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Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-1-
Differential Equations
A differential equation is an equation that involves one or more derivatives, or differentials. Differential equations are classified by:
1. Type: Ordinary or partial. 2. Order: The order of differential equation is the highest order derivative that occurs in
the equation. 3. Degree: The exponent of the highest power of the highest order derivative.
A differential equation is an ordinary D.Eqs. if the unknown function depends on only
one independent variable. If the unknown function depends on two or more independent variable, the D.Eqs. is a partial D.Eqs..
2
22
2
2
xya
xy
¶¶
=¶¶ is a partial D.Eqs..
Ex1:
35 += xdxdy 1st order-1st degree
Ex2: 522
3
3
÷÷ø
öççè
æ+÷÷
ø
öççè
ædx
yddx
yd 3rd order-2nd degree
Ex3:
05sin4 2
2
3
3
=++ xydx
ydxdx
yd 3rd order-1st degree
Exercise: Find the order and degree of these differential equations.
1. 0cos =+ xdxdy ans:1st order-1st degree
2. 043 2 =+ dyydx ans:1st order-1st degree
3. 22
2
yydx
yd=+
4. 22 2)( xyy =¢+¢¢ 5. xyyy =¢¢+¢¢¢ 2)(2
Solution
The solution of the differential equation in the unknown function y and the independent variable x is a function y(x) that satisfies the differential equation. Ex: Show that y=c1 sin 2x+c2 cos 2x is a solution of y''+4y=0 sol:
y=c1 sin 2x+c2 cos 2x y'= 2c1 cos 2x. - 2c2 sin 2x. y''=-4 c1 sin 2x-4 c2 cos 2x -4 c1 sin 2x-4 c2 cos 2x+4(c1 sin 2x+c2 cos 2x)=0 \ y is a solution
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-2-
Note:
The solution in example above is called general solution since it's contain an arbitrary constant c1 and c2, i.e. the general solution of differential equation is the set of all solutions, and the particular solution is any one of these solutions. Exercise:
1. Show that y=3e2x-e-2x is a solution to y''-4y=0 2. Determine whether y(x)= 2e-x+xe-x is a solution of y''+2y'+y=0 3. Determine whether y= x2-1 is a solution of (y')4+y2=-1
Ordinary Differential Equations:
Ordinary Differential Equations are equations involve derivatives. A. First Order D.Eqs.
1- Variable Separable. 2- Homogeneous. 3- Linear. 4- Exact.
1- Variable Separable:
A first order D.Eq. can be solved by integration if it is possible to collect all y terms with dy and all x terms with dx, that is, if it is possible to write the D.Eq. in the form
0)()( =+ dyygdxxf
then the general solution is:
cdyygdxxf =+ òò )()(
where c is an arbitrary constant. Ex.1:
Solve yxedxdy +=
Sol.: yx ee
dxdy
×=
dxeedy x
y =
dxedye xy òò ×=-
ceedxedye xyxy +=Þ=-×- -- òò - )( Ex.2:
Solve )1()1( 2 +=+ yxdxdyx
Sol.:
dxx
xy
dyò ò +
=+ 1)1( 2
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-3-
cxxy
dxx
dxy
++-=+
-=
-
- ò ò1lntan
11tan
1
1
Ex.3: Solve (1) )( 2 Lxydxdy
-=
Sol.: 1ddy 1
ddy ,u +=Þ=-=-
dxdu
xdxdu
xxyput ….. (2)
òò =-
Þ=+ dxudxdu
1udu 1 2
2
[ ]
cxeuu
cx
cx
dxduuu
+=+-
+=+
+=+
=úûù
êëé
+-
+-
\ òò
2
11
1u1-u ln
21
1)(uln -1)-(u ln21
12/1
12/1
Exercise: Separate the variables and solve.
1. x(2y-3)dx+(x2+1)dy=0 ans: (x2+1)(2y-3)=c
2. dy=ex-y dx ans: ey=ex+c
3. sin xdxdy +cosh 2y=0 ans: sinh 2y-2cosx=c
4. xeydy+ 012
=+ dxy
x ans: ey(y-1)+ 2
2x +ln |x|=c
5. 12 =dxdyxy ans:
cxy += 2
123
32
2- Homogeneous:
Some times a D.Eq. which variables can't be separated can be transformed by a change of variables into an equation which variables can be separated. This is the case with any equation that can be put into form:
)(
xyf
dxdy
= …(1)
Such an equation is called homogenous.
Put uxuxy
=Þ= y , dxduxu
dxdy
×+= and (1) becomes
)( ufu
dxdux =+×
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-4-
Ex.1:
Solve xy
yxdxdy 22 +
=
Sol.:
homo. 1 2
2
Þ+
=
xyxy
dxdy Put u
dxdux
dxdyu
xy
+×=Þ=
uuu
dxdux
21+=+× Þ
uuu
dxdux
221 -+=×
udx
dux 1=× , òò =×
xdxduu
cxcxu+=Þ+= ln
2xy ln
2 2
22
Ex.2: Solve the homogenous D.Eq 02 =+ ydxxdy
Sol.: xy
dxdyydxxdy 22 =Þ= put u
dxdux
dxdyu
xy
+×=Þ=
uudxdux 2=+× cux =- ||ln||ln c
yxc
ux
=Þ=Þ2
Exercise: Show that the following differential equations are homogenous and solve.
1. (x2+y2)dx+xy dy=0 ans: x2(x2+2y2)=c
2. x2dy+(y2-xy)dx=0 ans: cx
xy-
=ln
3. 0)( =-+ xdydxyxe xy
ans:
cex xy
=+-
||ln 3 - Linear
The equation of the form Qypdxdy
=×+ where P and Q are functions of only x or
constant is called linear in y and .dxdy
Find integrating factor ò=Pdx
efI .).( , then the general solution is
ò=× dxQfIfIy . .).(.).(
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-5-
Ex.1: Solve xexxy
dxdy
×=-
xexxQx
xP .)(,1)( =-=
xeefI xdx
x 1.).( ln1
==ò= --
Solution is
ò ××=× dxxexx
y x11
cexy x +=
Ex.2:
Solve xyxdxdy
=+ .
P=x, Q=x
2
2
.).(x
xdxeefI =ò=
Solution is
solution theis 1 222
22
222
22
xxx
xx
ceyceey
dxxeey-
+=Þ+=×
××=× ò
Exercise:
1. xeydxdy -=+ 2 ans: y=e-x+ce-2x
2. 2sin3
xxy
dxdyx =+ ans: x3y=c-cosx
3. ydyydxxdy =+ ans: y
cyx +=2
4- Exact
The equation 0),(),( =+ dyyxNdxyxM is said to be exact if xN
yM
¶¶
=¶
¶
General Solution is
)dy containsnot do ( xNintermsMdxc ò ò+=
Ex.1:
Show that the following D.Eq. are exact D.Eq. a) 0)2()23( 232 =++++ dyyxxdxxyyx
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-6-
xN
yM
xxxNxx
yM
¶¶
=¶
¶
+=¶¶
+=¶
¶ 23 , 23 22
\ The D.Eq. is exact. b) 0)cos(()]sin()cos([ =+++++ dyyxxdxyxyxx
)cos()sin(
)cos()sin(
yxyxxxN
yxyxxy
M
+++-=¶¶
+++-=¶
¶
\ the D.Eq. is exact.
Ex.2: Is the D.Eq.
2
)( 22
xyyx
dxdy +
-= exact or not?
Sol. dxyxxydy )(2 22 +-=
exact is theD.Eq.,
2 , 2
\¶¶
=¶
¶
=¶¶
=¶
¶
xN
yM
yxNy
yM
Q
Ex.3:
Solve the exact D.Eqs. in Ex.1(a) above 0)2()23( 232 =++++ dyyxxdxxyyx Sol.
c
y
ydxxyyxc
=++
×+×+×=
++= ò ò
23
223
2
yyxy xissolution the3y 2
2 x2
3 x3y
dy2)23(
2
Ex.4:
Solve 0)()( 2 =+++ dyyxdxyx Sol.
exact is theD.Eq.
