Multivariable Calculus Notes. Faraad Armwood. Fall: Chapter 1: Vectors, Dot Product, Cross Product, Planes, Cylindrical & Spherical Coordinates

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1 Multivariable Calculus Notes Faraad Armwood Fall: 2017 Chapter 1: Vectors, Dot Product, Cross Product, Planes, Cylindrical & Spherical Coordinates Chapter 2: Vector-Valued Functions, Tangent Vectors, Arc Length, Curvature, Motion Chapter 3: Graphing function of two or more variables, level curves/traces, limits & continuity, partial derivatives, differentiability & tangent planes, gradient & direction derivatives, chain rule, second derivative test, lagrange multipliers Chapter 4: Double Integrals, Triple Integrals, Change of Variables Formula, Integration in different coordinate systems Chapter 5: Vector Fields, Line Integrals, Conservative Vector Fields, Parametrized Surfaces and Surface Integrals, Surface Integrals of Vector Fields Chapter 6: Green s Theorem, Stokes Theorem, Divergence Theorem 1

2 Definition 1 (vector): A vector is a quantity which denotes displacement. It tells us where a particular particle will move after a small period of time. Hence, if we wish to mathematize this definition, we need a way of denoting something that not only has direction, but length corresponding to that direction. We will define a vector v = v R n to be the ordered n-tuple, v 1 v 2 v =. The distinction between a vector and a point in euclidean space is a completely abstract one. We think of v as emanating from a point p E n. Thus, attaching v to p, should mean, from p, go in the direction v. With this in mind, we will think of vector v as an arrow. v n Definition 2 (vector): Given two points x, y R n, we define y x to be the vector extending from x to y. Comment: The above definition makes sense due to the fact that vectors denote direction i.e if q is in direction v from p, then v + p should be q. With that in mind, we see (y x) + x = y and so we are justified. Due to the fact that we can parallel transport (i.e carry vectors) vectors in R n without changing their length and direction, we can assume that all vectors have the same tail. This means that we won t distinguish between x p and x q where q p. In addition, for extra simplicity, we will take p = 0 and so v p = v 0 = v. Definition 3 (dot product): The dot product of two vectors a, b R n is the scalar quantity given by, a b = n a j b j = a b cos ν a,b j=1 Comment: cos ν a,b = adj hyp = b a a b a b cos ν a,b = b a 2

3 Definition (projection): Given two vectors a, b, one may wish to know how much of vector b is in the direction of a? With this in mind we have, ( ) b a proj a b = a a 2 Definition (cross product): Given two vectors a, b R n we define the cross product, i j k a b = det a 1 a 2 a 3 b 1 b 2 b 3 Comment: Some important properties of the cross product is that a b = ±V where V is the area of the plane spanned by a, b, a b = b a, a a = 0 and a b is perpendicular to both a, b. Definition (Scalar Triple Product): Let u, v, w be a set of linearly independent vectors. Then the parallelepiped spanned by u, v, w has volume ± u v proj u v w. The scalar triple product is the quantity which gives this volume i.e, w w (u v) = det u v Comment: The equality above needs to be shown. To get the result, one just dots the expanded cross product with w (given in the previous definition) and compare it to the matrix determinant. Definition (Orientation): Since we will be studying surfaces in E 3, it would be nice if we could determine how it sits in space. We will give the mathematical definition here and a geometric summary of what this definition captures. w {u, v, w} positively oriented det u > 0 v 3

4 Comment: Essentially, orientation is a way to determine which side do we wish to look at an object from. Observe that a two dimensional object has its directions determined by linearly independent vectors say u, v. By the above we know that, ( (u, 0) (v, 0) = u v) det = ±A where A is the area of the plane spanned by u, v. So we will define {u, v} to be equivalent to {u, v } the area of a region in the coordinates of {u, v} maintains the same sign in the coordinates of {u, v }. We give the same equivalence for frames {u, v, w}, replacing area with volume. Our last observation is that, u v det u v = (u v) (u v) = u v 2 > 0 {u, v, u v} positively oriented Definition (plane): Given a point p E 3, we wish to define a plane passing through p. Now there are many, so to specify one in particular, we orient the plane with a normal vector n. Hence, denoting our plane by X we have, X = {x : (x p) n = 0} Comment: If we let x = (x, y, z), p = (p 1, p 2, p 3 ) and n = (a, b, c) then any point on the plane X is must satisfy the following equation, a(x p 1 ) + b(y p 2 ) + c(z p 3 ) = 0, where x y E 3 z 4

