Calc3 cheat sheet onesheet - University of Utah

Derivatives

x= x

Dxe e

sin( ) = cos( )

Dx x

x

cos( ) = sin( )

Dx

x tan( )

=

sec2 (

x )

Dx x cot( ) =

csc2x( )

Dx x

x

sec( ) = sec( ) tan( )

Dx x

xx

csc( ) = csc( ) cot( )

Dx x

xx

sin Dx

1

=

p1

1

2 x

,

x

2

[

1 1] ,

cos Dx

1

=

p

1

1 2,x 2 [

x

1 1] ,

tan Dx

1

=

1 1+

2,

2

x

2

x

sec 1 = p1

|| 1

Dx

| | 2 1, x >

xx

sinh( ) = cosh( )

Dx

x

x

cosh( ) = sinh( )

Dx

x

tanh( ) =

2x( )

Dx

x sech

coth( ) =

2x(

)

Dx

x

csch x

( )=

( ) tanh( )

Dxsech x

sech x

x

( )=

( ) coth( )

Dxcsch x

csch x

x

sinh 1 = p 1

Dx

2 +1

x

cosh 1 = p 1

1

Dx

2 1,x >

x

tanh Dx

1= 112

1

1

~v

= <

v1, v2,

v3

>

^^^

ijk

= ~u ~v

u1

u2

u3

v1 v2 v3

~u ~v = ~0 means the vectors are paralell

Lines and Planes

Equation of a Plane (x0, y0, z0) is a point on the plane and

is a normal vector < A, B, C >

A(x x0) + B(y y0) + C(z z0) = 0

< A, B, C > ? < x

x0, y

y0, z

z0

= >

0

+ + = where

Ax By Cz D

D = Ax0 + By0 + Cz0

Equation of a line

A line requires a Direction Vector

~u =< u1, u2, u3 > and a point (x1, y1, z1) then,

a parameterization of a line could be:

x = u1t + x1 y = u2t + y1 z = u3t + z1

Distance from a Point to a Plane

The distance from a point (x0, y0, z0) to a plane Ax+By+Cz=D can be expressed

by the formula:

= |Axp0+By0+Cz0 D|

d

2+ 2+ 2

ABC

Coord Sys Conv

Cylindrical to Rectangular

= cos( )

xr

= sin( )

yr

=

zz

Recp tangular to Cylindrical

=

2+ 2

r xy

tan( ) = y

=

x

zz

Spherical to Rectangular

= sin( ) cos( )

x

= sin( ) sin( )

y

= cos( )

z

Recp tangular to Spherical

=

2+ 2+ 2

xyz

tan( ) = y

cos( ) = px z

2+ 2+ 2 xyz

Spherical to Cylindrical

= sin( ) r

=

= cos( ) z

Cylipndrical to Spherical = 2+ 2

rz =

cos( ) = p z

2+ 2 rz

Surfaces

Ellipsoid

2 x

2

+

2

y 2

+

2 z

2

=1

a

b

c

Hyperboloid of One Sheet

2 x2

+

2

y 2

2 z2

=1

(aMajorb Axisc: z because it follows - )

Hyperboloid of Two Sheets

2 z2

2 x2

2

y 2

=1

(cMajoar Axisb: Z because it is the one not

subtracted)

Elliptic Paraboloid

z

=

2 x2

+

2

y 2

(Majaor Axbis: z because it is the variable

NOT squared)

Hyperbolic Paraboloid

(Major Axis: Z axis because it is not

squared)

2

z

=

y 2

2 x2

b

a

Elliptic Cone

(Major Axis: Z axis because it's the only

one being subtracted)

2 x2

+

2

y 2

a

b

2 z2

=0

c

Cylinder

1 of the variables is missing

OR

(

)2 + (

2) =

xa yb c

(Major Axis is missing variable)

Partial Derivatives

Partial Derivatives are simply holding all other variables constant (and act like constants for the derivative) and only taking the derivative with respect to a given variable.

