January
Notes
9
Area between
Washer Methods
curves
f(x), gly);cont.
-
[a,b],
over
Yxc(a,b)
9(x),f (x),
:
Then,
area
between these
curves
=
i)
((+
A
=
-
(ix
=
3
-
+
Let f(x)
(a, (a)
d,f1b)), which
to
subintervals,
[a,b].
over
xo, X, a, Xn,
solid obtained
volume
between them
area
g(x)3dx
by revolving
of
around
axis, region
asset
Cylindrical Shells
54]! =
(around y-axis)
Step 1) divide (a,b] into
n
Ox-
equal subdivisions,
Step 2) calculate V ofrevolution by assuming heightis
of the
consider section of
curve
boundary values,
2
35 revolved subinterval has volume V. xy*
4:
=
find volume,
V
is
-
xX 3* k3*x(2x:*)
=
2x,x3dx
=
below
above x-axis,
#.
DV=ax!
n
Xi *-=X,
Xnx,
volume
mean
+
Step
total
approx.
y*= f(x), x*=(X: 4:-)/2
ie
Step
from
divide (a,b] into
call L. To find L,
we
Ilabelled
ox=
-
[a,b].049(x)(f(x)
bobetevereinin
Example:
be differential
=
=
5
-
Arc Length
-
over
by
=
x)2x
fle), g(x), continuous
x!(f(x)
DV
around X-axis
Examples find
jx3)! 4
-
+
g(x1]dx
-
=
=
(x 2 x
x
?(f(x)
(a,b] DA
over
consider
for xe[a,b]. To find volume/v) created by rotating
3=2
x2, o,5Jaround
ax?dx-afy
yaxis
syst
I
x=
I)
a,xnx b.
=
enough, I approximated by
small
is
ex
strc
k=
derived
derivation
result
examinable,
*
examinable
not
Disrence
then
Examples
exampletolength ofcareinCromas
-flan).
B
-
Letf(x)
x axis and
some
-
between
=
can
sphere: *=r2.
la,b). Consider region R,
on
xa
familiar solids
and
be
x
-
V,
given
Suppose within (X:.,
lowest and greatestvalues
↳ Hence,
↳
summing
im?x
over
<
a
washer
mi
over
y=rx,
volume
Alx),
X,and
f(x) <
M,
=
a
Xi.1.
/m.
,
M:are
region).
πMx- m
subintervals, V
iE,f(x*)WX, continued
=
revolved around y-axis, V
↳.
v
is
Integrating
distance from
=
Examples
volumes
cross
axis
+
use
2x (h
+
+
-
4)
-
x))dx
gradients==
triangles
2x() ( i).
-
-
+
-
+
+
-
didn'tsolve
Indy,
=
section
areas
rotation to the
of
revolution when
of
curve
revolving around y-axis
circle (art), where
of
curve
x, se, is rotated
=
y
around (a)x-axis, and (b) y-axis.
aex.htm.
-
from the
each
at
and
Example.
in
class, he
will postresults
d
is
cross
section
area:
in the interval
point
later
then V
suppose and find its
section
area,
of
YAlx)dx
=
Calculate the volume
a
of
solid which has
=
xsinede=(xcosx sinx]"
=
cross
[a,b]. The total volume the solid between
A(x) xsinx between 2 smallest roots (0,
v
x M:
Visionssuchrevolvingarovedthe state
is
R,
-
=
xxix) Ex (n
-
-
2] 2x( )) (E)(n r))
2x[
method, cylindrical shells
WX=
subintervals,
satisfies
f(x)
of
fle),
cones y= kx, cylinders
in
disc,
by revolving between
Vi
under
x+h)dx
=
=
I
above
2x(x)
=
2x() =h]
as it
flexthoouter
slope, diff y-int8h -4,g(x)
to find
x
hi(* [fix) -g(x)] be
=
b. Then revolve R around X-axis
generated,
Xi), f(x)
v
V
=
We will use
Disk formulas divide (a,b) into
Consider volume,
b
L
cont., positive
be
Same
is
of
(ras.*x=-*, e-intron,
nir
of Revolution
(ix), glut, Plak, glx,
suppose
conical shell with radius r, heighth, thickness i
calculate volume
h
inner:
set
Volume
shalls formulas
of
volume enclosed by flel, gle), (a,b] around y axis
+
-
=
cross section
area
i)
xcock+ sins
=
↑
January
Area
17Notes
of Surface of
Arc length
Revolution
and SAof parametric
curves
Lexsinbecontinuousbitsovercabsuchthat noclaim) suppose xegir. Weantenlengthbetweentake
the
a
subintervals,
line
-
length x,
[X1x1, Xis].
section has length
(1x+189,
Six =2x)f(x,x
then
->
because
Xi x 1 _,
=
-
Sx
↳then
Total SA
is
[7xx+ x/2
average value
of
8x, jy f(xx) f(x,-), put into to get
=
+
-
0x))wx)1 (+)xxx -144
+
x
wx
i) S
the
ofsurface revolution
parabola y=12x from
arc
,
fly)
=
f(y)=9
its inbyBounds,
2du,
LoL
-
1)
ii)
as
(i)
point on
of
is
3rd variable,v.
x
parametric equations
·
le
I far
*
B
-
X2
Equations
centered
↳
X
-
x
-
-
=
-
Derivatives Parametric
8
at
of
as
function
f(U), y g(v)
Example:
&
-
DV
=
10,y 4
=
=
let
v
1+
=
3
dus
ydy, Gydy-du
bu 37
=
=
of Mass
consider
w/variable
object
ID
mass
↳positionsf( M
Nastiest
density g(x)
M=pdx
4dx
bMXm=
=
Xcm
=
for xt[a,b]
i?Xdx
=
exampleare
=
=
,x cs
0
-
,x cos E,y4sin() 2
=
=
=
=
0,y 4sin)) 4
=
=
=
2
+
+
=
case,
the
Average
y, 4
+
=
4x4, -41-4,
so
-
=
a
want
circle centered at
w
1a,b), radius,
y-b rsint=
Curves letx=f(u),
-
y-glu),
Use chain
rule:
(l/*(iiva:
derivatives
fx,
Value of
x
Examples
same
of
x=etcost,
=etcost-etsint=et(east-sint)
=esint+ ecest= ellsint+cost)
-itissi
-
S
ilat- 3rdt.at"at
a total
X=V2058, y rsine: circle radius
second
for
=
-
Find firsta nd
4
L:
of
are
=
sine,
=
=
at
aised
-, () (8)
=
8 0,y0, x 1
at
origin /0,0). We
rcosE,
=
3 +2
the
curve.
+
they're
=
fi-i ts
1 0,
x=
Ide
f(u),b f(vz)
=
3
4x4 y, 4528 4Sin8=4,4x
circles:
=
g(u), then
a
=
=
H
expressed
=
=
-
at
G
be
eneresx.coss, a
+,yz 4
of
y
of the
=
ox
-
f(u),
=
can
=
Side des full-sets-1)
curves
a curve
=
=
=
↳=
-
-
V,x b nu v,
=
-
find length ofcurve between t 0, t
=2+;
Centre
Parametric Representation of
=
Yatte'dt, let ty
e
is r e
=
Nl*du
=
fel
x a bv
=
y0
x=
and
=
by revolving about
to
x 8
of
2x?fy)/+fildy,
=
-)
+
given
of
-/xx
1 (fast
othersatledoraround,
x-axis
-
+
(+Exn+x))
dx
-
Examples
+
2x)f(xx -) f(x)
areas
SA.
f(x))/(x 109).