1 , 1
\
=¶¶
=¶
¶xN
yM
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-7-
x)dycontainsnot do (ò ò+= NintermsMdxc
c
xy
ydxyx
=++
++=
++= ò
3yxy
2x issolution the
3y
2 x
dy)(
32
32
2
Exercise:
1. (2+yexy)dx+(xexy-2y)dy=0 ans: c=2x+exy-y2 2. (tanx+tany)dy+(ysec2x+secx tanx)dx=0 ans: c=y tanx-lncosy+secx 3. (2xy+y2)dx+(x2+2xy-y)dy=0 ans: x2y+y2x-y2/2=c
Problems: Solve the following differential equations: 1- 0)1(ln 2 =++ dyxydxy 2- 022 =- -+ dxedye xyyx 3- 0)2()2( =-++ dyyxdxyx
4- dxxyxydyx ))(cos ( 2+=
5- dxxyydyxyx )lnln1()ln(ln -+=- 6- 0)12( 2 =--+ dxxydyx 7- 0)cos-siny ( cos 2 =+ dyyxdxy 8- 0)12()1( 22 =++++ dyyxydxy
9- 0)()ln( =+
++ dyy
yxdxyex
10- 0)(21)1( 22 =+++ dyeyxdxex yy
References:
1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت) 4- Modern Introduction Differential Equations, Schaum's Outline Series.
Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:
-8-
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-9-
B. Second Order Differential Equations: The second order linear differential equations with constant coefficient has the genral
form is:
)(xFcyybya =+′+′′ …(1),
where a, b and c are constants.
If 0)( =xF then (1) is called homogenous.
If 0)( ≠xF then (1) is called non homogenous.
Ex:
1) y''-x2y'+sinx y=0 is linear, 2nd order, homo.
2) y''-(y')2+ y=sinx is non linear, 2nd order, non homo.
3) y''+2yy'=lnx
a) Homogeneous.
b) Nonhomogeneous.
- Undeterminant coefficients.
- Variation of parameters.
a) The Second order linear homogenous D.Eq. with constant coefficient:
The general form is
0=+′+′′ cyybya …(2)
where a, b and c are constants.
The general solution
Put y'=Dy and y''=D2y in eq. (2) (D is an operator)
⇒ a D2y+bDy+cy=0
⇒ 0y)cbDaD( 2 =++ (using D-operator)
now substitute D by r and leave y then
02 =++ cbrar
is called characteristic equation of the differential equation and the solution of this equation
(the roots r) give the solution of the differential equation where
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-10-
a
acbbr2
42 −−=
m
There are two values of r :
1- real (equal and not equal). 2- complex.
Case 1: If 042 facb − then r1 and r2 are distinct (r1≠ r2) and real roots, and the general
solution is xrxr ececy 2121 +=
Case 2: If 042 =− acb then r 21 == rr , and the general solution is:
rxexccy )( 21 +=
Case 3: If 042 pacb − then the roots are two complex conjugate roots βα ir ±= , 1−=i ,
and the general solution is:
)sincos( 21 xcxcey x ββα +=
Ex.1: Solve 032 =−′−′′ yyy
Solution:
032 =−′−′′ yyy
3 031 01
0)3)(1(y , y , 1y , 032 22
=⇒=−−=⇒=+
=−+=′′=′==−−
rrrr
rrrrrr
the general solution is xx ececy 3
21 += −
Ex.2: Solve 096 =+′−′′ yyy
Solution:
096 =+′−′′ yyy
3 0)3( 096
212
2
==⇒=−
=+−
rrrrr
xexccy 321 )( +=∴
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-11-
Ex.3: Solve 0=+′+′′ yyy
Solution:
0=+′+′′ yyy
231-
23-1-
1.21.1.41
1c 1,b 1,a 012
i
br
rr
±=
±=
−±−=
====++
23 ,
21-
23
21
==±−
= βαir
)23sin
23cos( 21
21
xcxceyx
+=∴−
Exercise: solve
1. 4y''-12y'+5y=0 ans:y=c1e(1/2)x+ c2e(5/2)x
2. 3y''-14y'-5y=0 ans:y=c1e5x+ c2e(-1/3)x
3. 4y''+y=0 ans:y=c1cos(x/2)+ c2sin(x/2)
4. y''-8y'+16y=0 ans:y=c1e4x+ c2xe4x
5. y''+9y=0 ans:y=c1cos3x+ c2sin3x
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-12-
b) The Second order linear non homogenous D.Eq. with constant coefficient:
The general form is: )(xFcyybya =+′+′′ …(3)
where a, b and c are constants.
The general solution
If yh is the solution of the homo. D.Eq. 0=+′+′′ cyybya , then the general solution of
eq. (3) is:
integral)r (porticula
function)tary (complemen
p
hph
yyyyy +=
hy )i is y homo.
py )ii (use the table)
Methods of finding py :
There are two methods:
1) Undetermined coefficients:
In this method py depends on the roots r1, and r2 of characteristic equation and on the
form of )(xF in eq. (3) as follows:
)(xF Choice of py M.R. nkx
nth degree polynomial 0
22
11 kxkxkxk n
nn
nn
n ++++ −−
−− L 0
pxke pxce p
xkorxk
ββ
cossin
xcxc ββ sincos 21 + βim
Note: For repeated term (root), multiply by x .
Ex.1: Use the table to write py
1) 2n , 3k , 3)( 2 === xxF
012
2 kxkxky p ++=
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-13-
2) c 21-k , e
21)( 3x- ⇒=
−=xF
xp cey 3−=
3) 3 , 2k , 3x cos 2)( === βxF
3xsin c 3x cos 21 += cyp
4) 2c , 3k , 2e - 53x-3)( 3x2 −=−=+= xxF
xp cekxkxky 3
012
2 +++=
5) sin 21 cos2)( xxxF −=
sin x c x cos 21 += cyp
6) 2 cos sin)( xxxF −=
2xsin B 2x cosA sin x c x cos 21 +++= cyp
Ex.2: Solve 242 xyyy =−′−′′ …. (1)
Solution:
y'' –y'-2y=0
the char. Eq. r2-r-2=0
2r ,1r 0)2r)(1r(
21 =−==−+
xxh ececy 2
21 += −
f(x)=4x2 is polynomial of second degree then
212
012
2
2 , 2
(2) ...
kykxkykxkxky
pp
p
=′′+=′⇒
++=
Substitution gives
4) (2)2(2 201
22122 xkxkxkkxkk =++−+−
-3k 022:2k 022:.2 42:.
0012
112
222
=⇒=−−=⇒=−−−=⇒=−
kkkconstkkxofcoeff
kkxofcoeff
3-2x2 2 +−= xy p
322)(yy 2221hg −+−+=+= − xxececy xx
p
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-14-
Ex.3: ey2yy x3=−′−′′
Solution:
(1) .... ey2yy x3=−′−′′
0y2yy =−′−′′
1,2 0)1)(2( 02
21
2
−==⇒=+−=−−
rrrrrr
)( 22
1xx
h ececy −+= , Put
3x3
3
9ce , 3
(2) ....
=′′=′
=
px
p
xp
ycey
cey
Substitute In (1)
9ce3x-3ce3x-2ce3x=e3x
9c-3c-2c=1⇒41c 14 =⇒=c
In (2)⇒ 41 3x
p ey =
xxxp eececy 3
22
1hg 41yy ++=+= −
Modification rule قاعدة التعديل
nkxxF) اذا كان 1 .x السابق في py يضرب ← 0 وكان احد جذري المعادلة القياسية = )(=2 ( - a اذا كان pxkexF .x السابق في py يضرب ← p وكان احد جذري المعادلة القياسية = )(=
b اذا كان - pxkexF .2x السابق في py يضرب ← p وكان جذري المعادلة القياسية = )(=
) اذا كان 3
=xsin k x cos
)(ββk
xF0 وكان , == αβir m .x السابق في py يضرب ←
Ex.4:Solve y''+y= sinx
Solution:
y''+y=0
r2+1=0, r2=-1 ⇒ r = ± i, α=0, β=1
yh=c1cosx+c2sinx
yp=x(c3cosx+c4sinx), y'p=x(-c3sinx+c4cosx)+(c3cosx+c4sinx)
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-15-
y''p=x(-c3cosx-c4sinx)+(-c3sinx+c4cosx)+(-c3sinx+c4cosx)
Substitution gives
-2c3sinx+2c4cosx=sinx
-2c3=1⇒ c3=-1/2, 2c4=0⇒c4=0
xxxcxcyg cos2
sincos 21 −+=
Exercise: Find the general solution
1) y''=9x2+2x-1
2) y''-y'-2y=sin2x
3) y''-5=3ex-2x+1
4) y''+2y'+y=3e-x
5) y''-y'-2y=x2-x
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-16-
2- Variation of parameters.