Given z=f(x,y), the partial derivative of

z with respect to x is:

(

)=

=

@z

=

() @f x,y

fx x, y zx

likewise for parti@axl with r@exspect to y:

(

)=

=

@z

=

() @f x,y

fy x, y zy

@y

@y

Notation

For

, work "inside to outside"

fxyy

fx

then , then

fxy

fxyy

3

= fxyy

@f

2,

@x@ y

3

For

@f 2

, work right to left in the

@x@ y

denominator

Gradients

The Gradient of a function in 2 variables is r =

f < fx, fy > The Gradient of a function in 3 variables is r =

f < fx, fy , fz >

Chain Rule(s)

Take the Partial derivative with respect

to the first-order variables of the

function times the partial (or normal)

derivative of the first-order variable to

the ultimate variable you are looking for

summed with the same process for other

first-order variables this makes sense for.

Example:

let x = x(s,t), y = y(t) and z = z(x,y).

z then has first partial derivative:

@z and @z

@x

@y

x has the partial derivatives:

@x and @x

a@nsd y ha@stthe derivative:

dy

Idnt this case (with z containing x and y as well as x and y both containing s and t), the chain rule for @z is @z = @z @x

@s @s @x @s

The chain rule for @z is

@t

@z = @z @x + @z dy

@t @x @t @y dt

Note: the use of "d" instead of " " with @

the function of only one independent variable

Limits and Continuity

Limits in 2 or more variables

Limits taken over a vectorized limit just

evaluate separately for each component

of the limit.

Strategies to show limit exists

1. Plug in Numbers, Everything is Fine

2. Algebraic Manipulation

-factoring/dividing out

-use trig identites

3. Change to polar coords

( ) ! (0 0) , ! 0

if x, y

,

r

Strategies to show limit DNE

1. Show limit is dierent if approached

from dierent paths (x=y, = 2 etc.)

x y, 2. Switch to Polar coords and show the

limit DNE.

Continunity

A fn, = ( ), is continuous at (a,b) z f x, y

if

( f a,

) b

=

lim(x,y)!(a,b)

( f x,

) y

Which means:

1. The limit exists

2. The fn value is defined

3. They are the same value

Directional Derivatives

Let z=f(x,y) be a fuction, (a,b) ap point

in the domain (a valid input point) and

^ a unit vector (2D). u The Directional Derivative is then the

derivative at the point (a,b) in the

direction of ^ or:

u

( )= ^?r ( )

Du~ f a, b u f a, b

This will return a

. 4-D version:

scalar

(

)= ^?r (

)

Du~ f a, b, c u f a, b, c

Tangent Planes

let F(x,y,z) = k be a surface and P =

(x0, y0, z0) be a point on that surface. Equation of a Tangent Plane:

r F

(x0 ,

y0 ,

z0 )?

<

x

x0, y

y0, z

z0 >

Approximations

let = ( ) be a dierentiable z f x, y

function total dierential of f = dz

dz = rf ? < dx, dy >

This is the

change in z

approximate

The actual change in z is the dierence

in z values:

= zz

z1

Maxima and Minima

Internal Points

1. Take the Partial Derivatives with

respect to X and Y ( and ) (Can use

fx

fy

gradient)

2. Set derivatives equal to 0 and use to

solve system of equations for x and y

3. Plug back into original equation for z.

Use Second Derivative Test for whether

points are local max, min, or saddle

Second Partial Derivative Test 1. Find all (x,y) points such that

rf

( x,

) y

=

~0

2. Let = ( ) (

)

2( )

D fxx x, y fyy x, y

f x, y

xy

IF (a) D 0 AND

0 f(x,y) is

>

fxx < ,

local max value

(b) D 0 AND ( ) 0 f(x,y) is

>

fxx x, y >

local min value

(c) D 0, (x,y,f(x,y)) is a saddle point <

(d) D = 0, test is inconclusive

3. Determine if any boundary point

gives min or max. Typically, we have to

parametrize boundary and then reduce

to a Calc 1 type of min/max problem to

solve.