+
=
find
Six
+
S=2xh
Example:
of
sum
y=
esint
a
function
(ab). Average
find average value
value
function: f= s)
of
f(x) I t
of
=
since
for
?**)
=
xe[0,2x]
de
January
24, 2023
Ordinary Differential
·
equations involving
A
general
·
solution of
general solution
A
·
of
(ex.g(x))
OD i s
is
Exponential Growth
function -(x) such thatO DIsatisfied
a
"n"will contain
of
Modelling
derivatives
and its
arbitrary constants
an
k constant
=
useful ODI's
integrated
with respectto
f(x), g(x)
↳n
=
(gly)*dx=(f(x1dx
x
x
p
=
+
tan(y)
variables"ODE
In
·
In most
given
form
30 for
These conditions
general solution. An
i) divide
integrates
iii)
(n(y)
+
v) 80 solution
its
have
↳ for
=
y
jx
=
=
-
y10)
2xy with
2
+
1
2
*
=
y
=
1.8
0
=
Aix,
=
f
=
u
set
t o extend
we want
now
(0)
=
0
initial form.
in
differentiable function. To make
any
=
or
and
z ox
=
=
f
if
=
#8, this
equation
is
therefore
x
=
=f(u)
separable
(,
.dv (7bx
=
Examplefind general solution
to ODS:
y y2.x
2xy.
=
i)
yy-x-14 17=243de
ii)
letv
-
and
yvX
=
(v. 1) x* U.
iii)
=
E E( ) 1
iv)
v)
+
=
=
Consider rights ide
*
d=x+ v(productree)
=
+
substract
)
xdX
v
=Ah(x)
v1
↳X E
=
Shape?
+
=
Sinde.
c
+
=
BAE
=
(v +1) x*x. then divide.
=
first). Findr-dv-tln(v+1)
=
=
0
=
h(x) 1 Eh(v+1) b e( x
+
+1
u
=
u 1.Recall
=
+
v
in+i=v
=
=
1.Esolate (or yz y=Ax
+
+ 10
as
was
is
(C(y)
living organism.
constant. When organism dies,
stops.
5715 years
is
15.
to
y Ae"+
+
=
1 5715,3
at
=
=
3.
(y0 y.2k15715) 1b 2((3715) p(n(t) 5715K 0 b
=
=
=
=
=
=
k
=
0.00012
h 10.0001+
+
=
-
a
tank thatc ontains
runs in
18
at
1
5300 years
in
water
18001
per
the tank
of
=
-
42
1
+
-
u
in
which
minute and each litre
is
kept
at1% ofthe volume
per minute. Find the
&y dx 0./productrule),recall y 7(1) f(u)
=
-
(j
to
and eating
(j
(5
of
dissolved salt. The mixture
of
x f(x,
And
=
roots and sign
⑭
10
-
this mummy is 52.5% ofthatofa
is dissolved. Brine
salt
1.8 = A
0
=
out
+
=
0,if((8,X
no
Dating
Chem. The half life of
where A eExample Consider
=
A20
=
useful tool to solve JDEs but
↳Ex f(u). v.
condition
Ae+
=
determines
Mixingbasic model, single tank
find solutions to JDC's thataren'tseparable
=
in
=
1.82*
progress with this ODEwe
=
((
1.8
1X
=
Also observe solution has
shares sign with
Csby breathing
=
=
=
example, consider
X
=
+
of
v(( 10.00012
=
a
some
mummly named Oatzif rom Meolithic period ofstone age
a
((5)
and
3. A
separable form
found
=
iv)
24
to solve; 3(0) 1.8
y
to
to
use
is
y
c
x
-
use
conditions
-We
.
decay.
iii) Att o, suppose Igo, and
(58x=-S2xdx
=
Reduction
Np:dy
following
=
ii)
in
solution to
38%d
by
constraints I known initial
a
i.e. kx,
for
the ice ofSetzal Alps. When did the 8etz, die is the ratio ofcarbon 14
Ky
ii)
in
unique particular solution.
a
as
is lety be ratio
arbitrary constants
of
will require
a
Background
values, typically
Xo,30CR
determine the value
of
ODE
order
values) to produce
ExampleFind
constants
some
used to
are
0,
=
p*
-
or
(n(x) k++c
of
in
=
in
its absorbation
problem (particular solution)
obtained fromgeneral solution and initial conditions, we given
de**
x=
proportionality (K)
Background Bio: In living organisms ratio (
a
is
3/4)
only ifA
to carbon 12
+
the unique solution of
cases
growth
have
September 1991
found
Initial Value Problem (IVP's)
-
we
Example Radiocarbon
+
1
=
=
size,
current
of
agrees with the sign of initial
=
y tan(x c)
=
+
to their
this ODE gives
1
t
constant
always
-x 0
y=
=
the
whether
-Rifk>8,X
=
x c
=
general
p edx 7dx
integrate both sides
iii)
a
of
1x 1
0DE:
general solution to
i) divide by1 yz
it
=
will obtain
=
Notice thatthe sign of
can
"separable"
thatcan be solved using called
Example find
which
Sclidy (f(xdx
*
are
solution of the ODE. Method called "separation
f(x),
=
c ontinuous, integral exists, evaluating
if
proportionality. Solving
of
-(h-=k p(7dX (kd=
gly).
be reduced to form
can
proportional
rate
a
at
constant A.
Separable Ordinary Differential Equations
many
Decay
and
Mostphysical systems
formula to find all solutions. Ageneral solution
a
order
ODE
an
Equations (ODC's)
function
uniform
initially
180kg
contains Skg
by stirring. Brine
ofsalt
in
amount
runs
the tank attime +
January 31,
2023
Consider
Limits w/ multiple Variables
Every path approaching
-
y-b
k(x a)
=
-
for
point(x,y)=(a1b)
like
Kow
some
Hence to determine ifa limit exists
a
-
I
wie
limitdoesn'trely
Exampled
on
i) consider the line
klx-all)
bikleal
replaces with
Second
-
case D
-
in
=
wise, ). DNS
-ook
i)
K.
-
Because I gives
limit
us
the
changes
as
path
we
we're
on,
on
limitD NE.
wish is (in +3)
Evaluate
+
I
Direct
substitution:
f(xiy)
A function
points
for
over
region R
Higher
keep other constant
one,
find both derivatives
f(x,3) xsing
of
=
Order Derivatives
rote that while variables treated as
variable after
a
constants during differenciation, they
process is
complete
it atateei
Examples
find the four second derivatives f(x)
of
xsing
=
=2siny
-xaxsiny
-
=xcosy
-
-
-
xsing
xlag) excosy ixs(x)= axcose
is ex
-
Observe that
Chain Rules
-consider
↳
z
z(x,y), where
x
=
x(t) and
=
y y(t)
=
-
Example:
i)
continuous
h
return to being
-
if
f(a,b)
=
=exsingcos
i)
-
continuous
** thigl-f(x,3)ox,3-fleie)
f(x,y),
Examples
=
is
tR,(x,y).-(a,)f(x,z)
=
Derivatives: diff
Partial
-
(x, 3) (a,b) if
at
function
and
R BX(a,b)
of
Letz xy+Sinx, where x=t2, y=cost.
=
=2t= -sint
ii)
**y cosx
iii)
a =(3 x)(at) (x))-sint)
iv)
clean up:
+
x+sinx
=
+
2+/cos c0sx)
x
+
2r(1)=
=
2u
↓) clean up8 rsincos8+
Zucos8sin8
O
=
Total Derivateset
the
total derivative ofz(x,y)
is
formed when
-
+
sint
Calculate the firstd erivative
i)
E=-ysinky)
ii)
--
iii) clean up8
13)(1) =
+
continuous
is
·xyla,)f(x,3)=f(ab)
all
at
=
clean up! Ercos+ zrsin
Example:
so
approach (8, %
Continuity
-
0,3 rcoso
y y(x), to
=
give
=
changing I will change the limit. This
differentpaths. So
and
sing
+ -(2x(using) +(2u)(rcoso)
is
es
rsing, find
rcoss, y=
x=
casso,
+
=
Example:
y(sit), then
=
i) firstc onsider ykx
ii)
=
-+E (2x(cos) (2y)(sing)
iii)
0
=
y2, where
+
x
=
he
sin
y kx
Evaluate
a
is
↳=
the linerateitsilolim-o
Example:
Letz
2x
=
ii)
es
and 3
=
-
Examples
Xa
K
t
wishla,
Evaluate
-
must examine the two limits:
we
ybf(a,y)
slope.
