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1 Math Notes of 12/03/2018 Math Fall 2018 Review Remaining Events Fr, 11/30: Starting Review. No study session today. Mo, 12/03: Review Tu, 12/04: 9:40-10:30, AEB 340, study session We, 12/05: Review, Q&A Th, 12/06 9:40-10:30, AEB 340, study session Fr, 12/07 Reading Day, no events, last hw closes Mo, 12/10, 10:30-12:30, JTB 110, study session Tu, 12/11, 1:00-3:00, JTB 110, study session We, 12/12, 8:00-10:00am, JTB 130, Final Exam, comprehensive, same format as the midterms. Math Notes of 12/03/2018 page 1

2 A few general points the important thing is not the answer but your ability to figure out the answer! You want to be aware always what kind of object you are dealing with. These may be scalars, vectors, points, functions, constants, variables, curves, surfaces, solids, line integrals, multiple integrals, iterated integrals, For example, if the problem asks for a minimum value then the answer is a scalar, not a point where the minimum occurs, but the value of the function at that point. It s the value of the dependent variable, not the values of the independent variables! If you don t understand a problem you won t be able to solve it. Before you start any calculations, make sure you understand the problem. Always think of expectations before embarking on a calculation. Keep checking for plausibility and consistency with what else you know about the current problem. If you can t solve a problem, or understand a concept, simplify the problem or the context of the concept. For example, when computing a definite in- Math Notes of 12/03/2018 page 2

3 tegral you cannot get a function whose independent variable is the integration variable. On the other hand, an indefinite integral gives you a function. The dot product is a scalar. The cross product is a vector. The gradient is a vector. You can t divide by a vector. An expression like (u v) w does not make sense whereas u (v w) does (and is a scalar). Check your final answers. Always know what you are doing! Math Notes of 12/03/2018 page 3

4 1 Subject Matter Note: This list is not complete, nor is it self contained. The following items should stir your memory and activate your understanding. If any of these items are obscure to you, or you are not sure about them, go back to your notes and the textbook, and study the relevant parts. Vectors. Geometrically you can think of them as arrows. Algebraically: u =< 1, 2 > or v =< 1, 2, 3 >. (1) You can add vectors to vectors and multiply vectors with scalars. Scalars are numbers. The scalars making up a vector are its components. We add vectors by adding pairs of corresponding components. Geometrically we attach the tail of one vector to the head of the other. Vectors with 2 or 3 components are particularly important because we live in three-dimensional space 1. However, mathematically there is no problem having n components, for example x =< x 1, x 2,...,x n >. (2) The set of all vectors with 2, 3, or n components, is denoted by IR 2, IR 3, or IR n, respectively. Disclaimer: Physicists might question that assertion. It s close enough for our purposes. Math Notes of 12/03/2018 page 4

5 The words vectors and points are often used interchangeably. If a point is identified with a vector it s the head of the vector starting at the origin. With that notion it makes sense, for example, to add a vector to a point. dot product: < u 1, u 2, u 3 > < v 1, v 2, v 3 >= u 1 v 1 +u 2 v 2 +u 3 v 3. (3) Similarly for vectors with 2 (or n) components. Norm, magnitude, length of a vector: u = u u. (4) unit vector, a vector whose norm is 1. If v is a non-zero vector then u = v v (5) is a unit vector in the same direction as v. Replacing v with u is called normalizing v. Multiplying a vector with a scalar changes the norm of a vector, but not its direction. Particular unit vectors are the standard unit vectors i =< 1, 0 > and j =< 0, 1 > (6) Math Notes of 12/03/2018 page 5

6 in IR 2, or i =< 1, 0, 0 >, j =< 0, 1, 0 >, and k =< 0, 0, 1 >, (7) in IR 3. Thus < x 1, x 2 > in IR 2 can be written as < x 1, x 2 >= x 1 i + x 2 j (8) and < x 1, x 2, x 3 > in IR 3 can be written as < x 1, x 2, x 3 >= x 1 i + x 2 j + x 3 k. (9) Whether i and j are in IR 2 or IR 3 depends on the context, and should be clear from the context. Angles: Let θ be the angle formed by two vectors u and v. Then u v = u v cosθ. (10) Two vectors u and v are orthogonal, perpendicular, normal, or at right angles, if u v = 0. (11) The zero vector is orthogonal to all vectors. If v =< v 1, v 2 > is a vector then a particular vector that is orthogonal to it is given by w =< v 2, v 1 >. Obviously, v w = 0. That vector is sometimes denoted by v (pronounced v-perp). < v 1, v 2 > =< v 2, v 1 >. (12) Math Notes of 12/03/2018 page 6

