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# Multivariable Calculus & Matrix Algebra (Herron): Exam 2 Study Guide (S16) MATH 2010

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This 6 page Study Guide was uploaded by creask on Sunday March 6, 2016. The Study Guide belongs to MATH 2010 at Rensselaer Polytechnic Institute taught by Isom Herron in Spring 2016. Since its upload, it has received 148 views. For similar materials see Multivariable calculus and matrix algebra in Mathematics (M) at Rensselaer Polytechnic Institute.

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Date Created: 03/06/16

MATH 2010 - Multivariable Calculus & Matrix Algebra Professor Herron - Rensselaer Polytechnic Institute Exam 2 Study Guide Important: These notes are in no way intended to replace attendance in lecture. For best results in this course, it is imperative that you attend lecture and take your own detailed notes. Please keep in mind that these notes are written speci▯cally with Professor Herron’ s sections in mind, and no one else’ s. These notes cover every topic in a smaller amount of detail than in the weekly lecture notes I have uploaded. If you truly don’t understand one of these topics, I recommend checking out one of my weekly notes. That said, this will give you a basic crash course in all of the topics we have covered since the ▯rst exam. I recommend doing practice problems from the book in addition to the assignments. These notes do NOT cover Green’s Theorem. Email me if you’d like free notes on Green’s Theorem before the exam: creask AT rpi DOT edu. Double Integrals ZZ m n XX ▯ ▯ f(x;y)dA = lim f(xijy ijA ij R m;n!1 i=1 j=1 If f(x;y) ▯ D, the integral is the volume that lies above a rectangle and below the surface f. Double Integrals for a Constant Function Suppose f(x;y) = k. ZZ ZZ XX m f(x;y)dA = k dA = lim k▯A R R n;m!1 ij i=1 j=1 n m X X = k n;m!1 ▯A ijkA(R) i=1 j=1 A(R) is the area of R. RR A(R) = (b ▯ a)(d ▯ c). Thus: Rk dA = k(b ▯ a)(d ▯ c) Fubini’s Theorem: { If f is continuous on rectangle R = ((x;y)ja ▯ x ▯ b;c ▯ y ▯ d), then: RR R Rd R R { f(x;y)dA = f(x;y)dy dx = f(x;y)dxdy R a c c a 1 Three Important Properties RR RR 1. If f(x;y) ▯ g(x;y) for all (x;y) in R, then f(x;y)dA ▯ g(x;y)dA. R R RR RR RR 2. R[f(x;y) + g(x;y)]dA = Rf(x;y)dA + Rg(x;y)dA. RR RR 3. R cf(x;y)dA = c R f(x;y)dA, where c is a constant. Equation 5 (Iterated Integrals) ZZ Z b Z d g(x)h(y)dA = g(x)dx h(y)dy;R = [a;b] ▯ [c;d] R a c More General Regions De▯ne F with domain R by the following: ( f(x;y)if(x;y)is inD F(x;y) = 0iff(x;y)is inRbut not inD If F is integrable over R, de▯ne the double integral of f over D by: ZZ ZZ f(x;y)dA = F(x;y)dA D R If f is continuous on a type I region D such that: { D = f(x;y)ja ▯ x ▯ b;g (x) ▯ y ▯ g (x)g (vertically simple), then: 1 2 RR R R2(x) { f(x;y)dA = f(x;y)dy dx. D a g1(x) { This means that D has a "top" and a "bottom" boundary curve. If D = D 1 D ,2where D an1 D don’2 overlap except, perhaps, on their boundaries, then: ZZ ZZ ZZ f(x;y)dA = f(x;y)dA = f(x;y)dA D D1 D2 Type II Regions (Horizontally Simple) RR R R2(y) { D f(x;y)dA = f(x;y)dxdy, where D is a type II region. c h1(y) Average Value of a Function on a Region D RR ▯ D f dA f = A(D) Tra▯c science: ▯nd the average distance of points in a disk of radius a from the origin. 