Let yh=c1u1+c2u2 be the homogenous solution of )x(Fcyybya =+′+′′ and the particular
solution has the form 2211p vuvuy += where v1 and v2 are unknown functions of x which
must be determined, first solve the following linear equations for v'1 and v'2:
v'1u1+ v'2u2=0
v'1u'1+ v'2u'2=F(x)
which can be solved with respect to v'1 and v'2 by Grammar rule as follows
)x(Fu0u
D,u)x(Fu0
D,uuuu
D1
12
2
21
21
21
′=
′=
′′=
and D
Dv,DDv 2
21
1 =′=′
by integration of v'1 and v'2 with respect to x we can find v1 and v2.
Ex.1:
Solve x3ey2yy =−′−′′ ……. (1)
xxh ececy 2
21 += − , hence
xx
xx
eueu
eueu2
22
2
11
2
=′⇒=
−=′⇒= −−
yp= v1u1+v2u2
)(
0
2211
2211
xFuvuv
uvuv
=′′+′′=′+′
x3x22
x1
x22
x1
e)e2(v)e(v
0)e(v)e(v
=′+−′=′+′
−
−
Solving this system by Cramer rule gives
x2x3x
x
2x5
x2x3
x2
1x
x2x
x2x
eee0e
D,ee2e
e0D,e3
e2eee
D =−
=−===−
=−
−
−
−
xx2
xx
x2
2
x4x41
x4x
x5
1
e31e
31ve
31
e3e'v
,e121e
31ve
31
e3e'v
∫
∫
==⇒==
−=−
=⇒−
=−
=
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-17-
x3x22
x1
x3x2xxx4p
e41ececy:issolutiongeneralthe
e41ee
31ee
21y
++=
=+−=∴
−
−
Ex.2: solve
y''+y=secx
Solution:
y''+y=0
r2+1=0 ⇒r2=-1 ⇒ r = ± i α=0, β=1
yh=c1cosx + c2sinx, u1=cosx, u2=sinx, f(x)=secx
yp= v1u1+v2u2
= v1cosx +v2sinx then
xsec)x(cosv)xsin(v
0)x(sinv)x(cosv
21
21
=′+−′=′+′
1xsecxcosxsecxsin
0xcosD
,xtanxcos
1xsinxsecxsinxcosxsecxsin0
D
,1xsinxcosxcosxsinxsinxcos
D
2
1
22
==−
=
−=−=−==
=+=−
=
∫∫
==⇒=
=−
=⇒−=−
=′
xdxv1'v
|xcos|lndxxcosxsinvxtan
1xtanv
22
11
yp = ln |cosx| cosx + x sinx
yg = c1cosx + c2sinx + ln |cosx| cosx + x sinx
Exercise
1. y''-2y'+y = ex lnx
2. y''-2y'+y = 5
x
xe
3. y''+4y=sin22x
Mathematics: Lecture 2 مدرس مساعد ازهار مالك Differential Equations:
-18-
Problems: Find the general solution
1- xeyyy x ln42 −=+′+′′
2- -1(0)y , 0y(0) cos15sin23204 =′=−=+′+′′ ttyyy
3- 3(0)y , 1-y(0) 434 3 =′==+′−′′ xeyyy
4- xxyy 2 +=+′′
5- 3
x
xe12yy2y =+′−′′
6- xyy 2 sec 44 =+′′
References:
1- calculus & Analytic Geometry (Thomas).
2- Calculus (Haward Anton).
3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
4- Modern Introduction Differential Equations, Schaum's Outline Series.
Mathematics: Lecture 3 مدرس مساعد ازهار مالك Differential Equations:
1
C. Higher order Differential Equations: How to find roots of an equation:
Let 022
11 =++++ −−
nnnn axaxax K be an eq. of degree n, we denote this eq. by 0)( =xf then:
1) r is a root of the eq. 0)( =xf if 0)( =rf .
2) r is repeated root of the eq. 0)( =xf if 0)( =′ rf .
3) If r is a root of the eq. 0)( =xf , then r must be a factor of na .
4) If r is a root of the eq. 0)( =xf , then )( rx − divides )(xf .
Ex.: Find all roots of 01834 23 =−−+ xxx
Solution: 18 9, 6, 3, 2, 1, : 18 mmmmmm=na
0186168)2( =−−+=f , 2 is a root of the eq. There are two methods to factorize f(x): long division & fast division.
First method: Fast division 1 4 -3 -18 2 ↓ 2 12 18 1 6 9 0
096 2 =++⇒ xx
3 ,3 ,2 0)3()3)(2( 0)96)(2(
321
2
−=−===++−=++−⇒
xxxxxx
xxx
The roots are 2, -3, -3
Second method: long division
01834 23 =−−+ xxx
0)3()3)(2( 0)96)(2( 2
=++−=++−⇒
xxxxxx
9x6x
18x9 18x9 x12x6
x3x6 x2x
18x3x4x)2x(
2
2
2
23
23++
±−
±
−
±
−−+−
0m
m
m
Mathematics: Lecture 3 مدرس مساعد ازهار مالك Differential Equations:
2
Higher order linear Differential Equations: The general form with constant coefficient is:
(1) ...... )(1)1(
1)( xFyayayay nn
nn =++++ −− K
If 0)( =xF then (1) is called homogenous, otherwise (1) is called nonhomogenous.
The general solution
The methods of solving second order homogenous D.Eqs. with constant coefficients can be
extended to solve higher order homogenous and nonhomogenous D.Eq. with constant coefficients.
a) Homogenous: the characteristic equation of nth order homogenous D. Eq.: 01
)1(1
)( =++++ −− yayayay nn
nn K is: 01
11 =++++ −
−nn
nn ararar K Let r1, nrrr , ... , , 21 be the roots of characteristic equation then: 1) If nrrr , ... , , 21 are all distinct then the solution is:
xrn
xrxr necececy +++= K2121h
2) If 1r repeated m times, then hy will contain the terms:
... 111 121
xrmm
xrxr excxecec −+++ 3) If some of roots are complex ( βα ir m= ) then yh will contain
xexcxc αββ ) sin cos( 21 +
Ex.1: solve y''' -3y''+2y'=0
Solution:
1 , 2 , 0 01)-2)(-( 0)23( 023
321
223
===⇒==+−⇒=+−
rrrrrrrrrrrr
are all distinct
xx
xrxrxr
ececcyecececy
32
21h
321h
321
++=
++=⇒
Mathematics: Lecture 3 مدرس مساعد ازهار مالك Differential Equations:
3
Ex.2:
y(4)-3y'''+3y''-y'=0
3m 1 , 0 01)-(
0)133( 033
4321
3
23234
=⇒====⇒=
=−+−⇒=−+−
rrrrrr
rrrrrrrr
xxx
rxmmmxr
excxececcyexcxcxcecy
24321h
14
23
321h
)(1
+++=
+++=⇒ −−−
Ex.3:
012223)4( =+′+′′−′′′− yyyyy
factor a is 2)-( root a is 2 012223 234
rrrrrr⇒=
=++−−
1 -3 -2 2 12 2 ↓ 2 -2 -8 -12 1 -1 -4 -6 0
064 23 =−−−⇒ rrr
factor a is 3)-( root 3 , 064)(2( 23 rrrrrr ⇒==−−−−⇒ )
1 -1 -4 -6 3 ↓ 3 6 6 1 2 2 0
022 2 =++⇒ rr
1 , 11 ,3 ,2 0)22( )3)(2(
21
2
=−=−====++−−
βαirrrrrrr
m
xxx excxcececy −+++=⇒ ) sin cos( 433
22
1h
b) Nonhomogenous: the general form of nth order nonhomogenous differential equation is:
(1) ...... )(1)1(
1)( xFyayayay nn
nn =++++ −− K
The general solution is py+= hg yy
Mathematics: Lecture 3 مدرس مساعد ازهار مالك Differential Equations:
4
Methods of finding yp:
1) Undetermined coefficients
We can extend the methods of solving second order non homogenous D.Eqs. with constant
coefficients to solve higher order nonhomogenous D.Eq. with constant coefficients.
Ex.1: y(4)-8y''+16y=-18sinx
Solution:
py+= hg yy
y(4)-8y''+16y=0
r4-8r2+16=0 ⇒ (r2-4)2=0 ⇒ r2=4 ⇒ r=±2
yh=c1e2x+c2xe2x+c3e-2x+c4xe-2x
let yp=Acosx+Bsinx, y'p=-Asinx+Bcosx, y''p= -Acosx-Bsinx
y'''p= Asinx-Bcosx, y(4)p= Acosx+Bsinx
Acosx+Bsinx+8Acosx+8Bsinx+16Acosx+16Bsinx=-18sinx
25Acosx+25Bsinx=-18sinx
25A=0⇒A=0
25B=-18⇒B=-18/25
xsin2518-xececxececy 2x-
42x-
32x
22x
1g +++=
2) Variation of parameters
In this method, the particular solution yp has the form yp=v1u1+v2u2+… +vnun
Where u1, u2, …, un are taken from yh=c1u1+c2u2+… +cnun.
