Cheat Sheet Calculus 3 Final Exam Review, Cheat Sheet of Calculus

Download Cheat Sheet Calculus 3 Final Exam Review and more Cheat Sheet Calculus in PDF only on Docsity! Multivariable Calculus Study Guide: A LATEX Version Tyler Silber University of Connecticut December 11, 2011 1 Disclaimer It is not guaranteed that I have every single bit of necessary information for the course. This happened to be some of what I needed to know this specific semester in my course. For example, Stokes’ Theorem is not even mentioned. 2 Vectors Between Two Points Given : P (x1, y1) & Q(x2, y2) −−→ PQ = ( x2 − x1 y2 − y1 ) 3 Vectors in the Plane let v = ( v1 v2 ) & u = ( u1 u2 ) 0 = ( 0 0 ) 3.1 Simple Operations cv = ( cv1 cv2 ) |v| = √ v21 + v 2 2 v + u = ( v1 + u1 v2 + u2 ) 1 3.2 Unit Vectors i = ( 1 0 ) & j = ( 0 1 ) v = v1i + v2j 3.3 Vectors of a Specified Length∣∣∣∣ cv|v| ∣∣∣∣ = |c| ± cv |v| ‖ v 4 Vectors in Three Dimensions 4.1 Notes Everything in the above section can be expanded to three dimensions. Simply add another component. k =  00 1  4.2 Random Equations xy-plane {(x, y, z) : z = 0} xz-plane {(x, y, z) : y = 0} yz-plane {(x, y, z) : x = 0} Sphere: (x− a)2 + (y − b)2 + (z − c)2 = r2 5 Dot Product 5.1 Definitions u · v = u1v1 + u2v2 + u3v3 = |u||v| cos θ u ⊥ v⇔ u · v = 0 u ‖ v⇔ u · v = ±|u||v| 2 9.3 Two-Dimensional Motion Given : v(0) = 〈|v0| cos θ, |v0| sin θ〉 & r(0) = 〈0, 0〉 Time = 2|v0| sin θ g Range = |v0|2 sin 2θ g MaxHeight = y ( T 2 ) = (|v0| sin θ)2 2g 10 Planes and Surfaces 10.1 Plane Equations The plane passing through the point P0(x0, y0, z0) with a normal vector n = 〈a, b, c, 〉 is described by the equations: a(x− x0) + b(y − y0) + c(z − z0) = 0 ax+ by + cz = d, where d = ax0 + by0 + cz0 In order to find the equation of a plane when given three points, simply create any two vectors out of the points and take the cross product to find the vector normal to the plane. Then use one of the above formulae. 10.2 Parallel and Orthogonal Planes Two planes are parallel if their normal vectors are parallel. Two planes are orthogonal if their normal vectors are orthogonal. 10.3 Surfaces 10.3.1 Ellipsoid x2 a2 + y2 b2 + z2 c2 = 1 10.3.2 Elliptic Paraboloid z = x2 a2 + y2 b2 It would be worth it to learn how to derive sections 9.2 and 9.3. 5 10.3.3 Hyperboloid of One Sheet x2 a2 + y2 b2 − z 2 c2 = 1 10.3.4 Hyperboloid of Two Sheets −x 2 a2 − y 2 b2 + z2 c2 = 1 10.3.5 Elliptic Cone x2 a2 + y2 b2 = z2 c2 10.3.6 Hyperbolic Paraboloid z = x2 a2 − y 2 b2 11 Graphs and Level Curves 11.1 Functions of Two Variables R2 → R z = f(x, y) F (x, y, z) = 0 11.2 Functions of Three Variables R3 → R w = f(x, y, z) F (w, x, y, z) = 0 11.3 Level Curves Imagine stepping onto a surface and walking along a path with constant eleva- tion. The path you walk on is known as the contour curve, while the projection of the path onto the xy-plane is known as a level curve. 6 12 Limits and Continuity 12.1 Limits The function f has the limit L as P (x, y) approaches P0(a, b). lim (x,y)→(a,b) f(x, y) = lim P→P0 f(x, y) = L If f(x, y) approaches two different values as (x, y) approaches (a, b) along two different paths in the domain of f , then the limit does not exist. 12.2 Continuity The function f if continuous at the point (a, b) provided: lim (x,y)→(a,b) f(x, y) = f(a, b) 13 Partial Derivatives 13.1 Definitions fx(a, b) = lim h→0 f(a+ h, b)− f(a, b) h fy(a, b) = lim h→0 f(a, b+ h)− f(a, b) h So basically just take the derivative of one (the subscript) given that the other one is a constant. 