounded above by a plane and below by a paraboloid. The intersection of the two surfaces is given by y = x + y which can be rewritten as x + (y 1) = 1, a circle. For this problem, use polar coordinates. The projection of the region onto the xyplane is the circle centered at (, 1) with radius 1. In polar coordinates, this circle has equation r = sin θ, where θ π. The plane z = y has equation z = r sin θ in cylindrical coordinates. The paraboloid has equation z = r. The volume of the region is computed by the following integral in cylindrical coordinates: π sin θ r sin θ r dz dr dθ. r The inner integral evaluates to r sin θ r 3. The next integral evaluates to (4/3) sin 4 θ. To integrate sin 4 θ dθ, use the double angle formula twice. Answer: 3π/4. Section 7.3, exercise # 13: Determine the volume of the region inside both x + y = 1 and x + y + z =. Solution: The figure is the region inside a cylinder of radius 1 and inside a sphere of radius. Cylindrical coordinates are most appropriate. Using symmetry, it suffices to double the volume of the region which lies above the xyplane. This region is bounded above by the sphere so z x y = r. The projection of the region onto the xyplane is the unit disk centered at the origin. Therefore the volume of the original region is compute by π r r dz dr dθ. The inner integral is equal to r r. The next integral is equal to ( 1)/3 (use u = r and du = r dr). So, the final answer is (4π/3)( 1). Section 7.3, exercise # 19: Find the volume of the region lying above the plane z = a and inside the sphere x + y + z = 4a by integrating in both cylindrical and spherical coordinates. Solution: The intersection of the two surfaces is the circle x + y = 3a lying in the plane z = a. If a =, then this is just a point. To simplify
12 matters, let s assume that a > since we can determine the values when a by using geometric arguments and the solution when a >. In cylindrical coordinates, the zcoordinate of a point in the region is bounded below by the plane z = a and above by the sphere z = 4a x y = 4a r. The projection of the region onto the xyplane is the disk r = x + y 3a. Therefore, if a >, the following integral computes the volume: π a 3 4a r a r dz dr dθ. In spherical coordinates, the plane z = a is expressed as ρ cos φ = a and the sphere is simply ρ = a. Thus, a sec φ a. (This assumes a > ). The value of φ ranges from to the angle determined by the triangle having height z = a and base r = a 3. Thus, φ tan 1 3 = π/3. Therefore, if a >, the following integral computes the volume of the region: Answer: 5a 3 π/3. π π/3 a a sec φ ρ sin φ dρ dφ dθ.
Solutions for Review Problems
olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector
More informationSolutions  Homework sections 17.717.9
olutions  Homework sections 7.77.9 7.7 6. valuate xy d, where is the triangle with vertices (,, ), (,, ), and (,, ). The three points  and therefore the triangle between them  are on the plane x +
More informationMULTIPLE INTEGRALS. h 2 (y) are continuous functions on [c, d] and let f(x, y) be a function defined on R. Then
MULTIPLE INTEGALS 1. ouble Integrals Let be a simple region defined by a x b and g 1 (x) y g 2 (x), where g 1 (x) and g 2 (x) are continuous functions on [a, b] and let f(x, y) be a function defined on.