7 The projection of a vector u onto a vector v is given by w = pr v u = u v v. (13) v v You should be able to derive this formula and interpret it geometrically. Note that you cannot cancel a factor v in this formula. It does not make sense to divide by a vector. The work W of moving an object along a vector d by applying a force f to the object is given simply by W = f d. (14) Again, you should be able to interpret this formula geometrically, and derive it, presumably by using a projection. The cross product of two vectors a =< a 1, a 2, a 3 > and b =< b 1, b 2, b 3 > in IR 3 is defined by a b = (a 2 b 3 a 3 b 2 )i+(a 3 b 1 a 1 b 3 )j+(a 1 b 2 a 2 b 1 )k. (15) The cross product is antisymmetric: a b = b a. (16) Suppose c = a b. Then c has these geometric properties: c is orthogonal to both a and b. Math Notes of 12/03/2018 page 7

8 c = a b sinθ where θ is the angle formed by a and b, and a, b, c form a right handed coordinate system. A convenient way to remember the cross product is to think of it as the determinant i j k a b = a 1 a 2 a 3 b 1 b 2 b 3. (17) The dot product of two vectors is a scalar. The cross product of two vectors is a vector, and it is defined only for vectors in IR 3. The area A of the parallelogram with sides u and v is given by A = u v. (18) A line can be written in many different ways, but also as x = p + td (19) where x is a general point on the line, p is a particular point on the line, d (direction) is a vector in the line, and t is a parameter. (We used this concept in the infamous Gandalf problem, and it caused a great many questions.) You can think of t as time, and x as traveling along the line. If d is a unit vector then t is the distance between p and x. Math Notes of 12/03/2018 page 8

9 A plane in IR 3 can also be be written in many ways. One way to describe it is in terms of a point P =< x 0, y 0, z 0 > in the plane, and a vector n =< A, B, C > that is orthogonal to the plane. Let < x, y, z > be a general point in the plane. We want an equation for the plane in terms of x, y, z. The vector < x x 0, y y 0, z z 0 > is in the plane and orthogonal to n. Hence < x x 0, y y 0, z z 0 > < A, B, C > = A(x x 0 ) + B(y y 0 ) + C(z z 0 ) = 0 (20) which can be rewritten as where Ax + By + Cz = D (21) D = Ax 0 + By 0 + Cz 0. (22) You can of course divide by any non-zero constant in (21). The coefficients of x, y, z in an equation of the form (21) are the components of the vector that s perpendicular to the plane. A line that s perpendicular to the given plane can be written as x = P + t < A, B, C >. (23) If < A, B, C > is a unit vector then t is the distance between P and x, and hence between x and the plane. These concepts were the second major source of queries about chapter Math Notes of 12/03/2018 page 9

10 11, but this should all be plain as daylight by now. To find the line of intersection of two planes find a point P on it, and its direction d. The direction is the cross product of the two normal vectors defining the planes. Write the line as x = P + td, as above in (19), normalizing d if necessary. Query: what happens to this procedure if the planes are parallel? If t denotes time then a vector valued function r(t) = f(t)i + g(t)j + h(t)k (24) describes curvilinear motion. The graph of r is a curve in IR 3. The velocity of r is v(t) = r (t) = f (t)i + g (t)j + h (t)k, (25) its speed is s(t) = r (t), and its acceleration is a(t) = r (t) = f (t)i + g (t)j + h (t)k. (26) The arc length of the curve r(t), a t b is given by L = b a s(t)dt. (27) This is a nice generalization from Calc 1: Distance is the integral of speed. Math Notes of 12/03/2018 page 10

11 The vector r (t) (28) is tangent to the curve defined by (24). We derived Kepler s second Law in class and the textbook has instructions and exercises for the other two laws. Vector Calculus is powerful! Kepler would have been thrilled to understand it. For a given curve (24) a local coordinate system can be constructed consisting of the following three vectors: The unit tangent vector T(t) = r (t) r (t), (29) the principal unit normal vector N(t) = T (t) T (t) (30) which is orthogonal to T(t) (why?), and the binormal vector B(t) = T(t) N(t) (31) which is of course orthogonal to T and N. Why do I say of course? The curvature κ of a plane curve y = g(x) is the reciprocal of the radius of the osculating Math Notes of 12/03/2018 page 11