=
=
z(x,y) wherex x/s,t)
now
-
either like
Where K local
x a.
labs)(xiel-Lafleib+
↳
locally behaves
Find
-
xsin(y)"
-
fixig)=cosky) where
of
-
ysin(xy)+)xsin(xa)
-ysin(x2)-sin(xy)
=
-sin(xn(x)(1 (nx)
+
y=
Inx
February
7Notes
Exactand Linear ODE's
general
y(ex 3) 1x(x
5)
N(x,3)x M(x,y)
writing 0D28
way
-most
+
of
+
0
=
it
then
↳n
function
a
=
+
+
=
+
Find UK, y). N
To start,
+
sin(x y)
+
m
its
↳
multiply
solved
↳i
Suppose
-
P(x,y) 8
ODE
exact
+
↳
which is exact
=
(x2)
The multiplication factor
=
i) p(x)
butif
we
is
(x,y)*=0, multiply it
↳
by f(x,3)
=
for itt o be
iii)
(((x,y)P(x,3)) (f(x,y)a(x,y))
we
↳
P(x,y) f(x,y) a(x,y) (x,y)(mostly useless]
=
+
=
+
-Iconsides aloneabolites we atDividers
A
-
notice [HS
is
function
only,
x
so
introduce R(x)
↳=R(x). Integrate both sides
5
->
-)
factor
exp((rx)dx)
=
and if I
function of
the integration factor
Example: Solve
fly)
M
then
R*3)= (-], and
exp()r*(y)dy]
=
e***ye+ (xe"
IVP
its exact.
1) Test if
-
10
=
et+ye, N xe"
with y(0)
=
=
-
-1
=
1
=ex+elytl), e
-
condition?
yonlu(f(y), then,
is
so
exact
not
=e+ ey 1)a e
2)
=
+
+
subtractfrom each others
( ) -1
=
-
3) find f
=
=
F
=
=
+
+
-
ex)
5) Testi f exact
by
0
=
which
is
justp
-
=
exp((R*dy] exp(-1ky exp(
=
4) multiply original equation
Lie 3 (x
-ex+ye,
R* R*
=
=
-
y)
e-es[e***ye
e
=
+
Y
(xe"
-
1)
0
=
as
tan(x),
splx). Integrate
=
solve.
=
220s(x)
1
enkossell
=
+
solution y(x)
yl0)=
1
1
=
+
((scx)
-
=
-
=
=
seckx) F(e)
=
-
2 sinxcosxSex=2sinx
-
=
=
-
200S(x)
+ 2
2c05(x) acos/x)=general solution.
+
2cx5k) ((0)(0)
+
-
-
2 c
+
100) 3
=
=
2cxs"(x)
to
+plxly=g(xy?
Linear Form
consider
-
1f
&
via Then, kift
↳ then tidy to
Basically
putis
=
Consider
+
Integrate
((p(x)f(x)dx c)
=
(p(x)de=(tan(eldx -In(cosx)
r(x) sin12x)
=
=
then integrate.
=
(r(x)f(x)dx 2/sinx dx
-
require
3
-
lvp*ytanx=sin(ax), with
=
=
we
x(f(x)y]
exp()p(x)dx),3(r(xf(x)dx)
=
a (1
Inf=(RIelde, use integration
-
f(x)
method
write LAS
exp((p(xdx]
(f(xy]= rel.
f(x)
can
we
=
Reduction
-
=
Aexp(-(p(xdx)
similar
a
can
which implies
=
Find f f
y
p(x)3
=
=
=
integrating factor.
is
=
-
X
use
we
and hence,
=
r(x). This,
( y 3c0s()
required
we
exact
↳
want
suppose
=
solve the
iv) y(0)
f(x,y)G(x,y) 0.
f(x,y)P(x,y)+
xy
=
=
-
=
factor
to make itexact.
↳
=
0
A
ii) r(x)f(x) sin(ax)sec(x)
integration
=
-
=
(r(x)f(e)de
f(x)
be exact
- and ,
* *** 8,
get
ycX.
c 1+e
=
and will any function
+p(xy 0 a
0
=
+
"
1. Definitions:F
equations that could potentially
derivative, play
and its
notes the inputofODE refers to function v(x), while
Examples
note xac tas
=
+
*3 ( x)(xy)
-
f(xy=
Li
conclude the solution
we
non exact
byxwe
non
so
=
=
get
constant,
[f(x)y).
=
RHS,
and
sin(x+y) constant
+
- y 0,
we
+
Examples
y
everywhere. To
0
is not
(f(x)y]
↳is
+
v=
+
=
=
soconstant, sof(x)=cons.
3),
+
8 nes
to find
common
consider
cxs(x
=
of
=+play
-
=
50(1) e 2 constant
-1
=
=(dyv((-d5 (p(x)dx
U
=
=
=
c.
+
+
y y2
f(x)
=
3)solve ODE? We know
-x
+
e
r(x)
case
only when
y 8
rix
Expand
woo
the
=
↳
+
+
+
Wanna Be Exact
its exact,
so
2y+cosk+y)dy
+
( ( x y)
y3 y2
sin(x+y),
y integral?(34+
+
+
=
satisfies
((x)
xy e
=
=
examine
+p(xy
=
y sin(x y)
=
=
mV
=
play w(x)
=
Observe
+
=
+
-u
dy
+
Sidy -(p(x)dx b(y (p(x)dx x
↳
y, M x
=
↳ 3y+by cosk+y)
y
-
=
↳is note that
only linear in terms
Consider
↳-- sin(x+ y):=
↳
e3
xy+e+et-constant
takes the form
-
DN 3y xy cs(x
y);M cos(x 3)
(B
-
Linear ODE's
a)y
exact consider
1) testi texact s
2)
its exact
so
(x
=
+et
Xy+ e+e=1
Conclude
icostofvanneverkeinenearasanaree
-
=
9)
UI,3)= constant
is
xy+e
v
7) form solution
statement
repeated below
same
1,
=
Dintegrate uldig
-
8)Apply initial condition y(0)
=
that
follows
i t ope is
check
exists
N(x,y)
M(x, y) and
=
↳ then general solution
6) thus,
if there
is exact
8
=
+
u(x, y) such that
atex
then
and bla),
Equation: N(x,3 M(x,y)
to
e
+
=
Recall itv ixiy)
-
e2)
-
-
L y . Recall32* =eGla)=ex
Exact IDE's
↳ Ifr,
=
x
-
0
=
or
y= 11-a)[s(x)-p(y)y'a],
get
a)p(x)v (1
=
-
1 its linear, otherwise
up respectto
4
Recall
v=y"
we
get
a)g(x)
using substitute variable VII to turn
non
fun
not
inya
linear to linear
stuff
the ODE
cosy
etysinx=
provided the domain doesnt include
x is
any
roots
linear since
Los X
of
*ytanx=xsex
February
Notes
14
Consider the
The
general
most
way
write
can
we
A(x,y)ax B(x,y) ((x,y)
+
+
linear ODE's the
=
homogenous only ifR(X) 0 everywhere,
non
Homogenous
homogenous
+
8
xy ay,
=
byz
+
+
+
↳
a
+
+
+
+
0
=
=
0
alt) beaxbz=
0
↳ This
p(x[u's, +4yi] 9(x)(vie,]
+
since
solution. Obsbecomes
y, is
by writing
↳
a
V y'
+
diy, vildel -a
reduce itt o just order,
separable
=
we
caveat
y
exp(-)(A B1)
=
simple
a
+
root!