To find v1, v2, …, vn, we must solve the following linear eqs. For v'1, v'2, …, v'n:
)(uv uvuv0uv uvuv
0uv uvuv0uv uvuv
)1(nn
)1(22
)1(11
)2(nn
)2(22
2)-(n11
nn2211
nn2211
xfnnn
nn
=′+…+′+′=′+…+′+′
=′′+…+′′+′′=′+…+′+′
−−−
−−
M
Mathematics: Lecture 3 مدرس مساعد ازهار مالك Differential Equations:
5
Ex2: solve y'''+y'=secx
Solution:
Let y'''+y'=0
r3+r=0 ⇒ r(r2+1)=0 ⇒ r=0, r2=-1 ⇒ r1=0, r=±i
yh=c1+c2cosx+c3sinx
u1=1, u2=cosx, u3=sinx, f(x)=secx
xxxxx
inxosx
sec)(sinv)cos(v)0(v0)(cosv)sin(v)0(v
0svcvv
321
321
321
=′−−′+′=+−′+′
=′+′+′
1xcosxsinxsinxcos0
xcosxsin0xsinxcos1
D 22 =+=−−
−=
xsec)xcosx(sinxsecxsinxcosxsec
xcosxsin0xsinxcos0
D 221 =+=
−−−=
1xsecxcosxsinxsec
xcos0
xsinxsec0xcos00xsin01
D2 −=−=−
=−
=
xtanxsecxsinxsecxcos
0xsin
xsecxcos00xsin00xcos1
D3 −=−=−−
=−−=
∫ +==⇒==′ )xtanxln(secxdxsecvxsecDDv 1
11
∫ −=−=⇒−==′ xdx1v1D
Dv 22
2
∫ =−=⇒−==′ xdxxvxDD
v coslntantan 33
3
yp= ln (secx+tanx)-x cosx-ln cosx sinx
yg= c1+c2cosx+c3sinx+ ln (secx+tanx)-x cosx - ln cosx sinx
Mathematics: Lecture 3 مدرس مساعد ازهار مالك Differential Equations:
6
Exercise: Solve 1) y'''-6y''+12y'-8y=0
2) y'''-y=0
3) y(5)-2y(4)+y'''=0
4) y'''-6y''+2y'+36y=0
5) y(4)+8y'''+24y''+32y'+16y=0
6) y(4)-4y''+4y=0
Problems: Find the general solution y'''-6y''+12y'-8y=0
1) y(4)+8y''+16y=0
2) y(4)+y=x+1
3) y'''-3y'+2y=ex
4) y(4)-16y=0
5) y'''-y'=4x3+6x2
References:
1- Calculus & Analytic Geometry (Thomas).
2- Calculus (Haward Anton).
3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
4- Modern Introduction Differential Equations, Schaum's Outline Series.
Mathematics: Lecture 4 مدرس مساعد ازهار مالك Matrices:
1
Matrices: When a system of equations has more than two equations, it is difficult to discuss
them without using matrices and vectors. The size of the matrix is described by the number of its row and columns. A
matrix of n rows and m columns is said to be mn × matrix.
[ ] m.1,2,...,j ,n 1,2,...,i ,
21
22221
11211
===
=
×
ij
mnnmnn
m
m
a
aaa
aaaaaa
A
L
M
L
L
Types of matrices: Square matrix: It is a matrix whose number of rows are equal to the number of columns ( mn = ). For example:
224251
×
=A ,
33081123031
×
=B
Diagonal matrix: It is a square matrix which all its elements are zero except the elements on the main diagonal. For example:
=
100090004
A
Identity matrix: It is a diagonal matrix whose elements on the main diagonal are equal to 1, and it is denoted by In. For example:
=
=
1001
,100010001
23 II
Transpose matrix: Transpose of A is denoted by )( TA , means that write the rows of A as columns in At. For example:
23
T
32 3-1172-3
A , 312
173
××
=
−−
=A
Mathematics: Lecture 4 مدرس مساعد ازهار مالك Matrices:
2
Matrix addition and multiplication If ][ and ][ ijij bBaA == and both A & B are mn × matrices, then ][][][ ijijijij babaBA +=+=+ Ex.1:
=
+
−7423
5432
2011
For any scalar (number) c , we can multiply A by c as follows:
]ca[]c[acA ijij ==
Ex.2:
−=
−6033
2011
3
A matrix with only one column, 1×n in size, is called a column vector, and one
of only one row, m×1 in size, is called a row vector.
Matrices multiplication Let A be an n× k matrix and B be a k×m matrix then C=AB is an n×m matrix,
where the element in the ith row and jth column of AB is the sum
.,...,2,1,...,2,1,...1
2211 pjandmibabababacn
kkjiknjinjijiij ===+++= ∑
=
Ex.3
Suppose
2332312
173
××
=
−−
=1-1
302-5
B , A then
2210131416
×
−
=AB
Determinants With each square matrix A we associate a number det(A) or |aij| called the
determinant of A, calculated from the entries of A as follows: For n=1, det(a)=a,
Mathematics: Lecture 4 مدرس مساعد ازهار مالك Matrices:
3
For n =2, 122122112221
1211det aaaaaaaa
−=
Minors To each element of a 33× matrix there corresponds a 22× matrix that is obtained by deleting the row and column of that element. The determinant of the 22× matrix is called the minor of that element.
For a matrix of dimension 3×3, we define
3231
222113
3331
232112
3332
232211
333231
232221
131211
333231
232221
131211
detaaaa
aaaaa
aaaaa
aaaaaaaaaa
aaaaaaaaa
⋅+⋅−⋅==
.
,,
3231
2221
3331
2321
3332
2322
13
1211
a of minor the is
a of minor the is a of minor the is where
aaaa
and
aaaa
aaaa
Ex.4:
Find the determinant of each matrix
a)
− 52
31
115231
=−
b)
12642
012642
=
Ex.5: Find the determinant of A where:
−−
−=
970642531
A
Sol.: By choosing the first column we get
62
6453
09753
)2(9764
1970642531
)det(
=
−⋅+
−−
⋅−−−
⋅=−
−−
=A
Mathematics: Lecture 4 مدرس مساعد ازهار مالك Matrices:
4
Ex.6: Evaluate the determinant of A if:
−−
−=
970642531
A
Solution: By choosing the second row we get
62
7031
69051
49753
)2(970642531
)det(
=
−⋅−
−+
−−
−−=−
−−
=A
Note that 62 is the same value that was obtained for this determinant in Example above. Note:
If a matrix A is triangular (either upper or lower), its determinant is just the product of the diagonal elements: Linearly Dependent and Linearly Independent Definition: the vectors v1, v2, …, vm are linearly dependent if | v1 v2 … vm|=0, and if | v1 v2 … vm| ≠ 0 then v1, v2, …, vm are linearly independent. Ex1: Let v1=(3, 6, -1); v2=(8, 2, -4); v3=(1, -1, 1), determine whether v1, v2, v3 are linearly dependent or not. Sol: Since
068)224()16(8)42(341
261116
81412
3141126
183≠−=+−+−−−=
−−+
−−
−−
−=
−−−
then v1, v2, v3 are linearly independent Ex2: Let v1=(2, 4, 6); v2=(1, 3, 3); v3=(1, 2, 3), determine whether v1, v2, v3 are linearly dependent or not. Sol: Since
0606)1812()1212()69(23634
3624
3323
2336234112
=−−=−+−−−=+−=
then v1, v2, v3 are linearly dependent
Mathematics: Lecture 4 مدرس مساعد ازهار مالك Matrices:
5
Exercise: 1) Determine whether the given vectors are linearly dependent or linearly independent.
a) (3,2);(1,-1) b) (4,-3,1);(10,-3,0);(2,-6,3)
2) Find determinant of the following matrices
a)
−
3001020000101014
b)
−−−
−
1231101
123
References: 1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
1
Solving a system of linear equations Let A be a matrix, X a column vector, B a column vector then the system of linear
equations is denoted by AX=B. The augmented matrix
The solution to a system of linear equations such as
6352
=+−=−
yxyx
Depends on the coefficients of x and y and the constants on the right-hand side of
the equation. The matrix of coefficients for this system is the 22× matrix
−1321
If we insert the constants from the right-hand side of the system into the matrix of
coefficients, we get the 32× matrix
−−65
1321
We use a vertical line between the coefficients and the constants to represent the
equal signs. This matrix is the augmented matrix of the system also it can be written as:
−=
−65
1321
yx
Note: Two systems of linear equations are equivalent if they have the same solution set.