5 Definition (cylindrical coordinates): To get these type of coordinates, we think of the cylinder S 1 [0, 1] placed in R 3, i.e the coordinates are ((r, θ), z). To each p = (x, y, z), there is a unique cylindrical coordinate namely, r = x 2 + y 2, θ = tan 1 ( y x ), z = z Defintion (spherical coordinates): To get these type of coordinates we think of a sphere with radius R centered at the origin of E 3. The coordinate functions are denoted (ρ, θ, ϕ). To each p = (x, y, z), there is a unique spherical coordinate namely, ρ = x 2 + y 2 + z 2, θ = tan 1 ( y x ) ( ) z, ϕ = cos 1 ρ 5

6 Definition (vector-valued function): A function of the form t (x(t), y(t), z(t)) is called a vector valued function where t R. The name is justified due to our discussion about parallel transport i.e we are assuming our origin in 0 and so (x(t), y(t), z(t)) R 3, as well as E 3. Definition (tangent vectors): Let γ(t) = (x(t), y(t), z(t)) as above. Suppose γ is also differentiable i.e each component function must be differentiable and, γ (t) = lim h 0 γ(t + h) γ(t) h = (x (t), y (t), z (t)) R 3 Hence, γ (t) is the vector at γ(t). We now list some very important properties of differentiable functions like γ. However, it may be useful to divulge that γ is also called a differentiable curve. Here we associate γ with its image since the two are in bijection with one another. (1) (2) (3) d dt (f(t)r(t)) = f(t)r (t) + f (t)r(t) d dt (r 1(t) r 2 (t)) = r 1 (t) r 2(t) + r 1(t) r 2 (t) d dt (r 1(t) r 2 (t)) = [r 1 (t) r 2(t)] + [r 1(t) r 2(t)] Definition (tangent line): Given a differentiable curve γ(t) we define the tangent line at γ(t 0 ) in the direction of γ (t 0 ) to be, L(t) = γ(t 0 ) + tγ (t 0 ), t R 6

7 Definition (arc-length): Let γ(t) R 3 be a differentiable curve as defined above. Then the length of the curve from γ(t 0 ) γ(t) is given by the integral, s = s(t) := t t 0 γ (u) du Comment: The above is due to the mean value theorem. However, for some quick geometric intuition, observe that if t is sufficiently close to t 0 then the path from γ(t) γ(t 0 ) is almost a straight line and γ(t) γ(t 0 ) = γ (t 0 ) (t t 0 ) + ɛ which is why you see this in the integrand. Definition (speed): By the fundamental theorem of calculus we can derive the speed. We simply take the time derivative of both sides. s (t) = ds dt = d dt t = γ (t) t 0 γ (u) du Definition (reparametrization): Let γ i : I i R 3 be differentiable curves where i = 1, 2. We say that γ 1 is a reparam. of γ 2 if there exists a diffeomorphism φ : I 1 I 2 s.t (γ 2 φ)(t) = γ 1 ( t). This implies that γ 2 (I 2 ) = γ 1 (I 1 ) i.e they describe the same curve. Comment: Given s(t) is monotone on [t 0, t], it is invertible. Observe that t s defines a diffeomorphism from [t 0, t] s[t 0, t]. Hence if we let g = s 1 then γ(t) = (γ g)(s). γ (t) = d dt γ(g(s)) = d γ(t) d g(t) = γ 1 (g(s)) dt t=g(s) dt t=s s (g(s)) Hence, if γ (t 0 ) 0 then γ has a unit speed reparametrization γ(s) := (γ g)(s) where s is the arc-length parameter. 7

8 Definition (curvature): If γ(u) R 3 is a differentiable curve and u is a unit speed parameter then we define the curvature at γ(u) to be κ(u) = γ (u). Comment: Given γ (t) is the instantaneous rate of change at γ(t), then γ (t) gives the rate at which the velocity vector γ (t) will change at γ(t). Hence the accelerator vector γ (t) gives the direction that the point γ (t) + γ(t) must move to arrive at γ (t + h) + γ(t + h) where h is a very, very small real number. Definition (curvature): Given γ(t) is a differentiable curve and t not a unit speed parameter then how do we define κ(s)? We adjust our class of curves to regular ones i.e ones in which γ (t) 0 for all t dom(γ). With this in mind, the arc-length function is invertible. We can then write γ(t) = γ(s(t)). Since γ is just a reparametrization of the domain for γ, we are safe to just write γ = γ. One can then show that, κ(t) = γ (t) γ (t) γ (t) 3 Definition (acceleration): The acceleration of a particle i.e it s change in direction is given by the vector a(t) := γ (t). Comment: The acceleration vector is a linear combination of both T and N i.e the tangential and normal vectors. γ(t) = γ(s(t)) γ (t) = γ (s(t)) ν(t) = T(s(t))ν(t) γ (t) = T(s(t))ν (t) + ν 2 (t)t (s(t)) = T(s(t))ν (t) + κ(s(t))ν 2 (t)n(s(t)) We define a T = ν and a N = κν 2 to be the tangential and normal components of the acceleration vector. In what is to follow, we will give simpler expressions for each such component. 8