12 circle. It is given by κ = g ( 1 + g 2 ) 3/2. (32) Surfaces in three dimensional space can be defined by equations of the form F(x, y, z) = 0. (33) If F is a polynomial of degree 2 the surface is a quadric surface, corresponding to a conic section in two variables. In the 3D case as in the 2D case, there are a bunch of different types. We also discussed polar, spherical, and cylindrical coordinate systems. That s basically old (Calc I or II) stuff. Much of the discussion here has obvious generalizations to the case of more than 2 or 3 independent variables. Much of what is said here requires that the functions involved be continuous or (sufficiently often) differentiable. We assume this without explicitly saying so each time. real valued functions of several variables, f(x, y), f(x, y, z). The big new item in Calc 3 is that we may have more than one independent variable. Math Notes of 12/03/2018 page 12

13 Graph of an equation F(x, y, z) = 0 is the set of all points whose coordinates satisfy the equation. Graph of a function f(x, y) is the graph of the equation z = f(x, y). It s a surface in three dimensional space. domain, range, natural domain. level curves, contour lines, graphs of f(x, y) = constant. contour plot. Partial Derivatives are derivatives with respect to one of the variables, keeping the other variables constant. For example, suppose z = f(x, y). Then we have several notations: f x, f x, f xy, 2 f x y, z x.... mixed partial derivatives commute if they are continuous. Implicit Differentiation works pretty much as it does in one variable. Continuity, interior points, boundary points, bounded sets, interior of a set, closed sets, open sets, neighborhood of a point. The gradient f of a function f is the vector of its partial derivatives. A function is differentiable at a point p if it can be approximated at that point by a linear Math Notes of 12/03/2018 page 13

14 function. Very subtle concept. If it exists that linear function is L(x) = f(p) + f(p) (x p). This is the first order Taylor expansion (or polynomial) of f about (or at) the point p. Higher order expansions are possible. Taylor series form a huge subject. Let u be a unit vector. The directional derivative of a function f in the direction u is f (p) = lim u h 0 f(p + hu) f(p) h = f(p) u. The directional derivative is maximized in the direction of the gradient, and minimized in the direction of the negative gradient. Thus the gradient is the direction of steepest ascent, and its negative is the direction of steepest descent. The gradient is orthogonal to contour lines, level sets, tangents, and tangent planes. The Chain Rule occurs in several forms. We discussed two of them. They become obvious if you appreciate that in one variable the chain rule says that the derivative of the composition is the product of the derivatives, and if you have a function of several variables than each change in one of the variables contributes a change in the function value. Math Notes of 12/03/2018 page 14

15 Let x = x(t), y = y(t), and z = f(x, y). Then dz dt = z dx x dt + z y dy dt. Let x = x(s, t), y = y(s, t), and z = f(x, y). Then z s = z x x s + z y y s and z t = z x x t + z y y t. Tangent planes can be defined for implicitly defined surfaces. For the surface defined by F(x, y, z) = k the equation of the tangent plane at (x 0, y 0, z 0 ) is F(x 0, y 0, z 0 ) < x x 0, y y 0, z z 0 >= 0. Partial derivatives can be used to approximate changes in function values. For example, if z = f(x, y), and denotes the changes in the corresponding variables we have z f x x + f y y. We defined local and global minimum, maximum, or extreme values much as we did in the case of one variable. Math Notes of 12/03/2018 page 15

16 Extreme values can occur only at critical points of which there are three kinds: Stationary points where the gradient is zero Singular Points where the function is not differentiable Boundary Points on the boundary of the domain. We may be able solve constrained extreme value problems by the method of Lagrange Multipliers. Specifically, we can identify critical points of the problem f(x) = min subject to g(x) = 0 by solving the system of equations f(x) = λ g(x) and g(x) = 0. λ is the Lagrange Multiplier. This is the tip of a very large iceberg. Math Notes of 12/03/2018 page 16

17 1 y x Figure 1. Lagrange Multipliers. Example: Find the extreme values of the function f(x, y) = x 2 y 2 on the ellipse x 2 + 2y 2 1 Math Notes of 12/03/2018 page 17

18 Multiple Integrals are integrals of functions of n variables over regions in IR n. For our purposes n = 2, or n = 3. For n = 2 we obtain a double integral, for n = 3, a triple integral. To obtain a multiple integral, think in terms of what happens in a part of the integration region that corresponds to small changes in the integration variables. Add up those contributions, take the limit as the number of contributions goes to infinity and their size goes to zero, and replace the sum by an integral. Double integrals. Let D be a region in IR 2. D f(x, y) dx dy = D f(x, y) da = lim x, y 0 f(xi, y j ) x i y i (34) where x = max i x i and y = max i y i Math Notes of 12/03/2018 page 18

