Derivatives
Dxex=ex
Dxsin(x)=cos(x)
Dxcos(x)=sin(x)
Dxtan(x)=sec
2(x)
Dxcot(x)=csc2(x)
Dxsec(x)=sec(x)tan(x)
Dxcsc(x)=csc(x)cot(x)
Dxsin1=1
p1x2,x2[1,1]
Dxcos1=1
p1x2,x2[1,1]
Dxtan1=1
1+x2,⇡
2x⇡
2
Dxsec1=1
|x|px21,|x|>1
Dxsinh(x)=cosh(x)
Dxcosh(x)=sinh(x)
Dxtanh(x)=sech2(x)
Dxcoth(x)=csch2(x)
Dxsech(x)=sech(x)tanh(x)
Dxcsch(x)=csch(x)coth(x)
Dxsinh1=1
px2+1
Dxcosh1=1
px21,x> 1
Dxtanh1=1
1x21<x<1
Dxsech1=1
xp1x2,0<x<1
Dxln(x)=1
x
Integrals
R1
xdx = ln |x|+c
Rexdx =ex+c
Raxdx =1
lnaax+c
Reaxdx =1
aeax +c
R1
p1x2dx = sin1(x)+c
R1
1+x2dx = tan1(x)+c
R1
xpx21dx =sec
1(x)+c
Rsinh(x)dx = cosh(x)+c
Rcosh(x)dx = sinh(x)+c
Rtanh(x)dx = ln |cosh(x)|+c
Rtanh(x)sech(x)dx =sech(x)+c
Rsech2(x)dx = tanh(x)+c
Rcsch(x)coth(x)dx =csch(x)+c
Rtan(x)dx =ln|cos(x)|+c
Rcot(x)dx = ln |sin(x)|+c
Rcos(x)dx = sin(x)+c
Rsin(x)dx =cos(x)+c
R1
pa2u2dx = sin1(u
a)+c
R1
a2+u2dx =1
atan1u
a+c
Rln(x)dx =(xln(x)) x+c
U-Substitution
Let u=f(x)(canbemorethanone
variable).
Determine: du =f(x)
dx dx and solve for
dx.
Then, if a definite integral, substitute
the bounds for u=f(x)ateachbounds
Solve the integral using u.
Integration by Parts
Rudv =uv Rvdu
Fns and Identities
sin(cos1(x)) = p1x2
cos(sin1(x)) = p1x2
sec(tan1(x)) = p1+x2
tan(sec1(x))
=(
px21ifx1)
=(px21if x 1)
sinh1(x)=lnx+px2+1
sinh1(x)=lnx+px21,x1
tanh1(x)=1
2lnx+1+x
1x,1<x<1
sech1(x)=ln[
1+p1x2
x],0<x1
sinh(x)=exex
2
cosh(x)=ex+ex
2
Trig Identities
sin2(x)+cos
2(x)=1
1+tan
2(x)=sec
2(x)
1+cot
2(x)=csc
2(x)
sin(x±y)=sin(x)cos(y)±cos(x) sin(y)
cos(x±y)=cos(x)cos(y)±sin(x) sin(y)
tan(x±y)= tan(x)±tan(y)
1⌥tan(x)tan(y)
sin(2x)=2sin(x) cos(x)
cos(2x)=cos
2(x)sin2(x)
cosh(n2x)sinh2x=1
1+tan
2(x)=sec
2(x)
1+cot
2(x)=csc
2(x)
sin2(x)=1cos(2x)
2
cos2(x)=1+cos(2x)
2
tan2(x)=1cos(2x)
1+cos(2x)
sin(x)=sin(x)
cos(x)=cos(x)
tan(x)=tan(x)
Calculus 3 Concepts
Cartesian coords in 3D
given two points:
(x1,y
1,z
1)and(x2,y
2,z
2),
Distance between them:
p(x1x2)2+(y1y2)2+(z1z2)2
Midpoint:
(x1+x2
2,y1+y2
2,z1+z2
2)
Sphere with center (h,k,l) and radius r:
(xh)2+(yk)2+(zl)2=r2
Vectors
Vect or: ~u
Unit Vector: ˆu
Magnitude: ||~u|| =qu2
1+u2
2+u2
3
Unit Vector: ˆu=~u
||~u||
Dot Product
~u·~v
Produces a Scalar
(Geometrically, the dot product is a
vector pro jection)
~u=<u
1,u
2,u
3>
~v=<v
1,v
2,v
3>
~u·~v=~
0 means the two vectors are
Perpendicular ✓is the angle between
them.