Rosts
of
get
Avoiding
.
=
general solution
solution is
thatImustnotbe
with
comes
32(x) expt) sin
Example: find
=
p2.) +y(3"+py: 43)
we
=
4b c0,
a
↳.8,
0
v,
+
4"y,v'(2yi +
to ODE gives
factor vivi, u"
yz.
+
+
↳ Applying this
=
w
e
y2
it
and
0
+
aly." play!q(xy) (23: dez)
L
-
+
+
solution
Case 3: Complex Pair
+
+
+
of
xexp)), and so general
v(x)y!(x) vxy,(x) v(x)y.x)
+
linearly independantwe employ reduction order, which
to find second
gives 3n1x):
=
vs, +2u'yi+un."
e
y'
(97, byz) a(x) x/97, byz) qxlan,+ byz)
b ple(a +be] exlan,+ byz)
↳
giei)
↳
->
=
-
=
=
known solution
is
=
v"(x)y,(x) zv(x)y/x) v(x)y,(x)
(
otherwise
linear second order ODE's
a p(x) +g(x)y
with y3, 1x)
=
+
+
y =v"(x)y,
+
=
+
linearly independant, write ya(x) 4(x)y,(X).
second
=
2 pasx+q(xy R(x)
form
of
↳is its
y" p(x)y' q(x)y 0
SDE
a
32 vx(y,(x) v(x)yi(x)
butwe will study
0
=
to find
->
order ODE
is
second
a
er
kwx), where
is
3
algebra
y,(x)=expl**)coskx)
find
we
wib-ia" (minus
the thing under square
root)
exp(*)[Acos(we) Bsin(wel]
=
+
general solution toy" 4y' 4y
-
= 0
+
0
woo
ODE
2
↳so
=
theory
below
suppose yy, (x) and
byn/x)
=
byz(x)
↳ yay, ()
=
+
is
y,1) and y21x)
↳ if
3ay, (x)
=
byz(x)
a
the
are
solution
linearly independant
(Vz
if
y=
proportional
is
8
↳ Hence
the
to
f, (x)
NOT
are
solution butnot in
v)
file)
are
right form
said to be
implies A 0 and B
=
3 a constant
y,(x)
=
-
=
3,
so
(X)
ratio
constant independant
not
-
(ratio]
so
constant,
its
not
so
+
ye
are
linearly
=
+
if y y, /x) and y y2(x)
=
=
general solution
are
y Ay,(x)
is
=
=
=
+
linearly independant
of
from
↳i general solution
i)y(x)
y
=
+
ii) y'(0) 8 A B
=
↳
A
0
=
with y101-7 and
y)=0
=
=
B
↳factor
-
1) Two
S
Ae*be
2)
itt o
general solution: y ek
ee*
=
cosh(e)
et
X,
2
cases
real
putting
Use
dx
(r (
=
=
log laws
4
into
-
-
=
=
0
b(x)n(x) 1)
are
a,b EIR. We found
Example? Solve
with the ODE
thats olution to y'+ ky=8
is
=
=
roots/X,, A2 tIR)
it
root
and general
O
its
a
quadratic
so
(a -43]
eX*
and
32=e**
a2x4b>0
0
<8
solution
is
y=
are
linearly
Ae**+ Be**
2: Real Double Root
term under square
Yu(x)
yi=
must be
-
both real and unique. Hence, solutions 3. (e) and ye/x)
independant
Case
are
-a
=
complexroots ifa? 4b
Two Real Roots
-
if
24h
a
of
X, Az
above
solutions to the ODC
3)
-
(
x=
and
+
Case 18
=
et*48, so+ aT+d
(-a /a-43]
are
↳ 3,(x)
exp(-)
=
vanishes,
root
we're leftw ith
linearly independant, butwe still have
not
1, 42, and hence bix)
one
solution
an
d
is gene
solution
solution thatsatisfies full equation.
sumotoptions
in
homogenous
in
r(x)
solution then
table, then
or
use
Choice
the
multiply by
Np
x/or it
Yo(x)
of
Let
ksin(x)
An nth
degree polynomial
120s(wx) Moin/wxl
+y=an", where
double
corresponding yp(x)
sum of
+
Ketcos(wx)orkeitsin(wx) ert((cos(wx)
constants, leaving us
solution
table below then try corresponding (p(x)
appears in
=
y' xe*; y" Nexx
and
a
Koslax)
Lekx
=
ifrix) is
yp(x)
ke**
3:
exq
if choice
+
+
p(x) and q(x)
+
then find
solve
one
y(x)=Y4(x) Yp/x). Here, I2(x)
apply following options:
(*) dx (n(x) 1
=
since
kx,n 0, 1,2..
+
some
so
we
if w(x) appears in
Term
In* InV
=
solution, up to the full non-home problem and
to solve,
=
to considers
double
real
A
Apair
0
=
b, so
=
Three
=
=
y=
+
are
here the
3
1A B1
=
+
-
above a
->
2
+
for
0
=
↳(A a i be**
solutions
Initial Value Problems
-
↳ De*** axe***bext=8. Factorize
boxed ODE,
Byz(x)
+
by
+
try
we
x
linear 8D Z v) constant
coefficients
exponential in
So
iii) general solution: y Ay,(x) Byz(x) y Ae Bex
=
coefficients
suppose that
y ay'
y, and
u
putting
solve by
A
et
=
=
h x+k(x- ).
=
-
To do
0
=
one
home problem,
sciated
to. Thus
xn(x) 1
32 x4 x
=
to
sides
(be:(dr=InV
=
-
know
=
+
solution using
=
x)
solution then becomes y a x
le*)-it=e*- et=0
*x=e
Example
ie) because
=
independant
(n(x 1)
+
. We
Non-Homogenous 8DC's. Reterning to " plxly'+ q(xy R(X). The general
ODE
x==x Integrate both
0
Aside:
=
Homogenous
=
=
-
Simplify
seperable
its
0. Note
=
+
21 x
=
↳ V
proportional
0.
=
-2) de 2) de- (dx by
vii) then hV =
viii) (nV
not
x)
x x
=
=
lex)-e=ex- et 0
Yz i *=
(du=).
solution.
+
=
-
=
-
x) v)-
+
v/ x) v12x x)
gen
Tidy
0.
=
+
vleit) v-2x)
v-
=
justone gene solution
solution find
8 8. Reduce it
-
↳ to find be x 1=PB
put
y2(x)
are
-
find
is
x
=
+
+
2x
-
(x)
can
=
+
xxvxv) x u
-
2.
+
y =xu" 2u!. Putinto equation
v
=
+
v
ODE
i f we
=
+
~itiby xei"
0
ii)
->
8
=
=
y =e*
=
v"( - x) w'( 2x)
let
linearly independant, then
definitions states thatt hey
i)
↳i
a
By substitution show thaty,=e* and
are solutions
y,*
ay
to
and
in)
=
+
↳ ratio constant:dependant,
Examples
is
3 acxs(x)+b(cos(x)
Af,(x) B72/x) 0
y2(x)
=
+
sink
ii)
=
Consider (22-xy" xy' y 5. Given
-
dy,(x))vdx
and 0 3(x) 93, (e)
find general solution ofa n
+
y=cos(e)
=
?exp(-(ple)dx]
can
↳v"( - x) v(22
↳
v'n, +v(23:+P(x1y,) 0
=
=
i) (x=x)/xu"+2u)
see
we
we
=
solution to the ODE4
=
suppose y, (x) and
AY, Byz
general
solutions. However,
↳ Formal definitions functions
↳
Using method
sufficiently differentthen
=
=
v
i)yz x.v(x) byi xu
ODE 4
solution to the IDEI
a
+ y 8, bu observation
y kask)
cannot
be
also solutions to the
are
are
is
+
the SDE
consider
-
ODE4
solutions to the
are
Examples
↳ yay, (x) and
and
Ja(x)
=
then
above,
-
solving gives
-
y
=
-
↳ then r=u" putinto initial equation
so itworks
8+
0 0
proof above,
->
Main(wall
+
yes=0, and
y'le)=Y
root)
February
Notes
28
Coordinate Systems
-
recall
is
we
↳is most
+
defines distance
a
rc058 rsin8
=
x y
pointto origin,
/x yz!,
r
D
=
and line segmentconnecting the
8
+
below
see
of
angle between
is
x
axis
point
taken
-s-tan
↳
arctan),
8-
by
however special
care mustbe
8
=
-
arctan()
=
cos
+
sin
=
↳
-
using
and
to find
in
We
Li applying
a.
