1 Derivatives sec(tan (x)) = p1+x2 Projection of ~u onto ~v: Surfaces Given z=f(x,y), the partial derivative of 1 u~ ~v x x pr ~u =( · )~v zwithrespecttoxis: Dxe = e tan(sec (x)) ~v ~v 2 Ellipsoid @f(x,y) 2 || || 2 2 2 @z Dx sin(x)=cos(x) =(px 1ifx 1) x y z fx(x, y)=zx = @x = @x 2 + 2 + 2 =1 Dx cos(x)= sin(x) =( px2 1 if x 1) Cross Product a b c likewise for partial with respect to y: 2 1  ~u ~v @z @f(x,y) Dx tan(x)=sec (x) sinh (x)=lnx + px2 +1 fy (x, y)=zy = = 2 Produces⇥ a Vector @y @y Dx cot(x)= csc (x) 1 p 2 Notation sinh (x)=lnx + x 1,x 1 (Geometrically, the cross product is the Dx sec(x)=sec(x)tan(x) 1 1 1+x tanh (x)= ln x + , 1 Hyperboloid of One Sheet fxyy = 2 , x x 2 @x@ y 1 1 e e x2 y z2 Dx cos = ,x [ 1, 1] sinh(x)= 2 ~v = + =1 @3f p1 x2 2 a2 b2 c2 For ,workrighttoleftinthe ex+e x @x@2y 1 1 ⇡ ⇡ cosh(x)= (Major Axis: z because it follows - ) Dx tan = , x 2 ˆ ˆ ˆ denominator 1+x2 2   2 i j k 1 1 D sec = , x > 1 ~u ~v = u1 u2 u3 x 2 ⇥ x px 1 | | Trig Identities v v v Gradients | | 1 2 3 Dx sinh(x)=cosh(x) sin2(x)+cos2(x)=1 The Gradient of a function in 2 variables 2 2 ~ Dx cosh(x)= sinh(x) 1+tan (x)=sec (x) ~u ~v = 0meansthevectorsareparalell is f = 2 ⇥ r Dx tanh(x)=sech (x) 2 2 The Gradient of a function in 3 variables 2 1+cot (x)=csc (x) Dx coth(x)= csch (x) sin(x y)=sin(x)cos(y) cos(x)sin(y) Lines and Planes is f = ± ± Hyperboloid of Two Sheets r Dxsech(x)= sech(x)tanh(x) 2 cos(x y)=cos(x)cos(y) sin(x)sin(y) Equation of a Plane z2 x2 y Dxcsch(x)= csch(x)coth(x) ± tan(x) tan(y)± 2 2 2 =1 Chain Rule(s) ± (x ,y ,z ) is a point on the plane and c a b 1 1 tan(x y)= 1 tan(x) tan(y) 0 0 0 Dx sinh = ± ⌥ (Major Axis: Z because it is the one not Take the Partial derivative with respect px2+1 sin(2x)=2sin(x)cos(x) is a normal vector 1 1 2 2 subtracted) to the first-order variables of the Dx cosh = ,x > 1 cos(2x)=cos (x) sin (x) px2 1 2 2 function times the partial (or normal) cosh(n x) sinh x =1 A(x x0)+B(y y0)+C(z z0)=0 1 1 derivative of the first-order variable to Dx tanh = 1 =0 1 x2 1+tan (x)=sec (x) · the ultimate variable you are looking for 1 1 2 2 Ax + By + Cz = D where Dxsech = , 0 Integrals tan (x)= 1+cos(2x) 2 x2 y zthenhasfirstpartialderivative: ~u = and a point z = + 1 dx = ln x + c sin( x)= sin(x) a2 b2 @z and @z x | | (x1,y1,z1) @x @y exdx = ex + c cos( x)=cos(x) (Major Axis: z because it is the variable xhasthepartialderivatives: R then, x 1 x tan( x)= tan(x) NOT squared) @x and @x a dx = ln a a + c aparameterizationofalinecouldbe: @s @t R ax 1 ax and y has the derivative: e dx = e + c x = u1t + x1 R a dy 1 1 Calculus 3 Concepts dx = sin (x)+c y = u2t + y1 dt R 2 p1 x z = u3t + z1 In this case (with z containing x and y 1 1 Cartesian coords in 3D R 2 dx = tan (x)+c as well as x and y both containing s and 1+x given two points: @z @z @z @x 1 1 Distance from a Point to a Plane t), the chain rule for @s is @s = @x @s R dx =sec (x)+c (x1,y1,z1)and(x2,y2,z2), xpx2 1 The distance from a point (x ,y ,z )to The chain rule for @z is Distance between them: 0 0 0 Hyperbolic Paraboloid @t R sinh(x)dx = cosh(x)+c 2 2 2 a plane Ax+By+Cz=D can be expressed @z @z @x @z dy (x1 x2) +(y1 