2 RR p 2 2 { d = N x +y dA ▯a2 { It is di▯cult to solve this in the Cartesian coordinate system, so convert the equation to polar coordinates. p RR 2 Z 2▯Z a ▯ ▯a d = N r r dr d▯ = 1 r dr d▯ = 1 [▯]2▯ 1 r3 = 2a ▯a 2 ▯a 2 ▯a 2 0 3 3 0 0 0 This means that there are "more points near the perimeter" (ring roads). Triple Integrals Useful for centroids, the "average" coordinates in three dimensions. RRR RRR RRR W x dV W y dV W z dV ▯ = V ,▯ = V , ▯ = V Other Coordinate Systems Cylindrical: x = r cos▯, y = r sin▯, z = z Spherical Gradient Vector Fields The gradient of f is de▯ned as: rf = hf ;f ix y Suppose that f = x ▯ y . Then: rf = h2x;▯2yi = F = F. ~ ^ ^ F = 2xi ▯ 2yj. This is called a 2D gradient vector ▯eld. A vector ▯eld F is called a conservative vector ▯eld if it is a gradient of some scalar function|that is, if there exists a function f such that F = rf. { In this instance, f is called the potential function for F. ~ Other Types of Gradient Vector Fields p 2 2 ^ ^ 1. Earlier, we had the function f(x;y) = 1 ▯ x ▯ y , and ~ r = xi + yj = r~ er. p▯xi▯yj^ p▯r rf = 1▯x ▯y2 = 1▯r2 This is called a radial gradient vector ▯eld, as it is only de▯ned for r < 1. 2. For the function f = ax + by: F = rf = ai + bj ^ This produces a constant vector ▯eld (all of the arrows are the same size and are pointing in the same direction). 3 Line Integrals Suppose you have a wire or chain, and you desire to know its mass. First, describe the wire as a curve that has arc length: R { l(C) = bk~r (t)k dt R a { s(t) = tk~r (t)k dt 0 Both of these equations measure distance along the curve. { s (t) = ds = k~r (t)k 2 dt ^ ^ { ~r(t) = x(t)i + y(t)j 0 p 0 2 0 2 ds { k~r (t)k = (x (t)) + (y (t)) = dt Next, de▯ne the line integral with respect to arc length as: R R p 0 2 0 2 { C f(x;y)ds = cf(x;y) (x ) + (y ) dt Rt=b p { = t=a f(x(t);y(t)) (x ) + (y ) dt If f is de▯ned on a smooth curve C, the line integral of f along C is: Z n X ▯ ▯ f(x;y)ds = lim f(x iy j▯s , if this limit exists. C n!1 i=1 R R R If f = k (constant), then f ds = k ds = k ds = k l(C): C C C Types of Line Integrals We’ve discussed the ▯rst type of line integral in the previous example, called line integrals of scalar functions. There are two additional types of line integrals: those over vector ▯elds and those with the generic/original form, also known as the "full de▯nition of the line integral." Line Integrals over Vector Fields R ~ R ~ ~ ~ C F ▯ dr = C(F ▯ T)dr, where T is the unit tangent vector. f = F ▯ T~ Z b 0 Z b ~ r (t) 0 ~ 0 F(r(t)) ▯ 0 kr (t)kdt = F(~r(t)) ▯r (t)dt a kr (t)k a Other Helpful Tips For a given arc such that C = C [ C 1 C : 2 3 4 { R = R + R + R ede C C 1 C 2 C 3 r(t) = acos ti + asin tj; 0 ▯ t ▯ 2▯ 2=3 2=3 2=3 { This creates an astroid: x + y = a Generic/Original Line Integral Form Full de▯nition of a line integral: R 0 { C F 1x;y)dx + F (x2y)dy, C : y = y(x), a ▯ x ▯ b, dy = y (x) { Before vectors, one would follow straight lines from point to point, hence line integrals. Orientation of a Curve R R Given a curve C : ~r(t) = hx(t);y(t)i, where a ▯ t ▯ b : C F ▯ dr = ▯ ▯C F ▯ dr ~ { F is the vector ▯eld. R R t F ▯ dr = F(~r(t)) ▯r (t)dt = 99, where F = hxy;y i2 C 1 4 R 99 ▯C F ▯ dr = ▯ 4 (opposite sign) Fundamental Theorem for Conservative Vector Fields Assume that F = rf on a domain D. 1. If r(t) is a path along C from point P to point Q, then: R { C F ▯ dr = f(Q) ▯ f(P) { In particular, the line integral is path independent H ~ 2. The circulation around a closed curve C is zero ( C F ▯ dr = 0) d 0 { Because dtf(r(t)) = rf(~ r(t)) ▯r (t) { If F(~r(t)) = rf(~ r(t)): R R * F ▯ dr = rf ▯ d~r CR C R * = t=brf ▯ r (t)dt = bd f(r(t))dt = f(~r(t))ja= f(Q) ▯ f(P) t=a a dt { If P = Q, then the curve is closed. Theorem ~ Suppose F is a vector ▯eld on an open, connected region D. R { If C F ▯ dr is independent of path in D, then F is a constant vector ▯eld in D. { That is, there exists a function f such that rf = F. ~ 5 Vortex Field 2D: F = F i + F j^ 1 2 3D: F = F 1 + F 2 + F 3; F(x;y;z) ~ ~ ~ { F = rf(x;y;z), with the criterion that rxF = 0 (see Section 16.5). ~ ^ ^ ^ ^ ^ ^ { F = F 1 + F 2 + F 3 = f x + fyj + fzk @F1 @F2 @F2 @F3 @1 @3 { @y = @x; @z = @y; @z = @x 2 ^ ^ ^3 ▯ ▯ ▯ ▯ ▯ ▯ i j k @F @F @F @F @F @F rxF = 4 @ @ @5 = 3 ▯ 2 i ▯ 3 ▯ 1 j + 2 ▯ 1 k @x @y @z @y @z @x @z @x @y F1 F2 F 3 A good example is Hill’s Spherical Vortex. End of Document 6

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