how they relate to each other
and
f(x,3)
geometric
s
terms
58
of
and
of
we
could
repeats, buti ts
f(r,8) 2 8w/
derivatives
of
x
=los*-**:(os8(28)-sin8(2r)
=
2008
=
↳
(orcosO-rsin8] x [arctan()
iii)
-+-s i n (28) cosozr)
2
=
ye
x
-
e] (scroll
+
(rosine
riosa)
=
extend
can
+
cylindricalpolar
2
and
-
↳ spherical
be expressed
as
Ia)
dot producti s
the other. This leads
of
a a
=
=
calculating the
of
amount
one
to a.b (lb) cost,
=
is
8
vector
acting
in
angle between,
a
original vectors
orthonormal to both
vector
thebeminarof amatix(hotneedtoit,
laybe.
As e
x
=
to initial
up
x,e,r,8)
20 sins
+
+
axb a x
=
+
=
D
2 cos
=
a(axb)
when combining,
+
coordinate
Two
systems
D can
-
in 3D,
similar to 2D
system
↳
where falt) Y. F(t)
F(t)=fx() + filt) felt
have direction tox8
↳D multiple
spherical polar system
system.
EEiji3
=
DX
A
-
,,,
are
vector from origin
to
point (X,3,z)
is
+
points, P (X,
=
to Pe
↳i shown graphically
is
31,z.)
and
Pa=/X2, 32, za), the vector that
F ri (x x,) (3 3,)5 (za z)
-
for for
= 2xcos(yz),
to
want
differentiate
gradiento f, written
is
as
it
al
p=xcss(yz)
=
-
sin(z1z= -sin(3z)3
2xcos(yz)Y xzsin(3z)5 xysin(3z)
=
-
importantt hatwe
can
-
isolate operator from
function,
see
below
siteteethcen
yy zz
have a
+
function 4(X, 3,z) and
+25+E. This
thus 50)
Lartesian unit vectors
points from PC
+
. ++
Example: find
i)
Vectors
+
=
+
F(x,3,z) fx(x,y,z) Fy(x,y,z5 f(x,y,z)
variables:
-> suppose we're given scalar
polar
0
=
Vector functions
almosta ll formulae will be given
Fx
i)
4.m(ax5) (ma) x5 ax(mb)
-
=
bx
=
DA
I
is
we
the result
+
ax(ma) 8
↳
if
+
+
by5 bzzyields
+
Gradient Operator
r,s
-
=
+
can
2 sing
+
systeme) is, derivatives
yk
=
b dx
ayby 42bz
3. ax(b
y(3artan(Y) x)
=
concepts to 3D.
I
cylindrical polar system
3
+
Productt akes a vectors and returns
ax5
1.
and y
=
=
=
-
=
+
+
+
conceptb ehind
direction
cocalculate
2.
ii)
rcost, y
respect to
=
using,* 28,8 a r
x
-
8,
+
ExampleCalculate
↳o
is
same
I
Cross
s
i)
↳n
the
b)(i d) 0(a.i a.2) B(b.c b.I)
+
9x4 9yx 92and
=
+
of vector
-
the
a
gives
=axbx
length
-
that
- cost - to-since *
↳is observe
-> we
outhonormal vectors
differenta nd I if they're
=
standard rules, in(
use
can
-
to do directa nalysis
nicer
other, dot producto f vector al
scalar productof 2
if
vectors is 8
=
rsing, and have
ross
of
oft he
direction
in
length vector squared. Then,
↳ . =1,7 8,.z 8
=
+
the abouto f I vector
I gives
returns
and scalar producto f 2 unit
=-rsinross as
-
-
+
Product
s elf
it
=coso,sing -- using reasses
↳D
-
y
+
=
P(quadrant4 )
x=rcos8,
-
=
+
geometrically,
a
8=
gotta understand derivatives
↳ recall
+
=
m(A B) mA m
stio: oaataralso tooartante
8 2x
c
+
=
=
(m+n)A mA nA
i
-> We
+
5.
different quadrants:
in
->
A (B
m(nA) (mm) A n(m)
outside the firstq uadrant.
is
if
(A B)
i)
2.
Scalar
is given
find
We
+
3.
+
scalar multiplation;
x
=
+
4.
to origin
point
"Tox
-
Ai
1.
=???
system
=
+
y(u,v))
parametrically by (x(4,v),
curve
a
polar coordinate
is
common
↳coordinate
vectors have properties for addition and
->
define
can
=
+
-
=
xPz
F
-
xp,
Tz
ri
+
i -
=
-
a
importantly I
is
vector operator,
a
not
vector, the order
matters
-
as d
March 7 Notes
Recall
++*
=
Consider F fx(x,y,z)
-
=
Gradient
f(x,y,z) fe(x,y,z)E.
+
+
Then
(x+5 z).(x(x,y,2) ((x,y,2)5+ fz(x,y,z)E)
expanding gives us,t,ip s. Called
5.5
to
=
,
Example:
My- xE
5.5 for F
find
=
x(xy) y(0) (x) 2x3
8.5
=
=
+
+
Curlof = x5
=
Examples find
2x3
=
+
curl of 7 xy
=
-
x=
diverneysecs teams
I
=(x) =0=xx)=-1
3=0=(x) x
-
=
(8 s) (8
5x
=
-
Properties
+
+
-
x4z 5 xz
=
↳
-
f
8.(A B)
+
-
3. 5x(x) x 8 x
+
.(f) (5 f). f15.Al
5.
x ) (8+)x
-
7.x(x):(5.3) B(5.) (A. 5B A(5.5)
-
-
+ +2.
5.8
-
=
↳ then
+
Note its
scalar operator
a
-eatthe trees.Is
lf (fE,e
-is
5.(x)
thatx(8t) 0,
note
=
5/5.5)
x/x )
-
0
=
vanishing
x F (x)x)
↳
is
is
.
0
=
D( xe)
=
=
called scalar potential
Examples
Observe
of
.(5x5)
=
and vector
ii)
4
+
rector
↳
=
+
2
x
=
f(y,z)
exist
is
potential
of
recall and
·.
5
We
-
(rf)
=
=
hence =
=
-
of
to
=
I4 x
+
5,
↳is
to
Then
classite
we
set =f
to get it:
s.18f):If
arr) it
=
wasting time wil spherical, formulae will be given
not
need it
you
want
+3
cylindrical, we're
were
2
8
&x1x,y
ex
=
cos0 +sin85
/
constant
respectto 0 =
8
not
E
+
as
+
+
divergence,
Classification
f(z)
+
because initial conditions
Any vector functionsc an be written
-
-
divergence
+ f fz).++f+ fz).
sin+ cost and
=
any time
=
=
8
importantly
Allthose
+
4 0504 xyxxz
iii)b xy
=
X
+
, see -
-A
fz)
+
Laplacians to getI
=
+
2xy
=
there
has vanishing curl and
x
+
i2xy x =
((3 2xz)dx xy xz
=
↳
(+f
r
-
to form
0
function,is
=(y2xz) x 2xy5
=
or
=
find the scalar potential ofE .
i) +2xz
= sin+c0sO8
*
=
=
-back to
=
vanishing dir everywhere then
I
8)
↳
0
-
Lavector functions has
-
then there exists
curl everywhere
=
=
same
..