19 Examples. area = mass = average value of f = da D δ(x, y) da D f(x, y) da D da D where δ is density for example, f is temperature, depth, or density (35) Similar formulas apply for triple integrals. Suppose S is a region in IR 3. Conceptually f(x, y, z) dx dy dz = f(x, y, z) dv S For example: S = lim x, y, z 0 volume of S = S f(xi, y i, z i ) x i y i z i. (36) dv. (37) The formulas for the center of mass familiar from Calculus 1 actually simplify, and be- Math Notes of 12/03/2018 page 19

20 come more general. Let δ(x, y) denote density. Then the center of mass of a subset D of IR 2 is ( x, ȳ) where x = D D xδ(x, y) da δ(x, y) da and ȳ = D D yδ(x, y) da δ(x, y) da (38) The formulas for the center of mass of a region S in IR 3 are straightforward generalizations of the 2D case: xδ(x, y, z) dv yδ(x, y, z) dv x = S δ(x, y, z) dv, ȳ = S δ(x, y, z) dv and S zδ(x, y, z) dv S (39) z = S δ(x, y, z)dv (40) S The practical way of computing multiple integrals is to convert them to iterated integrals which are nested one variable integrals. Math Notes of 12/03/2018 page 20

21 In general the limits of integration of the inner integrals depend on the variables of the outer integrals. For example, to integrate the function f over the unit sphere W (defined by x 2 +y 2 +z 2 1) we can write 1 1 z 2 1 z 2 y 2 f(x, y, z) dv = 1 f(x, y, z) dx dy dz. 1 z 2 1 z 2 y 2 W (41) For example, to obtain the volume of the unit sphere we set f(x, y, z) = 1 and obtain V = = = 1 1 z z z z 2 y 2 1 z 2 y 2 dx dy dz 1 z z 2 y 2 dy dz (and, since this is the area of a circle of radius 1 z 2 ) π(1 z 2 )dz ) = π (z z3 1 3 = 4π 3 1 which of course is no surprise. (42) For iterated integrals, the sequence of the integration variables matters! To Math Notes of 12/03/2018 page 21

22 get the limits of integration, and the sequence of the integration variables, figure them out geometrically. There is no set rule or simple recipe. Math Notes of 12/03/2018 page 22

23 Example: Suppose you want to integrate a function f over the tetrahedron T with the vertices (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c). What are the limits of integration and the sequence of integration in I = T f(x, y, z) dv = β1 α 1 β2 (y) β2 (y,z) α 2 (y) α 3 (y,z) f(x, y, z) d?d? d? Math Notes of 12/03/2018 page 23

24 In many applications, integrals get much simplified by using appropriate coordinates. To figure out an integration formula in a certain coordinate system, consider a part of the region in which the coordinates change by some small value. Don t try to memorize formulas. However, in polar coordinates, letting S be a region in IR 2, and S the same region expressed in terms of polar coordinates, we obtain S f(x, y) dx dy = S Note the additional factor r! ( ) f r cos(θ), r sin(θ) r dr dθ. (43) Similarly, letting W be a region in IR 3 we obtain in cylindrical coordinates f(x, y, z) dx dy dz = f ( r cos(θ), r sin(θ), z ) r dr dθ dz, W W (44) and in spherical coordinates f(x, y, z) dx dy = f ( ρcos(θ) sinφ, ρsin(θ) sinφ, ρcos φ ) ρ 2 s W W (45) The surface area of the graph of f(x, y) defined on a domain D in R 2 is Area = 1 + fx 2 + fy 2 dx dy. (46) D Math Notes of 12/03/2018 page 24

25 This is the natural analog of the one variable formula for arc length. The individual integrals in an iterated integral can of course be computed using any single variable integration technique. However, it is also possible to replace a whole set of variables with another set. Here is one example of such a substitution formula (called Change of Variables for Double Integrals in the textbook.) Suppose G is a one-to-one transformation from IR 2 to IR 2 which maps the bounded region S in the uv-plane onto the bounded region R in the xy-plane. If G is of the form then R G(u, v) = (x(u, v), y(u, v)) (47) f(x, y) dy dx = S f ( x(u, v), y(u, v) ) J(u, v) du dv where the Jacobian J(u, v) is defined by J(u, v) = x y u v x v (48) y u. (49) Variables can be changed to simplify the integrand, or simplify the region of integration. There are versions of the above formulas for all values of n (the number of variables). The Math Notes of 12/03/2018 page 25