The following only apply only if a

boundary is given 1. check the corner points 2. Check each line (0 x 5 would give x=0 and x=5 ) On Bounded Equations, this is the global min and max...second derivative test is not needed.

Lagrange Multipliers

Given a function f(x,y) with a constraint

g(x,y), solve the following system of

equations to find the max and min

points on the constraint (NOTE: may

need to also find internal points.):

r=r

f

g

( ) = 0(

)

g x, y

orkif given

Double Integrals

With Respect to the xy-axis, if taking an iRntRegral, R R dydx is cutting in vertical rectangles,

is cutting in horizontal dxdy rectangles

Polar Coordinates When using polar coordinates,

= dA rdrd

Surface Area of a Curve

let z = f(x,y) be continuous over S (a

closed Region in 2D domain)

Then the surface area of z = f(x,y) over

S is: R R q

=

2+ 2+1

SA

f f dA

Sx y

TR RriRple Integrals

(

)=

R

a2 a1

Rs

f

2

x,

() x

yR ,

1 (x)

z dv

2

() x,y

1 (x,y)

(

)

f x, y, z dzdydx

Note: can be exchanged for

in

dv

dxdydz

any order, but you must then choose

your limits of integration according to

that order

JR aRcobian Method

R

RG

( ( ) ( ))| ( )| f g u, v , h u, v J u, v dudv

=

()

f x, y dxdy

R

@x @x

( ) = @u @v

J u, v

@y @y

@u @v

Common Jacobians: Rect. to Cylindrical: Rect. to Spherical: 2rsin( )

Vector Fields

let (

) be a scalar field and

f x, y, z

~(

)=

F x, y, z

(

)^ + (

)^ + (

)^ be

M x, y, z i N x, y, z j P x, y, z k

a vector field,

Grandient of f = r = @f @f @f f< , , >

@x @y @z

Divergence of ~ : F

r ? ~ = @M + @N + @P

F

@x

@y

@z

Curl of ~ : F^ ^ ^

i jk

r ~ = @

@

@

F

@x @y @z

MN P

Line Integrals

CR

given

by

x

=R

() xt,

y

=

() yt,

t

2

[ a,

] b

c

f

( x,

) y ds

q=

b ( ( ) ( )) f x t , y t ds

a

where = ( dx )2 + ( dy )2

q ds

dt

dt

dt

or 1 + ( dy )2

q

dx

dx

or 1 + ( dx )2 dy

dy

To evaluate a Line Integral,

? get a paramaterized version of the line

(usually in terms of t, though in

exclusive terms of x or y is ok)

? evaluate for the derivatives needed

(usually dy, dx, and/or dt)

? plug in to original equation to get in

terms of the independant variable

? solve integral

Work

Let ~ = ^ + ^ + ^ (force)

F Mi j k

=(

) =(

)=

M M x, y, z , N N x, y, z , P

(

)

P x, y, z

(Literally)dR~r

=

^ dxi

+

^ dyj

+

^ dzk

Work = ~ ?

w F d~r

(Work done cby moving a particle over

curve C with force F~ )

Independence of Path

Fund Thm of Line Integrals

C is curve given by ( ) 2 [ ];

~r t , t a, b

0( ) exists. If ( ) is continuously

~r t

f ~r

dierentRiable on an open set containing

C, then r ( ) ? = (~) ( )

f ~r d~r f b f ~a

Equivalecnt Conditions

~ ( ) continuous on open connected set F ~r D. Then,

( ) ~ = r for some fn f. (if ~ is

aF f

F

conserRvative)

,

() b

Rc

F~

() ~r

?

d~risindep.of pathinD

, ( ) ~ ( ) ? = 0 for all closed paths

c F ~r d~r

in D. c

Conservation Theorem ~ = ^ + ^ + ^ continuously F Mi Nj Pk dierentiable on open, simply connected

set D.