13.2 Notation for Higher-Order Partial Derivatives ∂ ∂x ( ∂f ∂x ) = ∂2f ∂x2 = (fx)x = fxx ∂ ∂y ( ∂f ∂y ) = ∂2f ∂y2 = (fy)y = fyy ∂ ∂x ( ∂f ∂y ) = ∂2f ∂x∂y = (fy)x = fyx ∂ ∂y ( ∂f ∂x ) = ∂2f ∂y∂x = (fx)y = fxy Note: fxy = fyx for nice functions. 13.3 Differentiability Suppose the function f has partial derivatives fx and fy defined on an open region containing (a, b), with fx and fy continuous at (a, b). Then f is differen- tiable at (a, b). This also implies that it is continuous at (a, b). 7 17.2 Critical Points A critical point exists if either • fx(a, b) = fy(a, b) = 0 • one (or both) of fx or fy does not exist at (a, b) 17.3 Second Derivative Test Let D(x, y) = fxxfyy − f2xy • If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum at (a, b). • If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum at (a, b). • If D(a, b) < 0, then f has a saddle point at (a, b). • If D(a, b) = 0, then the test is inconclusive. 17.4 Absolute Maximum/Minimum Values Let f be continuous on a closed bounded set R in R2. To find absolute maximum and minimum values of f on R: 1. Determine the values of f at all critical points in R. 2. Find the maximum and minimum values of f on the boundary of R. 3. The greatest function value found in Steps 1 and 2 is the absolute maxi- mum value of f on R, and the least function value found in Steps 1 and 2 is the absolute minimum values of f on R. 18 Double Integrals 18.1 Double Integrals on Rectangular Regions Let f be continuous on the rectangular region R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}. The double integral of f over R may be evaluated by either of two iterated integrals: ∫∫ R f(x, y) dA = ∫ d c ∫ b a f(x, y) dx dy = ∫ b a ∫ d c f(x, y) dy dx 10 18.2 Double Integrals over Nonrectangular Regions Let R be a region bounded below and above by the graphs of the continuous functions y = g(x) and y = h(x), respectively, and by the lines x = a and x = b. If f is continuous on R, then∫∫ R f(x, y) dA = ∫ b a ∫ h(x) g(x) f(x, y) dy dx Let R be a region bounded on the left and right by the graphs of the continuous functions x = g(y) and x = h(y), respectively, and by the lines y = c and y = d. If f is continuous on R, then∫∫ R f(x, y) dA = ∫ d c ∫ h(y) g(y) f(x, y) dx dy 18.3 Areas of Regions by Double Integrals area of R = ∫∫ R dA 19 Polar Double Integrals 19.1 Double Integrals over Polar Rectangular Regions Let f be continuous on the region in the xy-plane R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β}, where β − α ≤ 2π. Then∫∫ R f(r, θ) dA = ∫ β α ∫ b a f(r, θ) r dr dθ 19.2 Double Integrals over More General Polar Regions Let f be continuous on the region in the xy-plane R = {(r, θ) : 0 ≤ g(θ) ≤ r ≤ h(θ), α ≤ θ ≤ β} where β − α ≤ 2π. Then.∫∫ R f(r, θ) dA = ∫ β α ∫ h(θ) g(θ) f(r, θ) r dr dθ If f is nonnegative on R, the double integral gives the volume of the solid bounded by the surface z = f(r, θ) and R. 11 19.3 Area of Polar Regions A = ∫∫ R dA = ∫ β α ∫ h(θ) g(θ) r dr dθ 20 Triple Integrals Let D = {(x, y, z) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x), G(x, y) ≤ z ≤ H(x, y)}, where g, h, G, H are continuous functions. The triple integral of a continuous function f on D is evaluated as the iterated integral∫∫∫ D f(x, y, z) dV = ∫ b a ∫ h(x) g(x) ∫ H(x,y) G(x,y) f(x, y, z) dz dy dx 21 Cylindrical and Spherical Coordinates 21.1 Definitions 21.1.1 Cylindrical Coordinates (r, θ, z) An extension of polar coordinates into R3. Simply add a z component. 