5.
=
almostt he
=
o
x5
↳ if
2.
vxnyz)-34+ x5). (cylindrical)
↳ Ifc ombine with E frr+708+fzz
is
turning points
define and classify (P for function
less straightforward. Startby defining
1.
(xo, 30)
is
local maxima if
2.
(40,30)
is
local
3.
(Xx,ys)
is
a
f(x,y). For f(xo,yo) to be
a
CP
is
10,30)
=
have the following
:
options?
20
and
and
5 48
minima if
saddle pointi f
4. No information obtained it
48
0
=
so
its
a8
(maximal
78 (minimal
(saddle point)
(l) tins,sol
a
=
-
saddle
4
point
-
Dx
-
.
function such thatI b
scalar
a
then itfollows that
-
Theorems
1. Ifvector function *
h as
yo using,
2=pose - focor-sino) +/sinceElling cose,
Hence
simpleseensichare
Potentials. 2
and
-
E
-
=
x=rewso
=
0
=
8
-
comande
=
We will find =2098-sin8
((a)(2) 101),p
=
a
Recall IS:
A.(x)
=
↳8
+32--since
f(x)
+
(Ax) 5.15x =)
Laplacian:
8
must have
()
iii) &
compare
=
Li (P=10,0)
2
20s8+s i n
8mustbe orthogonal to 5, and pointin positive 8direction (counter
clockwise),
since
2x, 23
=
=
↳=2 = =
cos! -8): sing
=
-
+
=
J.
=
spherical
=
4.
=
,
~3
0. 8 B
=
+
=
yz, hence -vay*vayz B(enlindricall
sin8=
Spherical Coordinates derivation
g
+
respectto
32
x
=
ICP a 3 a
normal to i
=
=
Vyz;
188:
we
=
+
6.
we
of
1.5( 9)
2.
1)5 (8
(
-
is
ii)
5. 20580,.
seer coso+sing,recall
=
↳y
8
a
18
0
away from z-axis,
with location atwhich they're being implied
())
=
Example8 f
Cartesian Coordinate
E
0 0
+
non
plane ofconstants, randoboth normal to 2. Note and 8change al
in
+
+
in
cylindrical coordinate system, I points directly
In
-
I
DY
March 14 Notes
Multiple Integrals (double integral DI)
reconsider volumenta cylinder
-
=
Consider function o fa variables, flx,y), defined
and
3, <3 <32. We
from ITC
I
form double
region
(Y)tx,yldedy.
↳ write
↳
=
(ef(3)dy
Note thatfor
-
rectangular
region you
true, only if limits in integral
can
switch order
integration, butn otgenerally
on
War ?
itw rite integrals V
(!)*edxdy
3].dy (integrate (de first)
↳v
Examples
-
sphere
a
() (rsinddadode
4. For
=
=
↳
1(c)de (.)-(i) do-do:*
=
third
6.
I?
h as
it
↳
I
is volume under function f(x,y) within rectangular region
(?)dedy;
volume abs,
a, b,
so
constants. Integral
c are
integral has value
=abc
(?)ded:1[x] dy (asdy abc
=
-> can
also
use
DI
to find
=
bounds. If
ofshape within
area
unity then height of object is 1, and
↳is but how to manipulate bounds to get the
shape
Examples
we
i ntegrand
set
to
When working with integrals alone
equal
to
area
want?
we
D if
Lines y m x b
=
+
Dy
=
proceed
we
-
-;8xxx2
Example? Find
1
=
6
Semi circle
with radius
a
↳
A
-
(?)**1 dadx
=
↳ tre
-
D
=
see*
change
Gloss
=
bounds?
=
↳ Ma =
(Ivar. elde
asin=
=
all-sinel= aco8=9c0s8
x=-a,8=*,
x=
x
Continuing,
volume
of
↳ suppose
region
is
then volume
Examples
is
object
a
triple integral
given by Ziligszelkiel, Yilelycyalel,
a
Jedededx
ds
a
cylinder of radius
↳
-
8
a<x
a
heighth
and
szch?
-axyax
yaBx
integral? (Y)ededede(need
V
↳e
(we
=
theorewingmist
to
polar coordinates w/
=
=
origin
3
parts
intercept
xyz
means
closed
8xlequationtor
8 3.=
=
3xlequation for line)
**
34:
-
(18.x1dx -(8x.7 (110.2) -140 1)
=B-
=
-
2
to
5 to
--(3xdx -(* -(0 s) 6
0
=
=
5
=
-
D-(xdx= -()*:
=
12
=
if
curve
isn't closed?
and
curve 2.
tangential distance elementaround the
We
can
(.tlx, ylds
write I=
Q(x,y)dy, where P(x,y)= flx, y and alx,y)= Hle,y)"
+
+
Evaluate integral along
+
8
- 1+ from
curve
and
Alo,1)
and
B12,5t,(x+ alde
t
=
=
+
x
62
=
+
8
+
+
=
+
3x
+
x).
+
16
=
integral along
2+x
curve y=
from Ale, 21 and
B13,5),
(,(( 3)dx (x ydz)
dededz=rdrdode
↑
cylindrical
it(,(x kxx))dx
and
dedydz=vsinddrdebd
spherical
where
curve
(,7(x,y)ds=([P(x,y)dx a(x,y)dy]
Example evaluate following
MI
path)
=
=
integral((xx3x)dx (ix 3(1 x)dx (i(x 3 3x4dx (=
-Write
if
we go
to find equations for line
segments
3ib 0
cuz
split itup? Elx, ylds=P(x,y)dx
↳hence
↳y
know?"Vax=
Bonk required to know results
from
do integral
we
+
Cartesian
x
↳ y 1 x x between
to do
o
the path,
is
d, 6-1112
from 5 to 2
e
x, x from
+6x
=
ach
=
=
=
-
=
=
Example:
< a
(Yadraxdedz:(**d==
↳
V
,b 0
28 4 3x,
the
is
0,ydx/c
=
=
Line integrals Consider function
flx,y)
-
Volumet
=
positive,
is
area
(14- ) =
38
Can
↳
v=
i) Bounds:
ii) write
A
found through
-
path
-
*
can be
=
↳ path
↳
(sin(28) 8)* **
+
4
2,8
=
-bydx. Break into
=
=
re
=
A
a,8:*(using asint)
when
thesenasolacetoadose
↳ evaluate: A
ii)
↳ Path 18 3.=8-x,
dx=acosOde
-
When
y=mxxb
i use
↳ m
BA=
=
Va2.x a?_
x aSin =
A
as
4
DX
-a
to allow
triangle my vertices (5.0), (5,3), (26)
"
1 ine , Rounds.Ocean en
way
of
⑧
I (m=
-
Examples,
area
a
negative
is
area
integral
curve
natural to arrange limits in
more
anticlockwise direction, resulting
in
the
curve
notation for closed
-
-area:1Y*de=(Ilxtde=(x-l? 2.=
path
around
clockwise around
1
=
a
A=-()"xdx+)cede
continuous path be taken, i.e.
a
rightangle triangle of lengths and
↳ 0<y<1
path its
-
volume is
↳(*12x
+
(,113-2)-deeesopotamonterseal
-
I
+
+(= -2.) (9
=
6
+
2) (1= 10.)-(2
+
+
-
-
4
-
54)
-
=
15
we
parallel to the
y axis, i.e.
parallel to the x axis, i.e.
is
divided into
a ais 1
=
arc
xy
x
integration
(_ isds
x
k
=
dx 0,
=
=
y= k
=bdy 0,
=
(,((x,y)ds (,f(x,k)dx
a
parts
es
joining
3
A to Brick. Then
+
=
is
k
single valued
not
for
length formulation should be
Example.