Two augmented matrices are equivalent if the systems they represent are equivalent. Ex.1: Write the augmented matrix for each system of equations.
a) 042
3 25
=+−=+
=−+
zyxzx
zyx
−
−
035
412102111
b) 0
6 1
==+=+
zzyyx
− 561
100110011
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
2
We'll take two methods to solve the system AX=B 1) Cramer's rule
The solution to the system
222
111
cybxacybxa
=+=+
Is given by where and DD
yDDx yx ==
and ,22
11
22
11
22
11
caca
Dbcbc
Dbaba
D yx ===
Provided that 0≠D Notes:
1. Cramer's rule works on systems that have exactly one solution. 2. Cramer's rule gives us a precise formula for finding the solution to an
independent system. 3. Note that D is the determinant made up of the original coefficients of y and x .
D is used in the denominator for both y and x . xD is obtained by replacing the first (or x ) column of D by the constants 21 c and c . yD is found by replacing the second (or y ) column of D by the constants 21 c and c .
Ex.1: Use Cramer's rule to solve the system:
32423
−=+=−
yxyx
Sol.: First find the determinants yD and ,, xDD :
-178--932
43 , -26-4
1324
7)4(31223
==−
===−
−=
=−−=−
=
yx DD
D
By Cramer's rule, we have
717y and
72
−==−==DD
DDx yx
Check in the original equations. The solution set is
−− )
717,
72( .
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
3
Ex.2: Solve the system:
532932
=−−=+
yxyx
Sol.: Cramer's rule does not work because
0)6(632
32 =−−−=
−−=D
Because Cramer's rule fails to solve the system, we apply the addition method:
140 532
932
==−−
=+yx
yx
Because this last statement is false, the solution set is empty. The original equations are inconsistent. Ex.3: Solve the system:
14106 753
=−=−
yxyx
Sol.: Cramer's rule does not apply because
0)30(3010653
=−−−=−−
=D
Multiply Eq.(1) by -2 and add it to Eq.(2)
00 14106
14106
==−
−=+−yxyx
Because the last statement is an identity, the equations are dependent. The solution set is{ }753),( =− yxyx . Ex.4: Use Cramer's rule to solve the system:
532
3)1(32 −=
−=+−xyyx
Sol.: First write the equations in standard form, CByAx =+
523
032 −=+−
=−yxyx
Find yD and ,, xDD :
-100--10
5302
, -1515-02530
5942332
==−−
===−
−=
−=−=−
−=
yx DD
D
Using Cramer's rule, we get
25
10 and 35
15=
−−
===−−
==DD
yDDx yx
Because (3,2) satisfies both of the original equations, the solution se is )}2,3{( .
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
4
2) The Gaussian Elimination method When we solve a single equation, we write simpler and simpler equivalent
equations to get an equation whose solution is obvious. In the Gaussian elimination method we write simpler and simpler equivalent augmented matrices until we get an augmented matrix in which the solution to the corresponding system is obvious.
Because each row of an augmented matrix represents an equation, we can perform the row operations on the augmented matrix. Elementary Row Operation: 1. Construct the augmented matrix (A:B). 2. Interchange two rows (Ri ↔ Rj). 3. Multiply any row by a constant different from zero (Ri ↔ kRi) 4. Add a constant multiply of any row to another row (Ri ↔ Ri + kRj) Ex.1:
Use Gaussian elimination method to solve the system (two equations in two variables):
12113
=+=−
yxyx
Sol.: Start with the augmented matrix:
−111
1231
222111
7031
RR +=′
−
−1-2R
22311
1031
RR7
1 =′
−
−
1213
21001
RRR +=′
−
3
This augmented matrix represents the system 2=x and 3−=y . So the solution set to the system is ( ){ }3,2 − . Ex.2: Use Gaussian elimination method to solve the system (three equations in three variables):
43
632
=−−=−+
−=+−
zyxzyx
zyx
Sol.:
−
−−−
−
463
113111
112
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
5
1
R
−
−−−
−↔
43
6
113112111
2R
33
22
RR
RR
+=′
+=′
1-3R
1-2R
14156
24033011
−−
−−
−1
2R3
1- 2R
−−−−
=′
1456
240110111
33
11
RR
RR
+=′
+=′
24R
2-R
651
20010
01
−−1
1
3R2
1-3R
−−=′
351
100110
001 3R2R
−+=′
321
100010001
2R
This augmented matrix represents the system 1=x , 2=y and 3−=z . So the solution set to the system is ( ){ }3,2,1 − .
Ex.3: Solve the system
433
1=+−
=−yx
yx
Sol.:
−
−41
3311
13R
−+=′→
71
0011
22 RR
R2 corresponds to the equation 0 = 7. So the equations are inconsistent, and there is no solution to the system. Ex.4: Solve the system
226
13=+
=+yx
yx
Sol.:
21
2613
2R12R-
→ +=′
01
0013
2R
In the R2 of the augmented matrix we have the equation 0 = 0. So the equations are dependent. For ordered pair that satisfies the first equation satisfies both equations. The solution set is { }13),( =+ yxyx
Exercises: Solve the following systems:
1) 13
3−=+−
=+yx
yx
2) 363
12=+
=+yx
yx
3) 22
142
=+−=−+
=++
zyxzyx
zyx
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
6
Matrix Inverse The matrix A has an inverse denoted by A-1 if |A|≠0 where A.A-1=I. We'll take two
methods to find A-1 where A is an n×n matrix. 1) By Gauss elimination method(Using row operations):
1. Construct the augment matrix (A:I) 2. Use row operation until we have (I:A-1)
Ex1: Use Row operation to find A-1 if A=
4112
1001
4112
→ 11 21 RR = →
10
021
41211
−→−=′1
21
021
270211
122 RRR
−=′
42
71
021
10211
72
22 RR
−
−
=−=′
72
71
71
74
1001
21
211 RRR
A-1 =
−
−
72
71
71
74
Ex2: Find A-1 if A =
−
−
204201
312
−
−
100010001
204201
312
11 21 RR =→
−
−
100010
0021
204201
23
211
→R2= R2-R1
→R3=R3-4R1
−
−
−
−
−
102
0121
0021
42027
210
23
211
→R2=2R2
−−
−−
−
102021
0021
420710
23
211
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
7
→R1= 1221 RR + →R3=-2R2+R3 33 10
1
140021010
1000710
2401
RR =→
−−−
−
−−−
−
101
520
021010
100710201
→ R1=2R3+R1→ R2=R2+7R3
−
−−
101
520
107
541
51
510
100010001
−
−−=∴ −
101
520
107
541
51
510
1A
2) By Cofactor Method (Using determinant of the matrix)
The cofactor of the element aij of the matrix A = (aij) is defined by cij = (-1)i+j Aij
where Aij is the determinant of the matrix that remains when the row i and the column j are deleted.
To find the inverse of a matrix whose determinant is not zero 1- construct the matrix of cofactors of A, cof (A) = cij 2- Construct the transposed matrix of cofactors called the adjoin of A = adj (A) =
cof (A)T
3- then A-1 = )(det
1A
adj A
4- to check your answer A.A-1 = I or A-1.A = I
Ex.: Use determinant to find A-1 where A =
4112
A-1 = )(1 AadjA
7184112
=−==A
Cof(A) =
−
−2114
C11 = (-1)1+1 |4| = 4 C12 = (-1)1+2 |1| = -1 C21 = (-1)2+1 |1| = -1 C22 = (-1)2+2 |2| = 2
Adj(A) =
−
−=
−
−2114
2114 T
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
8
−
−
=
−
−=∴ −
72
71
71
74
2114
711A
Ex2: Find A-1 if A =
−
−
204201
312
Solution:
−−
−=
=+=−−
=
172482
0100)(
1082204201
312
Acof
A
1,7)7)(1(,2,4,8,2)2(,0,1010)1(,0
333231
232221
131211
==−−==−=−==−−==−=−==
cccccc
ccc
−−−==
1407810220
tcof Adj(A)
−
−−=∴ −
101
520
107
541
51
510
1A
Mathematics: Lecture 5 مدرس مساعد ازهار مالك Matrices:
9
Problems: 1) write the augment matrix to the following systems then find the solution:
a) 3333
22221
−=−+−=+−
=+−
zyxzyx
zyx
b) 8333
122
=−+=+−
=−+
zyxzyx
zyx
c)
5102
2212
41
41
32
2
=+−=−
=−=
xxxxxx
x
2) Find the inverse of each following matrix
a)
−
−
425253112
b)
−−
−
121112311
References: 1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
1
Vector: A vector is a matrix that has only one row – then we call the matrix a row vector
– or only one column – then we call it a column vector.