More informationTriple Integrals in Cylindrical or Spherical Coordinates
Triple Integrals in Clindrical or Spherical Coordinates. Find the volume of the solid ball 2 + 2 + 2. Solution. Let be the ball. We know b #a of the worksheet Triple Integrals that the volume of is given
More informationDouble Integrals in Polar Coordinates
Double Integrals in Polar Coordinates. A flat plate is in the shape of the region in the first quadrant ling between the circles + and +. The densit of the plate at point, is + kilograms per square meter
More informationPROBLEM SET. Practice Problems for Exam #1. Math 1352, Fall 2004. Oct. 1, 2004 ANSWERS
PROBLEM SET Practice Problems for Exam # Math 352, Fall 24 Oct., 24 ANSWERS i Problem. vlet R be the region bounded by the curves x = y 2 and y = x. A. Find the volume of the solid generated by revolving
More informationFundamental Theorems of Vector Calculus
Fundamental Theorems of Vector Calculus We have studied the techniques for evaluating integrals over curves and surfaces. In the case of integrating over an interval on the real line, we were able to use
More informationx 2 + y 2 = 1 y 1 = x 2 + 2x y = x 2 + 2x + 1
Implicit Functions Defining Implicit Functions Up until now in this course, we have only talked about functions, which assign to every real number x in their domain exactly one real number f(x). The graphs
More informationThe Math Circle, Spring 2004
The Math Circle, Spring 2004 (Talks by Gordon Ritter) What is NonEuclidean Geometry? Most geometries on the plane R 2 are noneuclidean. Let s denote arc length. Then Euclidean geometry arises from the
More informationSolutions to Practice Problems for Test 4
olutions to Practice Problems for Test 4 1. Let be the line segmentfrom the point (, 1, 1) to the point (,, 3). Evaluate the line integral y ds. Answer: First, we parametrize the line segment from (, 1,
More informationAngles and Quadrants. Angle Relationships and Degree Measurement. Chapter 7: Trigonometry
Chapter 7: Trigonometry Trigonometry is the study of angles and how they can be used as a means of indirect measurement, that is, the measurement of a distance where it is not practical or even possible
More informationMath 2443, Section 16.3
Math 44, Section 6. Review These notes will supplement not replace) the lectures based on Section 6. Section 6. i) ouble integrals over general regions: We defined double integrals over rectangles in the
More informationSolutions to old Exam 1 problems
Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections
More informationSAT Subject Math Level 2 Facts & Formulas
Numbers, Sequences, Factors Integers:..., 3, 2, 1, 0, 1, 2, 3,... Reals: integers plus fractions, decimals, and irrationals ( 2, 3, π, etc.) Order Of Operations: Arithmetic Sequences: PEMDAS (Parentheses
More informationEstimating the Average Value of a Function
Estimating the Average Value of a Function Problem: Determine the average value of the function f(x) over the interval [a, b]. Strategy: Choose sample points a = x 0 < x 1 < x 2 < < x n 1 < x n = b and
More information4 More Applications of Definite Integrals: Volumes, arclength and other matters
4 More Applications of Definite Integrals: Volumes, arclength and other matters Volumes of surfaces of revolution 4. Find the volume of a cone whose height h is equal to its base radius r, by using the
More informationPYTHAGOREAN TRIPLES KEITH CONRAD
PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient
More informationMath 1B, lecture 5: area and volume
Math B, lecture 5: area and volume Nathan Pflueger 6 September 2 Introduction This lecture and the next will be concerned with the computation of areas of regions in the plane, and volumes of regions in
More informationwww.mathsbox.org.uk ab = c a If the coefficients a,b and c are real then either α and β are real or α and β are complex conjugates
Further Pure Summary Notes. Roots of Quadratic Equations For a quadratic equation ax + bx + c = 0 with roots α and β Sum of the roots Product of roots a + b = b a ab = c a If the coefficients a,b and c
More informationPractice Final Math 122 Spring 12 Instructor: Jeff Lang
Practice Final Math Spring Instructor: Jeff Lang. Find the limit of the sequence a n = ln (n 5) ln (3n + 8). A) ln ( ) 3 B) ln C) ln ( ) 3 D) does not exist. Find the limit of the sequence a n = (ln n)6
More information11.1. Objectives. Component Form of a Vector. Component Form of a Vector. Component Form of a Vector. Vectors and the Geometry of Space
11 Vectors and the Geometry of Space 11.1 Vectors in the Plane Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. 2 Objectives! Write the component form of
More informationSouth Carolina College and CareerReady (SCCCR) PreCalculus
South Carolina College and CareerReady (SCCCR) PreCalculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know
More informationAlgebra 2 Chapter 1 Vocabulary. identity  A statement that equates two equivalent expressions.
Chapter 1 Vocabulary identity  A statement that equates two equivalent expressions. verbal model A word equation that represents a reallife problem. algebraic expression  An expression with variables.
More informationBiggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress
Biggar High School Mathematics Department National 5 Learning Intentions & Success Criteria: Assessing My Progress Expressions & Formulae Topic Learning Intention Success Criteria I understand this Approximation
More informationStudent Performance Q&A:
Student Performance Q&A: 2008 AP Calculus AB and Calculus BC FreeResponse Questions The following comments on the 2008 freeresponse questions for AP Calculus AB and Calculus BC were written by the Chief
More informationReview Sheet for Test 1
Review Sheet for Test 1 Math 26100 2 6 2004 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And
More informationUnderstanding Basic Calculus
Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other
More information1.2 GRAPHS OF EQUATIONS. Copyright Cengage Learning. All rights reserved.
1.2 GRAPHS OF EQUATIONS Copyright Cengage Learning. All rights reserved. What You Should Learn Sketch graphs of equations. Find x and yintercepts of graphs of equations. Use symmetry to sketch graphs
More informationSection 11.1: Vectors in the Plane. Suggested Problems: 1, 5, 9, 17, 23, 2537, 40, 42, 44, 45, 47, 50
Section 11.1: Vectors in the Plane Page 779 Suggested Problems: 1, 5, 9, 17, 3, 537, 40, 4, 44, 45, 47, 50 Determine whether the following vectors a and b are perpendicular. 5) a = 6, 0, b = 0, 7 Recall
More informationMATH 132: CALCULUS II SYLLABUS
MATH 32: CALCULUS II SYLLABUS Prerequisites: Successful completion of Math 3 (or its equivalent elsewhere). Math 27 is normally not a sufficient prerequisite for Math 32. Required Text: Calculus: Early
More informationSolutions to Homework 5
Solutions to Homework 5 1. Let z = f(x, y) be a twice continously differentiable function of x and y. Let x = r cos θ and y = r sin θ be the equations which transform polar coordinates into rectangular
More informationIf Σ is an oriented surface bounded by a curve C, then the orientation of Σ induces an orientation for C, based on the RightHandRule.
Oriented Surfaces and Flux Integrals Let be a surface that has a tangent plane at each of its nonboundary points. At such a point on the surface two unit normal vectors exist, and they have opposite directions.
More informationVector surface area Differentials in an OCS
Calculus and Coordinate systems EE 311  Lecture 17 1. Calculus and coordinate systems 2. Cartesian system 3. Cylindrical system 4. Spherical system In electromagnetics, we will often need to perform integrals
More information4. How many integers between 2004 and 4002 are perfect squares?
5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started
More informationAlgebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123
Algebra Eponents Simplify each of the following as much as possible. 1 4 9 4 y + y y. 1 5. 1 5 4. y + y 4 5 6 5. + 1 4 9 10 1 7 9 0 Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from
More informationThis makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5
1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy,
More informationMA107 Precalculus Algebra Exam 2 Review Solutions
MA107 Precalculus Algebra Exam 2 Review Solutions February 24, 2008 1. The following demand equation models the number of units sold, x, of a product as a function of price, p. x = 4p + 200 a. Please write
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationSAT Subject Test Practice Test II: Math Level II Time 60 minutes, 50 Questions
SAT Subject Test Practice Test II: Math Level II Time 60 minutes, 50 Questions All questions in the Math Level 1 and Math Level Tests are multiplechoice questions in which you are asked to choose the
More informationTwo vectors are equal if they have the same length and direction. They do not
Vectors define vectors Some physical quantities, such as temperature, length, and mass, can be specified by a single number called a scalar. Other physical quantities, such as force and velocity, must
More informationAP CALCULUS AB 2008 SCORING GUIDELINES
AP CALCULUS AB 2008 SCORING GUIDELINES Question 1 Let R be the region bounded by the graphs of y = sin( π x) and y = x 4 x, as shown in the figure above. (a) Find the area of R. (b) The horizontal line
More informationLecture 8 : Coordinate Geometry. The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 20
Lecture 8 : Coordinate Geometry The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 0 distance on the axis and give each point an identity on the corresponding
More informationSAT Subject Math Level 1 Facts & Formulas
Numbers, Sequences, Factors Integers:..., 3, 2, 1, 0, 1, 2, 3,... Reals: integers plus fractions, decimals, and irrationals ( 2, 3, π, etc.) Order Of Operations: Aritmetic Sequences: PEMDAS (Parenteses
More informationAB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss
AB2.5: urfaces and urface Integrals. Divergence heorem of Gauss epresentations of surfaces or epresentation of a surface as projections on the xy and xzplanes, etc. are For example, z = f(x, y), x =
More informationGRAPHING IN POLAR COORDINATES SYMMETRY
GRAPHING IN POLAR COORDINATES SYMMETRY Recall from Algebra and Calculus I that the concept of symmetry was discussed using Cartesian equations. Also remember that there are three types of symmetry  yaxis,
More information10 Polar Coordinates, Parametric Equations
Polar Coordinates, Parametric Equations ½¼º½ ÈÓÐ Ö ÓÓÖ Ò Ø Coordinate systems are tools that let us use algebraic methods to understand geometry While the rectangular (also called Cartesian) coordinates
More informationDerive 5: The Easiest... Just Got Better!