cuboid-axbxa
is
-
=
Ifthe path
the
=
path ofintegration
a
Is
then
). (evaluate (dyl
LD I =
+
change sign ofresult
(,f(x,y)ds=(_1k,3)dy
5. Ifthe path
=
↳=
Whati s
(,f(x,y)ds=([P(x,y)dx a(x,y)dy]
path ofintegration
a
↳
Integrals
Path
=
line integral
length:
(_fkigds
3. For
0<2+90x 4 x
I
a
2. reversing direction of
↳
then second
?(3) dy (find/dx valuel
↳
arc
Splitting
(*).sind dod levabrate firstintegrall
↳V
variables of integration
of
1.
ah
volume of
Bounds:
Properties
=
of
constanta nd don'td epend
are
=
Examples
such that
=
("ordrdobz=Jede-Ibobshe
integral
andz
V
0 8 2 ; 8zch
Dr
i)
Ikelde=f(y),
fty)
know there exists
we
Bounds
-
X,X< Xa
=
integral
I"))k,elde) di
↳i with brackets: I
-
can
rectangular
on
part ofi ts extent, the path
is
divided,
Evaluate the following integral from Alo, 1) to Blo,-1) along the
1frxx,0I
9 (,(x 3)dx
=
=
+
or
used
semi circle
March
Notes
21
interesting
-
where integral is
case
consider(,(yde
↳a
+
independantof path taken.
xd3] from 1801
to (1,1)
the 3
over
examined scalar function of
~
↳ consider
curves
(8yx,(28y x,2,84y 0:0,x<1u3x 180xyx)
i sixxx edy=
=
=
I"iorderisi
a
horizontal line vertical
(,(3dx xdx] (,(3dx xxx) ((y4x xxx)
↳
key part
to this
a d.
is
total derivative. Suppose there exists
required for
=
-
+
connecting
any curve
Back
↳
Pay a
=
=
=
=
U((x,y):(1,11) U((x,y) (0,0))
1
=
0
-
4
x
=
=
integrand
is
a
a
function
is
=
1
=
if
exists
-then P,[Pdx+ Qdy]
Importantto
verify integrand
Green's Theorem
8 for
is a
allows us to
same so
Iamso
total derivative,
closed
examine
closed
any
=
curve
is
zero
curve
(
alterative
true
not
integral when integrand
by
curve
C. Then Pis always
thedx,
a
the
e
(Sn(-)dxdy -4,(P(x,y)dx a(x,y)dy)
=
lexil around borde se
-(2x y) 10 x(2y x) 1
-
=
=
-
-
↳
()p=
area
=
+
=
-
(a) a)dxdy 2)(ndxdy
=
region sort 2(127=8 i
of
we
integral
common
have scalar function
(,f()kx
f(x)
of
=
to see
x,y, z in terms
Suppose
by having X-30,
i)
find
↳
V
xyz, evaluate vector path integral
3 zu2,z=u
=
between
A (8,8,8) and
=
=
V
of
along
curves
B(3,2,1) from so for:
=4u, =Sucuz de + du
3
=
+
+
=
+
+
x,3,z, for
in
1218
↳(!(u)(4")(304)B2 405 302/d
+
+
2360°dr 5!482u 2/36udu
=
↳y 24
=
x(+1, yy(t),z z(t)
=
dx 1 4u5 3uz)bu
ii)(,V ((x5z)(2 4u5 3uz)du.sub
↳
x
di=(***-) dt
=
of
so
dxxydytdz. In practice its
=
further parameteric
of
↳ transform directional diff. elementusing
Example.
44,
=
+
+
y levaluate
+
+
integral,
directions)
put
their
u
terms
y 272,z=4 fort e
=
-
-
B
transform integral
4)(+)]dt
4+
find
-
to
in
terms of t
ofx,3,z
put into, simplify
4dt=evaluate
integral), Fode
vector
independantof path taken
is
function is
↳o thatm eans there exists
Hence, function is
F
integral
conservative i)
x
3. The vector path
↳ is
integral
the
curve
c,
fady+ fadz
=
We
conclude that
0
=
function I along
of conservative vector
I, Sedx P(B) -0(1)
B is
on
such thatF . recall itE =st,
=
1. Vector function
points
and
a
sides gives x I 0.8.
2. The vector path
is
2
function such thatS pode= fidx+
a
conservative ift here exists
curl of both
between
independent
d
conservative. Recall path integrals, must be total derivative
said to be
a
=
function I along
of conservative vector
path which joins
where I $
=
a
closed
path
vanishes
of, Fode 0
=
Integrals
earnedbetterteethingiens
Examples find the
a
vector
quantity
volume integral of
F xyx 25
=
+
x 0, x 2,3 0,y
-Z
"ish
-
=
xz
over
the volume oft he cuboid
=
?!(x 25- zdxdude
+
1?rededebe,1??dedude, "??*dedsda
jedide yyndede +2!!!Eide
2,9dz 3Y62d2 2! 8dz
·ation:
↳
-
3,z 8,t 4
=
=
=
=
-
+
+
+
-
=
position vector , flx). The line
of the
where di
vector
integral by
-
( 536
=
Vector Path Integrals
suppose
+
+
↳
iii) I,[(2x 3)dx (2y x)dy]
+
=
+
a
.=
+*
bounded by the planes
+
aple,snaraeoznin
i)
is
+
1
=
↳ Note this will return
Consider region R bounded
-
([128 " 4+3-12
Volume
8
=
total derivative.
-
↳
=
+
pointAto point
derivative, then closed path integral
total
↳ because startp ointa nd end
pointa re
a
isn't
=
in
↳
b50v xy
=
curve x
=
Recall thatsuch
if
+
then taking
-
-
If the
x
=
=
↳ then
and
to to C..
example =DP Y and G
to
VI(o) where
- 2yzz along
=
+
=4 4 4 342
then
·8),(P(x,y)dx a(x,3)dy) 1, dx v(c.)
=
+
((116+4(2+)+(4)(z 1
If line
xdx=(***)=P(x,y)dx+axis) dy
is
find line
Conservative Vector fields
Preandasareas
↳ We see that
we
+
+
xz
=
+
a
function vixis) such that
a
=
f(x) xy
=
↳
*condition
taxl+ fall,
+
(,f-dx (_(f,dx fady tndz]
-
consider
(s.x= (_(*dx xzdz (-2yzdz]
i)
+
example
7. (e)
=
=
ii)
=
vector line. Butw hati ffunction
over a
to t 1
(,[P(x,y)dx a(x,y)dy)
path integral
Recall
-
-x
+
+
=
Examples
↳
line
1
=
+
hot product
=
=
=
vector
a
consider vector field (x)
548 2)
+
+
-
32) 4(9
=
124-82)
+
=
ISSav
March 28 Notes
Surface Integrals
Example.
and theorems from vector
Evaluate the vector surface integral
ity=4 between
curved surface
=
4.y 8
of t= xyzove r the
5.x 1 ;i x
2:i
3n
and
z 0
the surface for
S is
the vector
in
to
28
=
differencial
surface
i
is
Adding
defined
and
the
using
4 In
unit vector
n ormal
unit
5 x y2 4,
is
iii) moststraightforward way
↳surface is then
given
=
-
+
***35,
is
5
=
thus
+
integral
our
88s*,
for
elementd '5
Example:
is
Here
=-z5-3 E.
of
integral
vector surface
I
vector function
a
of
(((
=
and vector
=
9 in
=
+
4z
a
-
surfaces which
is
we're
curve
see
We
integrating
thatt his
case we see
22, 0.