A row vector is of the form: [ ]n21 a ... a aa =
A column vector is of the form:
=
mb
bb
bM
2
1
A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment
A vector in the plane is directed line segment. The directed line segment AB
has initial point A and terminal point B; its length is denoted by AB . Two vectors
are equal if they have the same length and direction.
Component form
If v is a two dimensional vector in the plane equal to the vector with initial
point at the origin and terminal point ),( 21 vv ,then the Component form of v is:
),( 21 vvv =
If v is a three dimensional vector in the plane equal to the vector with initial
point at the origin and terminal point ),,( 321 vvv , then the Component form of v is:
),,( 321 vvvv =
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
2
The numbers 321 and , vvv are called the components of v .
Given the points ),,( 111 zyxP and ),,( 222 zyxQ , the standard position vector
),,( 321 vvvv = equal to PQ is
) , , ( 121212 zzyyxxv −−−=
The magnitude or length of the vector PQv = is the nonnegative number
212
212
212
23
22
21 )( )( )( zzyyxxvvvv −+−+−=++=
The only vector with length 0 is the zero vector )0,0(0 = or )0,0,0(0 = . This
vector is also the only vector with no specific direction.
Ex.: Find a) component form and b) length of the vector with initial point )1, 4, 3(−P
and terminal point )2, 2, 5(−Q
Solution:
a) )1-2 , 4-2 , 35( +−=v
The component form of PQ is 1) , 2- , (-2=v
b) The length or magnitude of PQv = is 3 9 )1( 2)( )2( 222 ==+−+−=v Vector Addition and Multiplication of a vector by a scalar
Let ),,( 321 uuuu = and ),,( 321 vvvv = be vectors with k a scalar.
Addition:
) , , ( 332211 vuvuvuvu +++=+
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
3
Scalar multiplication: ),,( 321 kukukuku =
If the length of ku is the absolute value of the scalar k times the length of u .
The vector uu −=− )1( has the same length as u but points in the opposite direction.
If ),,( 321 uuuu = and ),,( 321 vvvv = , ) , , ( 332211 vuvuvuvu −−−=−
Note that uvvu =+− )( and the difference vu − as the sum )( vu −+
Ex.:
Let )1,3,1(−=u and )0,7,4(=v , find
a) vu 32 + b) vu − c) u21
Solution:
a) 2) 27, (10, 0) 21, (12,2) 6, ,2(32 =+−=+ vu
b) 1) 4,- ,5(−=− vu
c) 1121
21 ,
23 ,
21
21
=
−
=u
Properties of vector operations:
Let wand v, u be vectors and b and a be scalars.
1) uvvu +=+ 2) )()( wvuwvu ++=++
3) uu =+ 0 4) 0)( =−+ uu
5) 00 =u 6) uu =1
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
4
7) uabbua )()( = 8) avauvua +=+ )(
9) buauuba +=+ )(
Unit vectors
A vector v of length 1 is called unit vector. The standard unit vectors are:
)1,0,0( , )0,1,0( , )0,0,1( === kji
kvjvivvvv
vvvvvvv
321
321
321321
)1,0,0()0,1,0()0,0,1(
),0,0()0,,0()0,0,(),,(
++=++=
++==
We call the scalar (or number) 1v the i-component of the vector v , 2v the
j-component of the vector v , and 3v the k-component. In component form,
),,( 1111 zyxP and ),,( 2222 zyxP is
kzzyyxxPP )( )j( )i( 12121221 −+−+−=
If 0≠v , then
vvu = is a unit vector in the direction of v , called the direction of the
nonzero vector v .
Ex.:
Find a unit vector u in the direction of the vector )1,0,1(1P and )0,2,3(2P .
Solution k-2j2i )1-0( 0)j-2( )i13(21 +=++−= kPP
3 9 )1(- (2) )2( 22221 ==++=PP
kjikjiPPPPu
31
32
32
322
21
21 −+=−+
==
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
5
The unit vector u is the direction of 21PP . Midpoint of a line segment
The Midpoint M of a line segment joining points ),,( 1111 zyxP and ),,( 2222 zyxP is the point
+++
2)( ,
2)( ,
2 )( 212121 zzyyxx
Ex.:
The midpoint of the segment joining )0,2,3(1 −P and )4,4,7(2P is
)2,1,5(2
40 , 2
42 ,2
73=
++−+
Product of vectors u & v are vectors,
There are two kinds of multiplication of two vectors:
1- The scalar product (dot product) u.v. The result is a scalar.
2- The vector product (cross product) u×v. The result is a vector.
1) The dot product In this section, we show how to calculate easily the angle between two vectors
directly from their components. The dot product is also called inner or scalar
products because the product results in scalar, not a vector.
Def.: The dot product vu ⋅ of vectors ),,( 321 uuuu = and ),,( 321 vvvv = is:
332211 vuvuvuvu ++=⋅
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
6
Note:
Ex.: a)
7)2()53(
scalar 7)2(5)1(3)2,1()5,3(=+−⋅+
=+−=−⋅jiji
b)
6)25()43(
scalar 68151)2,5,1()4,3,1(−=++⋅+−−=+−=⋅−
kjikji
Angle between two vectors
The angle θ between two nonzero vectors ),,( 321 uuuu = and ),,( 321 vvvv = is given by
θcos⋅⋅=⋅ vuvu
⋅
⋅= −
vuvu cos 1θ where )(0 πθθ ≤≤
Ex.: Find the angle between two vectors in space
98cos
98cos
441414422cos
1-=⇒=
++⋅++++
=⋅
⋅=
θθ
θvuvu
0 , 11.1 =
⋅⋅⋅
==
⋅⋅⋅
jkkjji
kkjj
ii
kjikjiurrrr
22v , 22 +−=+−=
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
7
Ex.: Find the angle θ in the triangle ACB determined by the vertices
⋅
=
=+−=
=−+−=
=+−−=⋅
=−−=
−
13294cos
13 )3()3(CB
29 )2()5(
4 (-2)(3))2)(5( CB
(-2,3)CB and )2,5(
1
22
22
θ
CA
CA
CA
Orthogonal vectors Vectors ),,( 321 uuuu = and ),,( 321 vvvv = are orthogonal (or perpendicular)
if and only if 0=⋅ vu
Ex.: a) )6,4( vand )2,3( =−=u are orthogonal because 0=⋅ vu b) kjkji 42 vand 23 u +=+−= are orthogonal because 0=⋅ vu
c) 0 is orthogonal to every vector u since
0
),,()0,0,0(0 321
=⋅=⋅ uuuu
Properties of the Dot product
If wand , vu are any vectors and c is a scalar, then
1) uvvu ⋅=⋅
2) )()()( vuccvuvcu ⋅=⋅=⋅
3) wuvuwvu ⋅+⋅=+⋅ )(
4) 2uuu =⋅
5) 00 =⋅u
C(5,2) and B(3,5) , )0,0(=A B(3,5)
C(5,2)
A
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
8
Vector projection Vector projection of u onto v
vv
vuuprojv
⋅= 2 …… (1)
uprojv ("The vector projection of u onto v")
Ex.:
Find the vector projection of kjiu 236 ++= onto kjiv 22 −−= and the
scalar component of u in the direction of v.
Solution:
We find uprojv from eq.(1):
kjikjikjivvvvuv
vvuuprojv 9
898
94- )22(
94- )22(
441466 2 ++=−−=−−
++−−
=⋅⋅
=
⋅=
We find the scalar component of u in the direction of v from eq.(2):
Problems: 1) Let )5,2( vand )2,3( −=−=u . Find the a) component form and b) magnitude
(length) of the vector.
