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Multivariable Calculus Cheat Sheet Reduced

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Multivariable Calculus Cheat Sheet Reduced 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 <x< 1 For fxyy ,work”insidetooutside”fx D csc(x)= csc(x)cot(x) 2 1 x area of a paralellogram with sides ~u x − − || || then fxy ,thenfxyy 1 − 1 1 1+p1 x2 and ~v ) Dx sin− = ,x [ 1, 1] @3f 2 sech− (x)=ln[ x− ], 0 <x 1 || || p1 x 2 − ~u =<u1,u2,u3 > 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 =<v1,v2,v3 > + =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 =<fx,fy > − 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 =<fx,fy ,fz > − ± ± 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) <A,B,C>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 <x<1 2 − 2 <A,B,C> <x x0,y y0,z z0 >=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 <x<1 1+cot (x)=csc (x) summed with the same process for other xp1 x2 1 cos(2x) D = Ax + By + Cz − 2 0 0 0 1 sin (x)= − 2 first-order variables this makes sense for. Dx ln(x)= x 2 1+cos(2x) Example: cos (x)= 2 Equation of a line 2 1 cos(2x) let x = x(s,t), y = y(t) and z = z(x,y). − AlinerequiresaDirectionVector Elliptic Paraboloid Integrals tan (x)= 1+cos(2x) 2 x2 y zthenhasfirstpartialderivative: ~u =<u1,u2,u3 > 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(φ)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(φ)sin(✓) ! , ! variable). vector projection) z = ⇢ cos(φ) Strategies to show limit DNE f(x) ~u =<u1,u2,u3 > 1. Show limit is di↵erent if approached Determine: du = dx dx and solve for Rectangular to Spherical Cylinder ~v =<v ,v ,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(φp)= 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(φ) 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(φ) 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(φ)= 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).
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