36/-18s8+Sin8y]=12(2+5)
Backo
=
·
Consider== 23* xxy
5 x+y z
+
=
+
=
-
+
=
+
+
+
-
X=
a
=
5 2xx 2y5 2z= (s)/2xx
2y5 2z=
bx
x
(),18xs)d5 (),z3 3z)-(x 3 zz)d1
↳ 18s 4(91 36
2 sindos8;y=2 sind
sin 8,
yx 2y z,
=
->
+
=
+
s n (x yy zz)
=
=
=
+
(x yy zz)-fon x3+23
+
2
+
=
+
=
+
spherical coordinates: dA =rsindd8dD 9 sind d8d6
=
↳ x 3sindcoS8, y 3 singsin8;2=3coS4
=
=
get(),8x F).d
=
the
over.
This
is
the xy plane
is
↳
↳
also need to
define
that bounds the surface. In
curve
and
I), Sodas(*g9sinsingcose+Ssindsin8+3058] 9 sind 460
IS, odA 911+l
-12x
We
*
Fide=2ydx-dytxydz.
on
y:2 sin
+
+
integration you'll
88*;0cr12;8(2<3
21058,
=
Asideocosscotolev2018,butundsee
bounded
=
I),18x stdA=((sinacossing sindcosaldod 8.
+
this
an open
Go, ode
-
coordinates:
=
+
-
spherical
in
so
x
=
36).sin8 cos8sin8y]d8
3z)dA
path integral?
y rsin8
-
defined by xyxe: 4 for
integral
=
/scsin+ scoesin]2zdzd8
9858sin8 8c08in8)(9]68
-
closed region
=
for
now
the
over
dA
+
+
()).sa= (islands
2c058; dA 4sindbd 8.
↳ after nasty
area
the firsto ctant.
-
=
I then
vector to.
iii)
=
-
to avoid icks, express
scalar
a
integral ofF y2 25 1
vector surface
the sprerical surface +
y+ z
over
integral
the vector surface
x
note that
we
Li
I), FodA 1), FondA
Evaluate the
is
=
easierisafree
firstoctant
in
y2 -4
+
2y5;(55) ( x (2y)2 4(y)
+
VdA= (*98cssin8z 8cos8sin25]2dade
=
(), 5xst
have
we
c,
curve
Example: Ahemisphere
ii)
=
=
=
=
=
functions defined everywhere
vector
a
closed
+
+
through dotproducto f
())5P= (),
= jel
=
=
i) firstc onsider the surface
36).sin8 cos8sin8y]d8
36(1084 sin8y] 12(2+3)
↳i achieved
X rc0S8,
-
()/j(x) (),x() .ndS
=
a
-
calculate
first
x
=
-
=
it
Bounds;
a
↳
(),5x E) ondA 8, where S
=
4 for
by
83.
integral becomes
our
=
we
theme
a
of
+
function. We could also consider the
on
differentvector function, for
a
Stoke's Theorem. Vector generlisation Green's theorem
curved surfaces
these coordinatesd)-rdOdz, hence
using
variation
a
a.dA
(95..saV= (), 7. A
-
ex/
expected
all
are
f x =
is
as
=
vs 2xx
D
functionsw ith
replace the vector
=
fo
2xx 2yy
=
A
CoVd=
(*98205sin8z 8cos8sin25]2dad8
In this example
DT, but
on
=
transform"into cylindrical coordinates
to
is
by r=2,
dA=2d8dz. Thus
case
I), F.dA= G,
get
we
5
x+y2 4
n
0
=
example, letF sa, then
to the
+
in this
=
=
these up
case we
L
=
=
Various rollaries
A
((, =
((,xz(x2 35)dA
importantly when
-
=
=
i
E1.
2
=
↳similarly, letI x , then
vector
-
=
=
Est.
unit
n ormal
=
6.x 0
the first octant.
=
ii) in this example surface
↳ Hence
=
-
dA=rdA,
normal to the surface
,
=
3
z=
areais
n5
=
of),d*where
firstl etus understand whatwe want to find
1.
b(),f-nd (.)-zdzdx
x(), -ndt (.)-dzdy 6
x
x(), -ndt 1.Jdzdy=
calculus
=
Note
integral
surface
23
-
satisfies Rx=4.
-
vector, vector gives scalar
ofs calar gives
Z
F x zy y=
2
=
+
+
↑
i) find outofsurface.
+
+
integrand
our
+
in
8 de=-2 sinzds; dy= 205848 from
=
integrate
integrate
sense to
8, and
=
curve we
I,J.dx ((4sin8(2sin8d 8) -2s8(28d]
+
want to
since we
z
=
=
ii)
polar coordinates. Along this
convertt o
y 2 sin8
becomes
for (x3 2z z)/3
+
we'll
iv)
5S=2xx+2y5 222. Here
(x yy zz). Hence
obtain
we
We see
over a
which
spherical surface, itmakes
spherical coordinates. In this
case we
most
get
dA rsindd8d$
we
can now
integrate
=
-
for, Jod=-12x,
to find
have x=2cost,
shall
098-2K.
irinee
i
Hence,
↳"Din
4)?(2in8 ces8)28
+
as we
expected
Deriving Green's Theorem
->
(), Gods,
other side for 6
see
5185 1z 4.
=
+
e
integrals
258,2=, x y2fans
=
+
-
=
dA 9sinddOdB,x 3 sinPcoS8,
=
=
↳
We
singsin8;2=3cos4
y3
=
I), Fode=(*g9sinsingcose+ Ssindsin8+3058] 9sindd468
To avoid doing all these nasty
Integral
our
For
a
3D
version
closed surface S, enclosing
of
a
normal vector
integration
in
then
on
-
in
particular
-
surface
z8
=
Verify the divergence theorem for
5 x zY +yz
=
+
over
the region
=
=
vector function
bounded by the planes
x 0,
=
x=1,
y0,y 3,z 0, z 2
=
-
=
=
=
this example the LIS is
easy to find
in
since
J.==2x, hence
((),805dV 6
=
-
1.
To find
z
Hence
2.
z
I), Fods
00n
=
=
mustconsider each
2. Following
=
3.3 3in 5
=
we
Eloutward.
on
6 surfaces
separately and add them
this surfaces= +
y and dA=dxdy
(),F-ndA=S!+yz)r)- Edydx =
2n
=
-
=
same
argument
b(),-ndt 1)2dzdx
=
=
as
earlier,
2
=
so
(),*-ndA=
be within 3D space
plane
z 5.
=
11)((pdx ady]
=
+
The outward
6), ob
=
538 =y
z
=
=
+
=
+
=
=
+
+
(4S x y 2 -4
=
x
+
y-)yddx:
).)dzdx 2
-
56°F x 25 42,
=
-
on
x
-n x 1
1
=
=
x8
=
for
-
=
:
-
x
=
x!)!1dydz 6
=
=
(?)dedx
0
=
=
=
=
+
=
+
+
=
Take
director the
curl, only need a
↳
spherical coordinates
↳o
do, od=12
a
=
55 2 x 2yy 2z2 (55) 14x 4y2 4z 4
-
↳ n x+35+2. See other site.
=
+
Fan
=
558 F x 25 y, on
iE
.dix F-(4xx dy) P(x,y)dx x/xig) dy
,
(SS.F. = V=S1, -d
Examples
choose itt o be the
we
E. Hence, by Stoke's I heurem-(((8x ).EdA
+
vector field
derive Green's theorem. Consider the vector function
↳ aspicksariel.
by parts
a
is
-y
(), odA=911+)
region V
to
=
for stoke's theorem,
Theorems for Vector Calculus
Divergence Theorems the
-
integral yields
stoke's theorem
I P +GY. We consider the plane for Green's theorem to
integrals, take these definitions:
-"siade-3: Isinpay: Inde
-Sinocosodo-t; Isinods-1 coseds-1
Applying these values to
can use
on
other side.
Bing integral
gives
-
12x
+
=