~u·~v=||~u|| ||~v|| cos(✓)
~u·~v=u1v1+u2v2+u3v3
NOTE:
ˆu·ˆv=cos(✓)
||~u||2=~u·~u
~u·~v= 0 when ?
Angle Between ~uand ~v:
✓= cos1(~u·~v
||~u|| ||~v|| )
Projection of ~uonto ~v:
pr~v~u=( ~u·~v
||~v||2)~v
Cross Product
~u⇥~v
Produces a Vector
(Geometrically, the cross product is the
area of a paralellogram with sides ||~u||
and ||~v||)
~u=<u
1,u
2,u
3>
~v=<v
1,v
2,v
3>
~u⇥~v=
ˆ
iˆ
jˆ
k
u1u2u3
v1v2v3
~u⇥~v=~
0 means the vectors are paralell
Lines and Planes
Equation of a Plane
(x0,y
0,z
0) is a point on the plane and
<A,B,C>is a normal vector
A(xx0)+B(yy0)+C(zz0)=0
<A,B,C>·<xx0,yy0,zz0>=0
Ax +By +Cz =Dwhere
D=Ax0+By0+Cz0
Equation of a line
AlinerequiresaDirectionVector
~u=<u
1,u
2,u
3>and a point
(x1,y
1,z
1)
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,y
0,z
0)to
a plane Ax+By+Cz=D can be expressed
by the formula:
d=|Ax0+By0+Cz0D|
pA2+B2+C2
Coord Sys Conv
Cylindrical to Rectangular
x=rcos(✓)
y=rsin(✓)
z=z
Rectangular to Cylindrical
r=px2+y2
tan(✓)=y
x
z=z
Spherical to Rectangular
x=⇢sin()cos(✓)
y=⇢sin()sin(✓)
z=⇢cos()
Rectangular to Spherical
⇢=px2+y2+z2
tan(✓)=y
x
cos()= z
px2+y2+z2
Spherical to Cylindrical
r=⇢sin()
✓=✓
z=⇢cos()
Cylindrical to Spherical
⇢=pr2+z2
✓=✓
cos()= z
pr2+z2
Surfaces
Ellipsoid
x2
a2+y2
b2+z2
c2=1
Hyperboloid of One Sheet
x2
a2+y2
b2z2
c2=1
(Major Axis: z because it follows - )
Hyperboloid of Two Sheets
z2
c2x2
a2y2
b2=1
(Major Axis: Z because it is the one not
subtracted)
Elliptic Paraboloid
z=x2
a2+y2
b2
(Major Axis: z because it is the variable
NOT squared)
Hyperbolic Paraboloid
(Major Axis: Z axis because it is not
squared)
z=y2
b2x2
a2
Elliptic Cone
(Major Axis: Z axis because it’s the only
one being subtracted)
x2
a2+y2
b2z2
c2=0
Cylinder
1 of the variables is missing
OR
(xa)2+(yb2)=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
zwithrespecttoxis:
fx(x,y)=zx=@z
@x=@f(x,y)
@x
likewise for partial with respect to y:
fy(x,y)=zy=@z
@y=@f(x,y)
@y
Notation
For fxyy, work ”inside to outside” fx
then fxy,thenfxyy
fxyy =@3f
@x@2y,
For @3f
@x@2y, work right to left in the
denominator
Gradients
The Gradient of a function in 2 variables
is rf=<f
x,f
y>
The Gradient of a function in 3 variables
is rf=<f
x,f
y,f
z>
Chain Rule(s)
Take the Partialderivative 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).