26 Jacobian is the determinant of the matrix of partial derivatives of the transformation. The concept of improper integrals can be generalized to the integration of functions of several variables. We saw one example: x e 2 y 2 da = π IR 2 which implies that e x2 dx = π. A scalar field is a function that associates a scalar (number) with a point. Natural examples include pressure and temperature. A vector field is a function that associates a vector with a point. Natural examples include force, flow, gradient of a scalar field. Associated with a vector field is its divergence, a scalar field, and its curl, a vector field. If the vector field is fluid flow then its divergence measures sources and sinks, and its curl measures the flows tendency to rotate. Specifically, if F = Mi + Nj + Pk then div F = M x + N y + P z Math Notes of 12/03/2018 page 26

27 and curl F = ( P y N ) ( M i+ z z P ) ( N j+ x x M ) k. y A convenient way to remember these formulas is div F = F and curl F = F. div and curl can be defined similarly for vector fields F = M(x, y)i + N(x, y)j as div F = M x + N y and curl F = N x M y. We defined the line integral of a function f along a curve C as the limit of a Riemann Sum. However, given a parameterization x = x(t), y = y(t), a t b of the curve the line integral can be evaluated by the formula C f(x, y) ds = b a f(x(t), y(t)) [x (t)] 2 + [y (t)] 2 dt. The value of the line integral is independent of the parameterization. Math Notes of 12/03/2018 page 27

28 A similar formula applies in three variables: C f(x, y, z) ds = b a f(x(t), y(t), z(t)) [x (t)] 2 + [y (t)] 2 + [z (t)] 2 dt. Suppose the force acting at a point (x, y, z) is given by a vector field F(x, y, z) = M(x, y, z)i+n(x, y, z)j+p(x, y, z)k. Suppose we move an object from a point A to a point B along a curve C parameterized as r(s) (where s = s(t) is arc length). Suppose T = dr/ds. Then the work required is given by W = C F Tds = C F dr dt dt = C F dr. A vector field F is conservative if it is the gradient of some scalar function f. The scalar function is the potential of the vector field. A vector field is conservative if and only if its curl is zero. We discussed how to compute the potential of a conservative field. Math Notes of 12/03/2018 page 28

29 Example 2, page 745: Determine whether F = (4x 3 + 9x 2 y 2 )i + 6x 3 y + 6y 5 )j is conservative, and if so determine its potential. Math Notes of 12/03/2018 page 29

30 The value of a line integral in a conservative field is independent of the path. Example: In the previous example, what is C F(r) dr for any curve C from (1, 2) to (3, 4). Math Notes of 12/03/2018 page 30

31 Suppose C is a curve starting at A and ending at B. The Fundamental Theorem for Line Integrals says f(r) dr = f(b) f(a). C Suppose C is a closed curve forming the boundary of a region S. Green s Theorem states ( N x M ) da = M dx + N dy. y C S If S is a region in the plane and the closed curve C its boundary, then the area A(S) of S is given by A(S) = y 2 dx + x 2 dy = 1 da C The second equality holds by Green s Theorem. Suppose F = Mi + Nj describes the flow of (an infinitely thin, or vertically homogeneous) fluid in the plane. S and C are as before, and n is the outward pointing unit normal vector along C. C is oriented counterclockwise, and parameterized by arc length. The total flow (flux) across the boundary of S is given by F nds = div F da. C Math Notes of 12/03/2018 page 31 S S

32 Thus the divergence of a vector field measures the local creation or destruction of the fluid. Assume the same circumstances and let T denote the unit tangent vector along C. Then the total flow in the direction of T is given by C F T = S curl F. Thus the curl denotes the local tendency of a fluid to rotate. A flowing fluid with zero curl is called irrotational. Cautionary Note: We have been casual about stating necessary assumptions. For example, functions may have to be continuous or differentiable, on regions with certain properties, such as being finite, simple (no holes), connected, or (x- or y)-simple. Assumptions can sometimes be relaxed, for example, a function might be piecewise differentiable or continuous. The reason for this that pedantic insistence on always stating necessary assumptions would slow us down and distract us from the heart of the subject. However, of course in any particular situation you should always wonder if you can actually do what you do, and you always have question, check, and double check, your answers. Math Notes of 12/03/2018 page 32

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