y2) +(z1 z2) (Major Axis: Z axis because it is not = + cosh(x)dx = sinh(x)+c by the formula: @t @x @t @y dt R Midpoint: squared) Note: the use of ”d” instead of ”@”with tanh(x)dx = ln cosh(x) + c Ax0+By0+Cz0 D 2 px1+x2 y1+y2 z1+z2 d = | | y x2 R | | ( , , ) 2 2 2 z = the function of only one independent tanh(x)sech(x)dx = sech(x)+c 2 2 2 pA +B +C b2 a2 R Sphere with center (h,k,l) and radius r: variable sech2(x)dx = tanh(x)+c R (x h)2 +(y k)2 +(z l)2 = r2 csch(x)coth(x)dx = csch(x)+c Coord Sys Conv Limits and Continuity R tan(x)dx = ln cos(x) + c Vectors Cylindrical to Rectangular Limits in 2 or more variables R cot(x)dx = ln sin(| x) +|c | | x = r cos(✓) Limits taken over a vectorized limit just R cos(x)dx = sin(x)+c Vector: ~u Elliptic Cone R Unit Vector:u ˆ y = r sin(✓) evaluate separately for each component sin(x)dx = cos(x)+c (Major Axis: Z axis because it’s the only of the limit. R 1 2 2 2 z = z 1 u Magnitude: ~u = u1 + u2 + u3 one being subtracted) dx = sin ( a )+c || || Strategies to show limit exists R pa2 u2 Rectangular to Cylindrical 2 2 2 Unit Vector:u ˆ = u~q x y z 1. Plug in Numbers, Everything is Fine 1 1 1 u u~ r = x2 + y2 2 + 2 2 =0 R 2 2 dx = a tan a + c || || a b c a +u tan(✓)= y 2. Algebraic Manipulation ln(x)dx =(xln(x)) x + c Dot Product p x -factoring/dividing out R z = z ~u ~v -use trig identites R · Spherical to Rectangular U-Substitution Produces a Scalar x = ⇢ sin()cos(✓) 3. Change to polar coords Let u = f(x)(canbemorethanone (Geometrically, the dot product is a if(x, y) (0, 0) r 0 y = ⇢ sin()sin(✓) ! , ! variable). vector projection) z = ⇢ cos() Strategies to show limit DNE f(x) ~u = 1. Show limit is di↵erent if approached Determine: du = dx dx and solve for Rectangular to Spherical Cylinder ~v = from di↵erent paths dx. 1 2 3 ⇢ = x2 + y2 + z2 1ofthevariablesismissing ~u ~v = ~0meansthetwovectorsare (x=y, x = y2, etc.) Then, if a definite integral, substitute tan(✓)= y OR Perpendicular· ✓ is the angle between x 2. Switch to Polar coords and show the the bounds for u = f(x)ateachbounds cos(p)= z (x a)2 +(y b2)=c them. 2 2 2 limit DNE. Solve the integral using u. px +y +z (Major Axis is missing variable) ~u ~v = ~u ~v cos(✓) Spherical to Cylindrical Continunity ~u · ~v = u|| v|| ||+||u v + u v r = ⇢ sin() Afn,z = f(x, y), is continuous at (a,b) Integration by Parts · 1 1 2 2 3 3 Partial Derivatives udv = uv vdu NOTE: ✓ = ✓ if uˆ vˆ = cos(✓) z = ⇢ cos() Partial Derivatives are simply holding all f(a, b)=lim(x,y) (a,b) f(x, y) R R ~u· 2 = ~u ~u Cylindrical to Spherical other variables constant (and act like Which means: ! Fns and Identities ~||u ||~v = 0 when· ⇢ = pr2 + z2 constants for the derivative) and only 1. The limit exists 1 sin(cos (x)) = p1 x2 Angle· Between ?~u and ~v: ✓ = ✓ taking the derivative with respect to a 2. The fn value is defined 1 2 1 u~ ~v z cos(sin (x)) = p1 x ✓ = cos ( · ) cos()= given variable. 3. They are the same value u~ ~v 2 2 || || || || pr +z