zthenhasfirstpartialderivative:
@z
@xand @z
@y
xhasthepartialderivatives:
@x
@sand @x
@t
and y has the derivative:
dy
dt
In this case (with z containing x and y
as well as x and y both containing s and
t), the chain rule for @z
@sis @z
@s=@z
@x
@x
@s
The chain rule for @z
@tis
@z
@t=@z
@x
@x
@t+@z
@y
dy
dt
Note: the use of ”d” instead of ”@” with
the function of only one independent
varia ble
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
if(x, y)!(0,0) ,r!0
Strategies to show limit DNE
1. Show limit is di↵erent if approached
from di↵erent paths
(x=y, x=y2,etc.)
2. Switch to Polar coords and show the
limit DNE.
Continunity
Afn,z=f(x,y), is continuous at (a,b)
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
ˆua unit vector (2D).
The Directional Derivative is then the
derivative at the point (a,b) in the
direction of ˆuor:
D~uf(a, b)=ˆu·rf(a,b)
This will return a scalar. 4-D version:
D~uf(a, b, c)=ˆu·rf(a,b, c)
Tangent Planes
let F(x,y,z) = k be a surface and P =
(x0,y
0,z
0)beapointonthatsurface.
Equation of a Tangent Plane:
rF(x0,y
0,z
0)·<xx0,yy0,zz0>
Approximations
let z=f(x,y)beadi↵erentiable
function total di↵erential of f = dz
dz =rf·<dx,dy>
This is the approximate change in z
The actual change in z is the di↵erence
in z values:
z=zz1
Maxima and Minima
Internal Points
1. Take the Partial Derivatives with
respect to X and Y (fxand fy)(Canuse
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 Testfor 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 D=fxx(x, y )fyy(x, y)f2
xy(x, y)
IF (a) D >0ANDfxx<0,f(x,y) is
local max value
(b) D >0ANDfxx(x,y)>0 f(x,y) is
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 x5would
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.):
rf=rg
g(x,y)=0(orkif given)
Double Integrals
With Respect to the xy-axis, if taking an
integral,
RRdydx is cutting in vertical rectangles,
RRdxdy is cutting in horizontal
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:
SA =RR
Sqf2
x+f2
y+1dA
Triple Integrals
RRR
sf(x,y, z )dv =
Ra2
a1R2(x)
1(x)R 2(x,y)
1(x,y)f(x,y , z)dzdydx
Note: dv can be exchanged for dxdydz in
any order, but you must then choose
your limits of integration according to
that order
Jacobian Method
RR
Gf(g(u,v),h(u,v ))|J(u, v)|dudv =
RR
Rf(x,y)dxdy
J(u,v)=
@x
@u
@x
@v
@y
@u
@y
@v
Common Jacobians:
Rect. to Cylindrical: r
Rect. to Spherical: ⇢2sin()
Vector Fields
let f(x,y, z )beascalarfieldand
~
F(x,y, z )=
M(x,y, z )ˆ
i+N(x,y, z )ˆ
j+P(x,y, z )ˆ
kbe
a vector field,
Grandient of f = rf=<@f
@x,@f
@y,@f
@z>
Divergence of ~
F:
r·~
F=@M
@x+@N
@y+@P
@z
Curl of ~
F:
r⇥~
F=
ˆ
iˆ
jˆ
k
@
@x
@
@y
@
@z
MNP
Line Integrals
Cgivenbyx=x(t),y=y(t),t2[a, b]
Rcf(x,y)ds =Rb
af(x(t),y(t))ds
where ds =q(dx
dt )2+(dy
dt )2dt
or q1+(dy
dx)2dx
or q1+(dx
dy)2dy
To evaluatea 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 ~
F=Mˆ
i+ˆ
j+ˆ
k(force)
M=M(x,y, z ),N =N(x, y, z),P =
P(x,y, z )
(Literally)d~r=dxˆ
i+dyˆ
j+dzˆ
k
Work w=Rc~
F·d~r
(Work done by movinga partic le over
curve C with force ~
F)
Independence of Path
Fund T hm of Lin e Integ ral s
Ciscurvegivenby~r(t),t 2[a, b];
~r0(t) exists. If f(~r) is continuously
di↵erentiable on an open set containing
C, then Rcrf(~r)·d~r=f(~
b)f(~a)
Equivalent Conditions
~
F(~r) continuous on open connected set
D. Then,
(a)~
F=rffor some fn f. (if ~
Fis
conservative)
,(b)Rc~
F(~r)·d~risindep.ofpathinD
,(c)Rc~
F(~r)·d~r= 0 for all closed paths
in D.