9 Definition (tangential and normal components): In the decomposition a = a T T + a N N we the following expressions, a T = a v v, a N = a 2 a T 2 a T T = ( a v ) v, a N N = a v v ( a v ) v v v Comment: If we consider a particle under uniform circular motion i.e constant speed then we can parametrize the trajectory as, γ(t) = R(cos(ωt), sin(ωt)) γ (t) = Rω( sin(ωt), cos(ωt)) a(t) = Rω 2 ( cos(ωt), sin(ωt)) a(t) = Rω 2 We can now give proof to the magnitude of the centripetal accelerator vector. Since ν = γ (t) where ν R is fixed, then ω = ν/r and so, a(t) = R ν2 R = ν2 2 R 9

10 Definition (graph): Let f : U R 2 R be a function. The graph of f denoted Γ(f) is defined Γ(f) = {(x, y, f(x, y)) : z = f(x, y)} R 3. Definition (trace): Let Γ(f) be the graph of some function f. We define the vertical trace of the graph to be the curve defined by γ(t) = (a, t, f(a, t)). The other vertical trace of the graph is the curve α(t) = (t, b, f(t, b)). These curves are the ones carved out by the planes x = a and y = b respectively. Definition (level curves): Let f again be a function of two variables. A level curve of Γ(f) is f 1 (c) = {(x, y, f(x, y)) : f(x, y) = c}. Definition (continuous): A function f(x, y) R is said to be continuous at (a, b) if given any ɛ > 0, there exists δ > 0 such that (x, y) (a, b) < δ f(x, y) f(a, b) < ɛ. Definition (limit): A function f(x, y) is said to have limit L if there exists a p U such that the following limit holds, lim f(x, y) = L (x,y) p Definition (partial derivative): Given f : U R 2 R, it s partial derivatives, if they exists are defined by the following limits, f f(a + h, b) f(a, b) (a, b) = lim x h 0 h f f(a, b + k) f(a, b) (a, b) = lim y k 0 k Comment: We will soon see that these partial derivatives are just the slopes γ(t) and α(t) i.e the vertical traces. 10

11 Comment: By Clairaut s theorem for the equality of mixed partial, if f xy and f yx are both continuous on a disk D, then f xy (a, b) = f yx (a, b) for all (a, b) D. Definition (differentiability): Let f : U R 2 R be a function. We say that f is differentiable at (a, b) U if there exists a linear function λ : R 2 R such that, lim (x,y) (a,b) ( x a f(x, y) f(a, b) λ y b) = 0 (x a)2 + (y b) 2 We write λ = Df(a, b). This definition is a bit cumbersome at first, but all it means is that Df(a, b) is the best linear approximation to f i.e, f(x, y) = f(a, b) + Df(a, b) ( ) x a ɛ(x, y) + ɛ(x, y), where lim y b (x,y) (a,b) (x a)2 + (y b) = 0 2 That is to say, since Df(a, b) is a linear map, and we can pick p 0 = (x 0, y 0 ) and p 1 = (x 1, y 1 ) such that the approximation holds and p 0 (a, b) and p 1 (a, b) are linearly independent. Thus differentiability of f at (a, b) implies that f has a tangent plane defined at f(a, b)! One can easily check that, Df(a, b) = ( f f (a, b) x ) (a, b) y Comment: A quick criterion for differentiability is that f x and f y exists and are continuous on a disk D. This is clear now due to the fact that if the partial exists and are continuous then we can define the linear map Df(a, b). Definition (linearization): Given f : U R 2 R is differentiable, we define it s linearization i.e linear approximation to be, L(x, y) = f(a, b) + f x (a, b)(x a) + f y (a, b)(y b) 11