Liverpool John Moores University, 115 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de Technologie Supérieure, Canada Email; mbeaudin@seg.etsmtl.ca 1. Introduction Engineering
More informationMODERN APPLICATIONS OF PYTHAGORAS S THEOREM
UNIT SIX MODERN APPLICATIONS OF PYTHAGORAS S THEOREM Coordinate Systems 124 Distance Formula 127 Midpoint Formula 131 SUMMARY 134 Exercises 135 UNIT SIX: 124 COORDINATE GEOMETRY Geometry, as presented
More information6. Define log(z) so that π < I log(z) π. Discuss the identities e log(z) = z and log(e w ) = w.
hapter omplex integration. omplex number quiz. Simplify 3+4i. 2. Simplify 3+4i. 3. Find the cube roots of. 4. Here are some identities for complex conjugate. Which ones need correction? z + w = z + w,
More informationChapter 17. Review. 1. Vector Fields (Section 17.1)
hapter 17 Review 1. Vector Fields (Section 17.1) There isn t much I can say in this section. Most of the material has to do with sketching vector fields. Please provide some explanation to support your
More informationAdditional Topics in Math
Chapter Additional Topics in Math In addition to the questions in Heart of Algebra, Problem Solving and Data Analysis, and Passport to Advanced Math, the SAT Math Test includes several questions that are
More informationWeek 13 Trigonometric Form of Complex Numbers
Week Trigonometric Form of Complex Numbers Overview In this week of the course, which is the last week if you are not going to take calculus, we will look at how Trigonometry can sometimes help in working
More informationG. GRAPHING FUNCTIONS
G. GRAPHING FUNCTIONS To get a quick insight int o how the graph of a function looks, it is very helpful to know how certain simple operations on the graph are related to the way the function epression
More information2 Integrating Both Sides
2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation
More informationMark Howell Gonzaga High School, Washington, D.C.
Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School,
More informationPRACTICE FINAL. Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 10cm.
PRACTICE FINAL Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 1cm. Solution. Let x be the distance between the center of the circle
More information3 Contour integrals and Cauchy s Theorem
3 ontour integrals and auchy s Theorem 3. Line integrals of complex functions Our goal here will be to discuss integration of complex functions = u + iv, with particular regard to analytic functions. Of
More informationName Class. Date Section. Test Form A Chapter 11. Chapter 11 Test Bank 155
Chapter Test Bank 55 Test Form A Chapter Name Class Date Section. Find a unit vector in the direction of v if v is the vector from P,, 3 to Q,, 0. (a) 3i 3j 3k (b) i j k 3 i 3 j 3 k 3 i 3 j 3 k. Calculate
More informationTechniques of Integration
CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an artform than a collection of algorithms. Many problems in applied mathematics involve the integration
More informationAnswer Key for the Review Packet for Exam #3
Answer Key for the Review Packet for Eam # Professor Danielle Benedetto Math MaMin Problems. Show that of all rectangles with a given area, the one with the smallest perimeter is a square. Diagram: y
More informationCHAPTER 8, GEOMETRY. 4. A circular cylinder has a circumference of 33 in. Use 22 as the approximate value of π and find the radius of this cylinder.