Conservation Theorem
~
F=Mˆ
i+Nˆ
j+Pˆ
kcontinuously
di↵erentiable on open, simply connected
set D.
~
Fconservative ,r⇥~
F=~
0
(in 2D r⇥~
F=~
0i↵My=Nx)
Green’s Theorem
(method of changing line integral for
double integral - Use for Flux and
Circulation across 2D curve and line
integrals over a closed boundary)
HMdyNdx=RR
R(Mx+Ny)dxdy
HMdx+Ndy=RR
R(NxMy)dxdy
Let:
·Rbearegioninxy-plane
·C is simple, closed curve enclosing R
(w/ paramerization ~r(t))
·~
F(x,y)=M(x, y )ˆ
i+N(x,y)ˆ
jbe
continuously di↵erentiable overR[C.
Form 1 : Flu x Acr oss B ounda ry
~n= unit normal vector to C
Hc~
F·~n=RR
Rr·~
FdA
,HMdyNdx=RR
R(Mx+Ny)dxdy
Form 2 : Cir cul ati on Alon g
Boundary
Hc~
F·d~r=RR
Rr⇥~
F·ˆudA
,HMdx+Ndy=RR
R(NxMy)dxdy
Area of R
A=H(1
2ydx+1
2xdy)
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
di↵erentiable in solid S
·S is a 3D solid ·@Sboundary of S (A
Surface)
·ˆnunit outer normal to @S
Then,
RR
@S~
F(x,y, z )·ˆndS =RRR
Sr·~
FdV
(dV = dxdydz)
Surface Integrals
Let
·R be closed, bounded region in xy-plane
·fbeafnwithfirstorderpartial
derivatives on R
·GbeasurfaceoverRgivenby
z=f(x,y)
·g(x,y, z )=g(x, y, f(x, y )) is cont. on R
Then,
RR
Gg(x,y, z )dS =
RR
Rg(x,y, f (x, y))dS
where dS =qf2
x+f2
y+1dydx
Flux of ~
Facross G
RR
G~
F·ndS =
RR
R[MfxNfy+P]dxdy
where:
·~
F(x,y, z )=
M(x,y, z )ˆ
i+N(x,y, z )ˆ
j+P(x,y, z )ˆ
k
·G is surface f(x,y)=z
·~nis upward unit normal on G.
·f(x,y) has continuous 1st order partial
derivatives
Unit Circle
(cos, sin)
Other Information
pa
pb=pa
b
Where a Cone is defined as
z=pa(x2+y2),
In Spherical Coordinates,
= cos1(qa
1+a)
Right Circular Cylinder:
V=⇡r2h,S A =⇡r2+2⇡rh
limn!inf(1 + m
n)pn =emp
Law of Cosines:
a2=b2+c22bc(cos(✓))
Stokes Theorem
Let:
·S be a 3D surface
·~
F(x,y, z )=
M(x,y, z )ˆ
i+N(x,y, z )ˆ
j+P(x,y, z )ˆ
l
·M,N,P have continuous 1st order partial
derivatives
·C is piece-wise smooth, simple, closed,
curve, positively oriented
·ˆ
Tis unit tangent vector to C.
Then,
H~
Fc·ˆ
TdS=RR
s(r⇥~
F)·ˆndS =
RR
R(r⇥~
F)·~ndxdy
Remember:
H~
F·~
Tds=Rc(Mdx+Ndy+Pdz)