1. vu 52 +−
2. vu54
53
+
2) Find the component form of the vector:
a. The vector PQ where Q(2,-1) and )3,1(=P .
b. The vector OP where O is the origin and P is the midpoint of
segment RS , where )3,4(S and (2,-1) −==R .
c. The vector from the point )3,2(=A to the origin.
d. The sum of AB and CD , where
, )0,2(B , (1,-1) ==A and (-1,3)=C )2,2(D −=
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
9
3) Let wand , uv as in the figure: find a) vu + , b) wvu ++ , c) vu − and
d) wu −
4) Find the vectors whose lengths and directions are given. Try to do the calculation
without writing:
Length Direction
a. i 2
b. k- 3
c. kj54
53
21
+
d. kji73
72
76 7 +−
5) Find a) the direction of 21PP and b) the midpoint of line segment 21PP .
a. )5,1,1(1 −P and )0,5,2(2P
b. )0,0,0(1P and )2,2,2(2 −−P
6) Find uv ⋅ , uv , , the cosine of the angle between uandv , the scalar
component of u in the direction of v and the vector uprojv .
a) kjiv 542 +−= , kjiu 542 −+−=
b) kiv )54()
53( += , jiu 125 +=
c) jiv +−= , kjiu 232 ++=
d) jiv += 5 , jiu 172 +=
e)
=3
1 , 2
1v ,
−=
31 ,
21u
v
u
w
Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:
10
7) Find the angles between the vectors:
a) kjiu +−= 22 , kiv 43 +=
b) jiu 73 −= , kjiv 23 −+=
c) kjiu 22 −+= , kjiv ++−=
8) Find the measures of the angles between the diagonals of the rectangle whose
vertices are D(4,1) and C(3,4) , B(0,3) , )0,1(=A
References: 1- Advanced Engineering Mathematics (Erwin Kreyszic)- 8th Edition. 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
11
2) The Cross product
The cross product is also called vector product because the product results a
vector.
Def.:The cross product n )sin u( θvvu =× , n unit vector (normal) perpendicular to the plane. Note: The vector vu × is orthogonal to both and vu
Parallel vectors
Nonzero vectors and vu are parallel if and only if 0 =× vu .
Properties of the cross product
If wand , vu are any vectors and r , s are scalars, then
1) ))(()()( uvrssvru ×=×
2) wuvuwvu ×+×=+× )(
3) uwuvuwv ×+×=×+ )(
4) ( )vuuv ×−=×
5) 00 =× u
Notes:
jkiikijkkj
kijji
=×−=×=×−=×
=×−=×
)()(
)(
0=
×××
kkjj
ii
n
vu ×
-n
uv ×
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
12
Calculating Cross product using determinants
If k321 ujuiuu ++= and k321 vjvivv ++= , then
321
321
vvvuuukji
vu =×
Ex.:
Find vu × and uv × if k2 ++= jiu and k34 ++−= jiv
Solution
10k6j2i )-( 10k6j--2i
34-12
14-12
- 1311
134112
−+=×=×+=
+=−
=×
vuuv
kjikji
vu
Ex.: Find a vector perpendicular to the plane of )0,1,1( −P , )1,1,2( −Q and )2,1,1(−R .
Solution
The vector PRPQ × is perpendicular to the plane because it is perpendicular
to both vectors.
kjikjiPQ −+=−−+++−= 2 )01()11()12(
kjikjiPR 222- )02()11()11( ++=−+++−−=
6k6i
22-21
22-1-1
- 221-2
222121
+=
+=−
−=× kjikji
PRPQ
Ex.: Find a unit vector perpendicular to the plane of )0,1,1( −P , )1,1,2( −Q and
)2,1,1(−R .
Solution
Since PRPQ × is perpendicular to the plane, its direction n is a unit vector
perpendicular to the plane
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
13
2
1 2
1 266k6i ki
PRPQPRPQn +=
+=
×
×=
Calculating the Triple scalar product (volume): also called Box product
321
321
321
)(wwwvvvuuu
wvu =⋅×
Ex.:
Find the volume of the box determined by k2 −+= jiu , k32 +−= iv and
k47 −= jw .
Solution
23- 470
302121
)( =−
−−
=⋅× wvu
The volume is 23 )( =⋅× wvu units cubed. Lines and Planes in Space
In the plane, a line is determined by a point and a number giving the slope of
the line. In space a line is determined by a point and a vector giving the direction of
the line.
Equations for a line
Suppose that L is a line in space passing through a point ),,( 0000 zyxP
parallel to a vector k321 vjvivv ++= . Then L is the set of all points ),,( zyxP for
which PP0 is parallel to v .
L
v
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
14
The standard equation of the line through ),,( 0000 zyxP parallel to
k321 vivivv ++= is:
∞∞+=+=+= pp t- , , , 000 tvzztvyytvxx
and ) , , (),,( 000 tvztvytvxzyx +++=
Ex.:
Find the equations for the line through )4,0,2(− parallel to k242 −+= jiv .
Solution
With ),,( 0000 zyxP equal to )4,0,2(− and k321 vjvivv ++= equal to
k242 −+= jiv
24 , 4 , 22 tztytx −==+−=
Ex.: Find the equations for the line through )3,2,3( −−P and )4,1,1( −Q .
Solution
The vector k734 +−= jiPQ is parallel to the line and equation with
)3,2,3(),,( 000 −−=zyx give
73 , 32 , 43 tztytx +−=−=+−=
We could have choose )4,1,1( −Q
74 , 31 , 41 tztytx +=−−=+=
An equation for a Plane in space
Suppose that plane M passes through a point ),,( 0000 zyxP and is normal to
the nonzero vector kCBjAin ++= . Then M is the set of all points ),,( zyxP for
which PP0 is orthogonal to n.
Thus, the plane through ),,( 0000 zyxP normal to kCBjAin ++= has equation:
0)()()( 000 =−+−+− zzCyyBxxA
or 000D e wher, CzByAxDCzByAx ++==++
n
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
15
Ex.:
Find an equation for the plane through )7,0,3(0 −P perpendicular to
k25 −+= jin .
Solution
2225072155
0)7)(1()0(2))3((50)()()( 000
−=−+=+−++
=−−+−+−−=−+−+−
zyxzyx
zyxzzCyyBxxA
Notice in this example how the components of k25 −+= jin become the
coefficients of zandyx , in equation 2225 −=−+ zyx . The vector
kCBjAin ++= is normal to the plane DCzByAx =++ .
Ex.:
Find an equation for the plane through )0,3,0( and )0,0,2( , )1,0,0( CBA .
Solution
We find a vector normal to the plane and use it with one of the point to write
an equation for the plane.
The cross product:
6 2 3 130102 kji
kjiACAB ++=
−−=× is normal to the plane.
6623
0)1(6)0(2)0(3=++
=−+−+−zyx
zyx
Lines of intersection
- Two lines are parallel if and only if they have the same direction.
- Two planes are parallel if and only if their normals are parallel.
- The planes that are not parallel intersect in a line.
Ex.:
Find a vector parallel to the line of intersection of the planes 15263 =−− zyx
and 522 =−+ zyx .
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
16
Solution
The line of intersection of two planes is perpendicular to both planes' normal
vectors 21 and nn and therefore parallel to 21 nn × . i.e. 21 nn × is a vector parallel to
the planes' line of intersection.
51 2 41 21226321 kji
kjinn ++=
−−−=×
Ex.: Find the point where the line , , tztytx +=−=+= 12238 intersects the plane
6623 =++ zyx
Solution
The point
+−+ 1 , 2 , 2
38 ttt
( ) ( )
-1t -88t
66t64t-6t8
6 6 2
==
=+++
=++−+
+ ttt 122
383
The point of intersection is 0) , 2 , 32( ),,( 1 =−=tzyx
Angles between planes
The angle between two intersecting planes is defined to be the angle
determined by their normal vectors.
Ex.:
Find the angle between the planes 15263 =−− zyx and 522 =−+ zyx
Solution
The vectors k2631 −−= jin and k222 −+= jin
are normals to the planes. The angle between them is
=
⋅=
−
−
214cos
cos
1
21
211
nnnnθ
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
17
Problems:
1) Sketch the coordinate axes and then include the vectors vuvu and , × as
vectors starting at the origin
a. j , i == vu
b. k j ,k -i +== vu
c. 2ji , j- i2 +== vu
d. ji , j i −=+= vu
2) In the triangle that determined by the points and , RQP , find a unite vector
perpendicular to plane PQR .
a. )1,1,3( and )3,1,2( , )1,1,1( −RQP
b. )2,2,1( and )1,1,0( , )0,2,2( −−−− RQP
3) Let k5 +−= jiu , k5−= jv and k3315 −+−= jiw . Which vectors, if any,
are:
a. Perpendicular?
b. Parallel?