Directional Derivatives Double Integrals Work Surface Integrals Other Information ~ ˆ ˆ ˆ pa a With Respect to the xy-axis, if taking an Let F = Mi + j + k (force) Let = Let z=f(x,y) be a fuction, (a,b) ap point pb b in the domain (a valid input point) and integral, M = M(x, y, z),N = N(x, y, z),P = Rbeclosed,boundedregioninxy-plane Where a Cone is defined as · p uˆ aunitvector(2D). dydx is cutting in vertical rectangles, P (x, y, z) fbeafnwithfirstorderpartial z = a(x2 + y2), ˆ ˆ ˆ · The Directional Derivative is then the dxdy is cutting in horizontal (Literally)d~r = dxi + dyj + dzk derivatives on R In Spherical Coordinates, RR derivative at the point (a,b) in the rectangles Work w = F~ d~r GbeasurfaceoverRgivenby p 1 a RR c · = cos ( ) direction ofu ˆ or: (Work done by moving· a particle over z = f(x, y) 1+a D f(a, b)=ˆu f(a, b) Polar Coordinates R ~ g(x, y, z)=g(x, y, f(x, y)) is cont. on R Right Circularq Cylinder: u~ curve C with force F ) · 2 2 · r When using polar coordinates, Then, V = ⇡r h, SA = ⇡r +2⇡rh This will return a scalar.4-Dversion: m pn mp g(x, y, z)dS = limn inf (1 + ) = e Du~ f(a, b, c)=ˆu f(a, b, c) dA = rdrd✓ G ! n · r Independence of Path g(x, y, f(x, y))dS Law of Cosines: RRR a2 = b2 + c2 2bc(cos(✓)) Tangent Planes Surface Area of a Curve 2 2 Fund Thm of Line Integrals RRwhere dS = fx + fy +1dydx let z = f(x,y) be continuous over S (a let F(x,y,z) = k be a surface and P = Ciscurvegivenby~r(t),t [a, b]; ~ Stokes Theorem closed Region in 2D domain) 2 Flux of F acrossq G (x0,y0,z0)beapointonthatsurface. ~r (t)exists.Iff(~r)iscontinuously ~ Let: Then the surface area of z = f(x,y) over 0 F ndS = Equation of a Tangent Plane: di↵erentiable on an open set containing G · S be a 3D surface S is: [ Mfx Nfy + P ]dxdy F (x0,y0,z0) C, then f(~r) d~r = f(~b) f(~a) RRR ·F~ (x, y, z)= r · SA = f 2 + f 2 +1dA c r · where: · S x y Equivalent Conditions RR~ M(x, y, z)ˆi + N(x, y, z)ˆj + P (x, y, z)ˆl R F (x, y, z)= Approximations RR q F~ (~r)continuousonopenconnectedset · M,N,P have continuous 1st order partial Triple Integrals M(x, y, z)ˆi + N(x, y, z)ˆj + P (x, y, z)kˆ · let z = f(x, y)beadi↵erentiable D. Then, derivatives f(x, y, z)dv = G is surface f(x,y)=z function total di↵erential of f = dz s (a)F~ = f for some fn f. (if F~ is · C is piece-wise smooth, simple, closed, a2 2(x) 2(x,y) r ~n is upward unit normal on G. · dz = f RRR f(x, y, z)dzdydx conservative) · st curve, positively oriented r · a1 1(x) 1(x,y) f(x,y) has continuous 1 order partial This is the approximate change in z Note: dv can be exchanged for dxdydz in ~ · Tˆ is unit tangent vector to C. R R R (b) c F (~r) d~risindep.ofpathinD derivatives · The actual change in z is the di↵erence any order, but you must then choose , · Then, (c) F~ (~r) d~r = 0 for all closed paths in z values: your limits of integration according to , Rc · F~c TdSˆ = ( F~ ) ndSˆ = in D. · s r⇥ · z = z z1 that order R ( F~ ) ~ndxdy Conservation Theorem H R r⇥ ·RR F~ = Mˆi + Nˆj + P kˆ continuously Remember: Maxima and Minima Jacobian Method Unit Circle RRF~ Tds~ = (Mdx + Ndy + Pdz) di↵erentiable on open, simply connected · c Internal Points f(g(u, v),h(u, v)) J(u, v) dudv = set D. (cos, sin) G | | H R 1. Take the Partial Derivatives with f(x, y)dxdy F~ conservative F~ = ~0 RRR ,r⇥ respect to X and Y (fx and fy )(Canuse (in 2D