F~ conservative , r F~ = ~0

(in

2D

r

~ F

=

~0

i

My

=

) Nx

Green's Theorem

(method of changing line integral for

double integral - Use for Flux and

Circulation across 2D curve and line

iHntegrals over a clRosRed boundary)

H M dy

N dx

=

R

RR

( Mx

+

) Ny dxdy

+

=

(

)

M dx N dy

Nx My dxdy

Let:

R

?R be a region in xy-plane

?C is simple, closed curve enclosing R

(w/ paramerization ( )) ~r t

? ~ ( ) = ( )^ + ( )^ be F x, y M x, y i N x, y j continuously dierentiable over R[C.

Form 1: Flux Across Boundary

~nH = unit nRorRmal vector to C

c F~H ? ~n = R r ? F~RdRA

,

=

( +)

M dy N dx

Mx Ny dxdy

Form 2: Circulation RAlong

BH oundary R R

c F~H ? d~r =

R

r

RF~R

?

^ udA

,

+

=

(

M dx N dy

Nx

AreaH =

A

of R

(

1

2 ydx

+

R

1 2

) xdy

) My dxdy

Gauss' Divergence Thm

(3D Analog of Green's Theorem - Use for Flux over a 3D surface) Let:

?F~

( x,

y,

) z

be

vector

field

continuously

dierentiable in solid S

?S is a 3D solid ?@S boundary of S (A Surface)

?^unit outer normal to

n

@S

TR hRen,

RRR

~(

)?^ =

r? ~

F x, y, z ndS

F dV

(dV@S= dxdydz)

S

Surface Integrals

Let

?R be closed, bounded region in xy-plane

?f be a fn with first order partial

derivatives on R

?G be a surface over R given by

=( )

z f x, y

?(

)= (

( )) is cont. on R

g x, y, z g x, y, f x, y

TR hRen,

R

RG

(

)

g x, y, z dS

=

R

( g x,

y,

fq(x,

)) y dS

where = 2 + 2 + 1 dS f f dydx

xy

FR lRux

of

~ F

across

G

R RG F~ ? ndS =

[

+]

M fx N fy P dxdy

wheRre:

?~(

)=

F x, y, z

(

)^ + (

)^ + (

)^

M x, y, z i N x, y, z j P x, y, z k

?G is surface f(x,y)=z

? is upward unit normal on G. ~n ?f(x,y) has continuous 1st order partial

derivatives

Unit Circle

(cos, sin)

Op thepr Information

pa = a

b

Wbhepre a Cone is defined as

= ( 2 + 2)

z ax y ,

In Sphericaql Coordinates,

= cos 1(

a 1+a

)

Right Circular Cylinder:

=2

= 2+2

V r h, SA r rh

lim !inf (1 +

n

m )pn

=

mp

e

Law of Cosinesn:

2 = 2 + 2 2 (cos( )) a b c bc

Stokes Theorem

Let:

?S be a 3D surface

?~(

)=

F x, y, z

(

)^ + (

)^ + (

)^

M x, y, z i N x, y, z j P x, y, z l

?M,N,P have continuous 1st order partial

derivatives

?C is piece-wise smooth, simple, closed,

curve, positively oriented

? ^ is unit tangent vector to C.

T

TH hen, R RF~c ?

^ T dS

=

R

R (r

s

~) F

?

^ ndS

=

(r ~ ) ?

F ~ndxdy

RH emRember: R

~?~ = (

+

+)

F T ds

M dx N dy P dz

c

Originally Written By Daniel Kenner for MATH 2210 at the University of Utah. Source code available at CheatSheet Thanks to Kelly Macarthur for Teaching and Providing Notes.

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