21.1.2 Spherical Coordinates (ρ, ϕ, θ) • ρ is the distance from the origin to a point P . • ϕ is the angle between the positive z-axis and the line OP . • θ is the same angle as in cylindrical coordinates; it measure rotation about the z-axis relative to the positive x-axis. 21.2 Rectangular to Cylindrical r2 = x2 + y2 tan θ = y x z = z 21.3 Cylindrical to Rectangular x = r cos θ y = r sin θ z = z 12 24.3 Line Integrals of Vector Fields 24.3.1 Definition Let F be a vector field that is continuous on a region containing a smooth oriented curve C parametrized by arc length. Let T be the unit tangent vector at each point of C consistent with the orientation. The line integral of F over C is ∫ C F ·T ds. 24.3.2 Different Forms F = 〈f, g, h〉 and C has a parametrization r(t) = 〈x(t), y(t), z(t)〉, for a ≤ t ≤ b∫ b a F ·r′(t) dt = ∫ b a (fx′(t)+gy′(t)+hz′(t)) dt = ∫ C f dx+g dy+h dz = ∫ C F ·dr For line integrals in the plane, we let F = 〈f, g〉 and assume C is parametrized in the form r(t) = 〈x(t), y(t)〉, for a ≤ t ≤ b. Then∫ C F ·T ds = ∫ b a (fx′(t) + gy′(t)) dt = ∫ C f dx+ g dy = ∫ C F · dr 24.4 Work F is a force field W = ∫ C F ·T ds = ∫ b a F · r′(t) dt 24.5 Circulation F is a vector field Circulation = ∫ C F ·T ds 24.6 Flux Flux = ∫ C F · n ds = ∫ b a (fy′(t)− gx′(t)) dt n = T× k, and a positive answer means a positive outward flux. 25 Conservative Vector Fields 25.1 Test for Conservative Vector Field Let F = 〈f, g, h〉 be a vector field defined on a connected and simply connected region D of R3, where f , g, and h have continuous first partial derivatives on 15 D. Then, F is a conservative vector field on D (there is a potential function ϕ such that F = ∇ϕ) if and only if • ∂f ∂y = ∂g ∂x • ∂f ∂z = ∂h ∂x • ∂g ∂z = ∂h ∂y For vector fields in R2, we have the single condition ∂f ∂y = ∂g ∂x . 25.2 Finding Potential Functions Suppose F = 〈f, g, h〉 is a conservative vector field. To find ϕ such that F = ∇ϕ, take the following steps: 1. Integrate ϕx = f with respect to x to obtain ϕ, which includes an arbitrary function c(y, z. 2. Compute ϕy and equate it to g to obtain an expression for cy(y, z). 3. Integrate cy(y, z) with respect to y to obtain c(y, z), including an arbitrary function d(z). 4. Compute ϕz and equate it to h to get d(z). Beginning the procedure with ϕy = g or ϕz = h may be easier in some cases. This method can also be used to check if a vector field is conservative by seeing if there is a potential function. 25.3 Fundamental Theorem for Line Integrals∫ C F ·T ds = ∫ C F · dr = ϕ(B)− ϕ(A) 25.4 Line Integrals on Closed Curves Let R in R2 (or D in R3) be an open region. Then F is a conservative vector field on R if and only if ∮ C F ·dr = 0 on all simple closed smooth oriented curves C in R. 26 Green’s Theorem 26.1 Circulation Form∮ C F · dr = ∮ C f dx+ g dy = ∫∫ R ( ∂g ∂x − ∂f ∂y ) dA 16 26.2 Area of a Plane Region by Line Integrals∮ C x dy = − ∮ C y dx = 1 2 ∮ C (x dy − y dx) 26.3 Flux Form∮ C F · n ds = ∮ C f dy − g dx = ∫∫ R ( ∂f ∂x + ∂g ∂y ) dA 27 Divergence and Curl 27.1 Divergence of a Vector Field div(F) = ∇ · F = ∂f ∂x + ∂g ∂y + ∂h ∂z 27.2 Divergence of Radial Vector Fields div(F) = 3− p |r|p F = r |r|p = 〈x, y, z〉 (x2 + y2 + z2)p/2 27.3 Curl curl(F) = ∇× F Just derive the curl by doing the cross product. 27.4 Divergence of the Curl ∇ · (∇× F) = 0 28 Surface Integrals 28.1 Parameterization 28.1.1 z is Explicitly Defined Use x = x, y = y, and since z is explicitly defined, you already have what z equals. 28.1.2 Cylinder Simply use cylindrical coordinates to parameterize the surface in terms of θ and z. 17