F~ = ~0i↵ My = Nx) gradient) RR @x @x r⇥ J(u, v)= @u @v 2. Set derivatives equal to 0 and use to @y @y @u @v solve system of equations for x and y Green’s Theorem 3. Plug back into original equation for z. Common Jacobians: (method of changing line integral for Use Second Derivative Test for whether Rect. to Cylindrical: r 2 double integral - Use for Flux and points are local max, min, or saddle Rect. to Spherical: ⇢ sin() Circulation across 2D curve and line integrals over a closed boundary) Second Partial Derivative Test Vector Fields Mdy Ndx = (Mx + Ny )dxdy 1. Find all (x,y) points such that let f(x, y, z)beascalarfieldand R Mdx + Ndy = (N M )dxdy ~ ~ H RRR x y f(x, y)=0 F (x, y, z)= Let: r 2 ˆ ˆ ˆ 2. Let D = fxx(x, y)fyy(x, y) fxy (x, y) M(x, y, z)i + N(x, y, z)j + P (x, y, z)k be HRbearegioninxy-planeRR · IF (a) D > 0ANDfxx < 0, f(x,y) is avectorfield, Cissimple,closedcurveenclosingR local max value @f @f @f · Grandient of f = f =< @x , @y , @z > (w/ paramerization ~r(t)) (b) D > 0ANDf (x, y) > 0f(x,y)is r xx Divergence of F~ : F~ (x, y)=M(x, y)ˆi + N(x, y)ˆj be local min value · F~ = @M + @N + @P continuously di↵erentiable over R C. (c) D < 0, (x,y,f(x,y)) is a saddle point r · @x @y @z Form 1: Flux Across Boundary[ (d) D = 0, test is inconclusive Curl of F~ : ˆ ˆ ˆ ~n = unit normal vector to C 3. Determine if any boundary point i j k F~ ~n = FdA~ ~ @ @ @ c R gives min or max. Typically, we have to F = @x @y @z · r · r⇥ Mdy Ndx = (Mx + Ny )dxdy parametrize boundary and then reduce MN P ,H RR R Form 2: Circulation Along to a Calc 1 type of min/max problem to H RR Boundary solve. Line Integrals F~ d~r = F~ udAˆ The following only apply only if a Cgivenbyx = x(t),y = y(t),t [a, b] c · R r⇥ · b 2 Mdx + Ndy = (N M )dxdy boundary is given f(x, y)ds = f(x(t),y(t))ds ,H RR R x y 1. check the corner points c a Area of R dx 2 dy 2 H 1 1 RR 2. Check each line (0 x 5would Rwhere ds = (R ) +( ) dt A = ( 2 ydx + 2 xdy)   dt dt give x=0 and x=5 ) dyq or 1+( )2dx H On Bounded Equations, this is the dx Gauss’ Divergence Thm global min and max...second derivative or q1+(dx )2dy test is not needed. dy Toq evaluate a Line Integral, (3D Analog of Green’s Theorem - Use get a paramaterized version of the line for Flux over a 3D surface) Let: Lagrange Multipliers (usually· in terms of t, though in F~ (x, y, z)bevectorfieldcontinuously · Given a function f(x,y) with a constraint exclusive terms of x or y is ok) di↵erentiable in solid S g(x,y), solve the following system of evaluate for the derivatives needed S is a 3D solid @S boundary of S (A Originally Written By Daniel Kenner for · · · equations to find the max and min (usually dy, dx, and/or dt) Surface) MATH 2210 at the University of Utah. points on the constraint (NOTE: may plug in to original equation to get in nˆunit outer normal to @S Source code available at · · need to also find internal points.): terms of the independant variable Then, https://github.com/keytotime/Calc3 CheatSheet f = g solve integral F~ (x, y, z) ndSˆ = FdV~ Thanks to Kelly Macarthur for Teaching and r r · @S · S r · g(x, y)=0(orkifgiven) (dV = dxdydz) Providing Notes. RR RRR