TEST A CHAPTER 8, GEOMETRY 1. A rectangular plot of ground is to be enclosed with 180 yd of fencing. If the plot is twice as long as it is wide, what are its dimensions? 2. A 4 cm by 6 cm rectangle has
More informationMetric Spaces. Chapter 7. 7.1. Metrics
Chapter 7 Metric Spaces A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y X. The purpose of this chapter is to introduce metric spaces and give some
More informationVector Calculus Solutions to Sample Final Examination #1
Vector alculus s to Sample Final Examination #1 1. Let f(x, y) e xy sin(x + y). (a) In what direction, starting at (,π/), is f changing the fastest? (b) In what directions starting at (,π/) is f changing
More informationTrigonometric Functions: The Unit Circle
Trigonometric Functions: The Unit Circle This chapter deals with the subject of trigonometry, which likely had its origins in the study of distances and angles by the ancient Greeks. The word trigonometry
More informationMATH 425, PRACTICE FINAL EXAM SOLUTIONS.
MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator
More informationGeorgia Department of Education Kathy Cox, State Superintendent of Schools 7/19/2005 All Rights Reserved 1
Accelerated Mathematics 3 This is a course in precalculus and statistics, designed to prepare students to take AB or BC Advanced Placement Calculus. It includes rational, circular trigonometric, and inverse
More information13.4 THE CROSS PRODUCT
710 Chapter Thirteen A FUNDAMENTAL TOOL: VECTORS 62. Use the following steps and the results of Problems 59 60 to show (without trigonometry) that the geometric and algebraic definitions of the dot product
More informationMathematics PreTest Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11}
Mathematics PreTest Sample Questions 1. Which of the following sets is closed under division? I. {½, 1,, 4} II. {1, 1} III. {1, 0, 1} A. I only B. II only C. III only D. I and II. Which of the following
More informationTrigonometric Functions and Triangles
Trigonometric Functions and Triangles Dr. Philippe B. Laval Kennesaw STate University August 27, 2010 Abstract This handout defines the trigonometric function of angles and discusses the relationship between
More information5.3 Improper Integrals Involving Rational and Exponential Functions
Section 5.3 Improper Integrals Involving Rational and Exponential Functions 99.. 3. 4. dθ +a cos θ =, < a
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form
ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form Goal Graph quadratic functions. VOCABULARY Quadratic function A function that can be written in the standard form y = ax 2 + bx+ c where a 0 Parabola
More informationMATH 275: Calculus III. Lecture Notes by Angel V. Kumchev
MATH 275: Calculus III Lecture Notes by Angel V. Kumchev Contents Preface.............................................. iii Lecture 1. ThreeDimensional Coordinate Systems..................... 1 Lecture
More informationof surface, 569571, 576577, 578581 of triangle, 548 Associative Property of addition, 12, 331 of multiplication, 18, 433
Absolute Value and arithmetic, 730733 defined, 730 Acute angle, 477 Acute triangle, 497 Addend, 12 Addition associative property of, (see Commutative Property) carrying in, 11, 92 commutative property
More informationAdding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors
1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number
More informationCOMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i 1 4 1 7 i 5 6i
COMPLEX NUMBERS _4+i _i FIGURE Complex numbers as points in the Arg plane i _i +i i A complex number can be represented by an expression of the form a bi, where a b are real numbers i is a symbol with
More informationHomework 2 Solutions
Homework Solutions 1. (a) Find the area of a regular heagon inscribed in a circle of radius 1. Then, find the area of a regular heagon circumscribed about a circle of radius 1. Use these calculations to
More informationALGEBRA 2/TRIGONOMETRY
ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Tuesday, January 8, 014 1:15 to 4:15 p.m., only Student Name: School Name: The possession
More informationAlgebra I Vocabulary Cards
Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression
More informationSection 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section 4.4 Using the Fundamental Theorem As we saw in Section 4.3, using the Fundamental Theorem of Integral Calculus reduces the problem of evaluating a
More informationFriday, January 29, 2016 9:15 a.m. to 12:15 p.m., only
ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Friday, January 9, 016 9:15 a.m. to 1:15 p.m., only Student Name: School Name: The possession
More informationCopyrighted Material. Chapter 1 DEGREE OF A CURVE
Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two
More informationEvaluating trigonometric functions
MATH 1110 0090906 Evaluating trigonometric functions Remark. Throughout this document, remember the angle measurement convention, which states that if the measurement of an angle appears without units,
More informationGeometry of Vectors. 1 Cartesian Coordinates. Carlo Tomasi
Geometry of Vectors Carlo Tomasi This note explores the geometric meaning of norm, inner product, orthogonality, and projection for vectors. For vectors in threedimensional space, we also examine the
More informationHigher Education Math Placement
Higher Education Math Placement Placement Assessment Problem Types 1. Whole Numbers, Fractions, and Decimals 1.1 Operations with Whole Numbers Addition with carry Subtraction with borrowing Multiplication
More informationReview B: Coordinate Systems
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of hysics 8.02 Review B: Coordinate Systems B.1 Cartesian Coordinates... B2 B.1.1 Infinitesimal Line Element... B4 B.1.2 Infinitesimal Area Element...