12 Definition (linear approximation): The linear approximation of f is given by L(x, y). Comment: If we let h = x a and k = y b then we have f(a + h, b + k) f x (a, b)h + f y (a, b)k. We take h = x and k = y if (x, y) is sufficiently close to (a, b). So we have an approximation for the infinitesimal change in f i.e, f f x (a, b) x + f y (a, b) y Hence, given any (x, y) near (a, b) for which f is differentiable, we can define the differential denoted df by the relation, df = f x (x, y) dx + f y (x, y) dy Remark: This is truly not just a shuffling of symbols and simple renamings. We are showing here that we can approximate sufficiently small change in coordinates by the differentials dx, dy. Definition (gradient): Given a differentiable function f : U R 2 R, then the gradient denoted f(a, b) is defined by, f(a, b) = ( f f (a, b) x ) (a, b) y Comment (chain rule): Given f : U R 2 R differentiable at (a, b) and γ : ( ɛ, ɛ) U differentiable with γ(t 0 ) = (a, b) we have, d f(γ(t)) = f(γ(t 0 )) γ (t 0 ) dt t=t0 = ( f x (γ(t 0 )) f y (γ(t 0 )) ) ( ) x (t 0 ) y (t 0 ) = f x (γ(t 0 ))x (t 0 ) + f y (γ(t 0 ))y (t 0 ) 12

13 Definition (directional derivative): Suppose f : U R 2 R is differentiable at p U. Let u = (h, k) be a unit vector such that γ(t) = p + tu U for small t. Then we define the derivative in the direction of u to be, D u f(p) = d f(γ(t)) = f(p) u dt t=0 Comment: Since f(p) u = f(p) cos ν, at f(p), the function f increases fastest in the direction of the gradient. Observe also that ν = 90 o implies f(p) is perpendicular to all of the coordinate lines f(p + tv) where v S 1. This implies that f(p) is a normal vector for the given tangent plane to Γ(f) at f(p). Definition (chain rule): Given f : U R n R and (x 1,..., x n ) = (x 1 (s 1,..., s k ),..., x n (s 1,..., s k )), letting x = (x 1,..., x n ) and p = (s 1,..., s k ) we have by the chain rule for paths, f ( (p) = f x s 1 (p) j x 1 ) s j (p) f x n (p). x n s j (p) Definition (local extremum): A function f(x, y) has a local extremum at p = (a, b) is there exists D r (p) such that, local minimum: f(p) f(x, y), (x, y) D r (p) local maximum: f(x, y) f(p), (x, y) D r (p). Definition (critical point): Let f : U R 2 R be differentiable at p U. We say p is a critical point of f if Df(p) = 0 or equivalently rank(df(p)) = 0. 13

14 Comment (Fermat s Theorem): If f(x, y) has a local minimum or maximum at p = (a, b) then p is a critical point of f. Sketch: If p = (a, b) is a local minimum then g(x) = f(x, b) has a local minimum at a and so g (a) = f x (a, b) = 0. Similarly h(y) = f(a, y) has a local minimum at y = b h (b) = f y (a, b) = 0. Definition (discriminant of of f C 2 ): Let f be a second differentiable function who s second order partials are also continuous in U. Suppose p U then we define the discriminant of f at p to be D(p) = f xx (p)f yy (p) f 2 xy(p). Comment (Second Derivative Test): Let p = (a, b) be a critical point of f C 2 (U p, R) then, if D(p) > 0 and f xx (p) > 0, then f(p) is a local minimum. if D(p) > 0 and f xx (p) < 0, then f(p) is a local maximum. if D(p) < 0, then f has a saddle at p. if D = 0, the test is inconclusive. Remark: The second derivative test comes about by approximating f(p) by the second degree polynomial in the taylor expansion of f about p i.e, f(x, y) f xx (a + th, b + tk)h 2 + 2f xy (a + th, b + tk)hk + f yy (a + th, b + tk)k 2 and the discriminant is precisely the one of this second degree polynomial. We won t give a sketch of this. We now come to optimizing functions via certain constraints and so this we have the method of Lagrange Multipliers. 14

15 Comment (Lagrange Multipliers): Suppose f(x, y) and g(x, y) are differentiable functions. If f(x, y) has a local minimum or a local maximum on a constraint curve g(x, y) = 0 at p = (a, b) and if Dg(p) 0 then there is a scalar λ such that Df(p) = λdg(p). Here λ is called the Lagrange multiplier. Sketch: Since Dg(p) = g(p) 0, we can define a differentiable curve γ(t) passing through p = γ(0). This curve γ g 1 (0) i.e g(γ(t)) = 0 and so g(p) γ (0) = 0 g(p) γ (0). And since p is a local extremum of f then f x (p) = f y (p) = 0 f(p) γ (0) = 0. Hence both f(p) and g(p) are normal vectors for the tangent plane at p f(p) = λ g(p). 15