More informationSolutions to Exercises, Section 5.1
Instructor s Solutions Manual, Section 5.1 Exercise 1 Solutions to Exercises, Section 5.1 1. Find all numbers t such that ( 1 3,t) is a point on the unit circle. For ( 1 3,t)to be a point on the unit circle
More information1. Introduction sine, cosine, tangent, cotangent, secant, and cosecant periodic
1. Introduction There are six trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant; abbreviated as sin, cos, tan, cot, sec, and csc respectively. These are functions of a single
More informationExam 1 Sample Question SOLUTIONS. y = 2x
Exam Sample Question SOLUTIONS. Eliminate the parameter to find a Cartesian equation for the curve: x e t, y e t. SOLUTION: You might look at the coordinates and notice that If you don t see it, we can
More informationTRIGONOMETRY Compound & Double angle formulae
TRIGONOMETRY Compound & Double angle formulae In order to master this section you must first learn the formulae, even though they will be given to you on the matric formula sheet. We call these formulae
More information2.1 Three Dimensional Curves and Surfaces
. Three Dimensional Curves and Surfaces.. Parametric Equation of a Line An line in two or threedimensional space can be uniquel specified b a point on the line and a vector parallel to the line. The
More informationThe Fourth International DERIVETI92/89 Conference Liverpool, U.K., 1215 July 2000. Derive 5: The Easiest... Just Got Better!
The Fourth International DERIVETI9/89 Conference Liverpool, U.K., 5 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de technologie supérieure 00, rue NotreDame Ouest Montréal
More information2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2
00 Problem If a and b are nonzero real numbers such that a b, compute the value of the expression ( ) ( b a + a a + b b b a + b a ) ( + ) a b b a + b a +. b a a b Answer: 8. Solution: Let s simplify the
More informationDetermine whether the following lines intersect, are parallel, or skew. L 1 : x = 6t y = 1 + 9t z = 3t. x = 1 + 2s y = 4 3s z = s
Homework Solutions 5/20 10.5.17 Determine whether the following lines intersect, are parallel, or skew. L 1 : L 2 : x = 6t y = 1 + 9t z = 3t x = 1 + 2s y = 4 3s z = s A vector parallel to L 1 is 6, 9,
More informationFactoring Polynomials
UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can
More informationUnified Lecture # 4 Vectors
Fall 2005 Unified Lecture # 4 Vectors These notes were written by J. Peraire as a review of vectors for Dynamics 16.07. They have been adapted for Unified Engineering by R. Radovitzky. References [1] Feynmann,
More informationSection 63 DoubleAngle and HalfAngle Identities
63 DoubleAngle and HalfAngle Identities 47 Section 63 DoubleAngle and HalfAngle Identities DoubleAngle Identities HalfAngle Identities This section develops another important set of identities
More informationCalculus AB 2014 Scoring Guidelines
P Calculus B 014 Scoring Guidelines 014 The College Board. College Board, dvanced Placement Program, P, P Central, and the acorn logo are registered trademarks of the College Board. P Central is the official
More information10.1. Solving Quadratic Equations. Investigation: Rocket Science CONDENSED
CONDENSED L E S S O N 10.1 Solving Quadratic Equations In this lesson you will look at quadratic functions that model projectile motion use tables and graphs to approimate solutions to quadratic equations
More information2 Session Two  Complex Numbers and Vectors
PH2011 Physics 2A Maths Revision  Session 2: Complex Numbers and Vectors 1 2 Session Two  Complex Numbers and Vectors 2.1 What is a Complex Number? The material on complex numbers should be familiar
More information