4) Find equations for the lines:
a. The line through the point )1 ,4 ,3( −−P parallel to the vector k++ ji .
b. The line through )1,0,1( and )1,2,1( −− QP .
c. The line through the origin parallel to the vector k2 +j .
d. The line through the point )1 ,2 ,3( − parallel to the line
3 , 2 , 21 tztytx =−=+=
e. The line through )1 ,1 ,1( parallel to the z-axis.
f. The line through )5 ,4 ,2( perpendicular to the plane 21573 =−+ zyx
g. The line through )0 ,7 ,0( − perpendicular to the plane 1322 =++ zyx
h. The line through )0 ,3 ,2( perpendicular to the vectors k32 ++= jiu
and k543 ++= jiv
i. The x - axis.
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
18
j. The z - axis.
5) Find equations for the planes:
a. The plane through )1 ,2 ,0(0 −P normal to k23 −−= jin
b. The plane through )3 ,1 ,1( − parallel to the plane 73 =++ zyx
c. The plane through )1,2,0( and )2,0,2( , )1,1,1( −−
d. The plane through )5 ,4 ,2(0P perpendicular to the line
4 , 31 , 5 tztytx =+=+=
e. The plane through )1 ,2 ,1( −A perpendicular to the vector from the
origin to A .
6) Find the plane determined by the intersecting lines:
∞∞−=+=−=
∞∞−=+=+−=pp
pp
s- 22 , 21 , 41 :2t- 1 , 2 , 1 :1
szsysxLtztytxL
7) Find a plane through )1 ,1 ,2(0 −P perpendicular to the line of intersection of the
planes 22 , 32 =++=−+ zyxzyx .
8) Find a plane through the points )1 ,2 ,3( , )3 ,2 ,1( 21 PP perpendicular to the
plane 724 =+− zyx .
9) Find the angles between the planes:
a. 222 , 1 =−+=+ zyxyx
b. 132 , 105 −=+−=−+ zyxzyx
10) Find the point in which the line meets the plane.
a. 632 , 1 , 3 , 1 =+−+==−= zyxtztytx
b. 12436 , 22 , 23 , 2 −=−+−−=+== zyxtztyx
References:
1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 7 مدرس مساعد ازهار مالك Vectors:
19
Double Integral : Lecture 8 مدرس مساعد ازهار مالك
1
Double Integral Definition: let R be closed region in the (x, y )- plane. If f is a function of two variables that is define on the region R, then the double integrals on R is written by
dAff y)(x, A)y,(x LimR
n
1rrrr
0An
r
∫∫∑ =∆=→∆
∞→
dydxاذا كانت المنحنيات بهذه الصيغة يؤخذ المقطع شاقولي
dx y)(x, y)(x,2
1R
dyfdAfb
a
y
y∫ ∫∫∫ =
dxdyاما اذا كانت المنحنيات بالشكل التالي يؤخذ المقطع افقيا
dxdy y)(x, y)(x,2
1R∫ ∫∫∫ =d
c
x
x
fdAf
R
R
a b
y1
y2
R
d
c
x2 x1
Double Integral : Lecture 8 مدرس مساعد ازهار مالك
2
Examples:
1) Evaluate )dydx8xy 1(3
0
2
1∫ ∫ +
i) sketch: since l verticadxdy ⇒ 2y , 1y == ii)
{ }
{ }
{ }
57=+=+=
+=
+=
++=
+−+=
+=+
∫
∫
∫
∫∫ ∫
54)(3(0)-6(9))(3
)2
12(
dx 12x1
dx 4x][1-16x][2
dx )]1(41[]4x(4)[2
)2y8x(y)dydx8xy 1(
3
0
2
3
0
3
0
3
0
3
0
2
1
23
0
2
1
xx
x
dx
2) Evaluate ∫∫ −
R
dAyx )2( 2 over the triangular R enclosed by
3y ,x 1y , 1 =+=−= xy
i) sketch:
(-1,0) & (0,1) , (1,0) & (0,1) 1 x 0y , 1 x 0y
1y 0 x , 1y 0 xx1y , 1
⇒⇒−=⇒==⇒=
=⇒==⇒=+=−=
ifififif
xy
Double Integral : Lecture 8 مدرس مساعد ازهار مالك
3
∫ ∫∫∫−
−
−=−3
1
1
1
22 )2()2(y
yR
dxdyyxdAyx
[ ] [ ]{ }
{ }
2112
)1()1()1()1( )(
3
1
322232
3
1
22223
1
1
1
22
∫
∫∫
−+−+−+−+−=
−−−−−−−=−=−
−
dyyyyyyyyy
dyyyyyyydyxyxy
y
624418 −=
−+−=+−
−+−
=
+−=
+−= ∫
32
21
28118)
32
21(18
218
)3
24y2(
)22(
3
1
34
3
1
23
y
dyyy
3) Evaluate dy dx e 2
0
1
2
x 2
∫ ∫y
Reverse the order of integration Since dxdy horizontal
2x y 2y x =⇒=
1 x = 2 0 fromy for →
1-e e e
)02(e
edx dy e dy dx e
011
0
x
1
0
x
2
0
1
0
x1
0
2
0
x2
0
1
2
x
2
2
222
=−==
−=
==
∫
∫∫ ∫∫ ∫
e
dxx
dxyxx
y
Double Integral : Lecture 8 مدرس مساعد ازهار مالك
4
4) Evaluate dy dx x
sin x 0∫ ∫π π
y
From left yx =
From right π=x
value of y , from x 0 ⇒
reverse the order
2 1)-(-1- cos - sin
sin sin
d dy sin dy d sin
00
000
0 00
====
⋅=⋅=
=⇒
∫
∫∫
∫ ∫∫ ∫
ππ
ππ
ππ π
xdxx
dxxx
xdxyx
x
xx
xxx
x
x
x
y
5)
[ ] [ ]
[ ] sin44 −=+=
+−=−=
=
∫∫
∫ ∫∫ ∫
20
22
2
0
22
00
2
0 0
22
0
22
ysin -
2y cos 2 xycos 2
dxdysin xy 2dydxsin xy 2
y
dyyydyy
yy
y
y
x
6) Write an equivalent double of integration reversed dx )dyy(x 22∫ ∫−
+
−
+0
1
2x
x
∫ ∫ ∫ ∫
∫ ∫
− −
−
+
−
+++=
+
1 0 2 0
2
2222
0
1
2
)()(0 1
22 dx )dyy(x
y y
x
x
dxdyyxdxdyyx
x=π
y=π
(-1,1)
Double Integral : Lecture 8 مدرس مساعد ازهار مالك
5
7) Draw the region bounded by y=ex, y=sinx, x=π, x=-π and evaluate its area.
π−ππ−π
π
π−
π
π−
π
π−
π
π−
−=π−−π+−
+=
−=
=
=
∫
∫
∫ ∫
eeee
xe
dxxe
y
dydxA
x
x
e
x
e
x
x
x
)cos(cos
cos
)sin(
sin
sin
8) Find the area bounded by y=-x, y=-3x and x=y+4. Solution:
262
)6
42
(
)3
()344(
0
2
222
3
22
0
2
2
3
0
2 34
2
3
4
34
2
3
4
3
0
23
=+−
+++
+−+++=
+=
+=
−
−
−
−
−
−
−
−−
−
−
+−
−
−
+
− −
−
−
∫∫
∫∫
∫ ∫ ∫ ∫
yyyyy
dyyydyy
dyxdyx
dxdydxdyA
yy
y
y
y
y
(1,-3)
(2,-2)
Double Integral : Lecture 8 مدرس مساعد ازهار مالك
6
Problems
1) dxdy 1)(xy
x 1
0
1
02∫ ∫ +
2) dxdy e xy 2 ln
0
1
0
y2
∫ ∫ x
3) dxdy xycos
x1
20
2
∫ ∫π
π
x
4) dydx e 8 ln
1
y ln
0
yx∫ ∫ +
5) dxdy yx
2
1
2
∫ ∫x
x
6) dydx xy cosy 1
0 0∫ ∫
π
7) dydx e 4
0
2x 3
∫ ∫y
8) θθ
πθ
ddr cosr 2
0
sin
0∫ ∫
9) Evaluate ∫∫R
dA , R: 1st quadrant bounded by x=2y & y=2x
10) Evaluate ∫∫R
xydA , R: the region bounded by x=2y & xy =
11) Evaluate ∫∫−
+R
dAyx 21
2 )1( , R: the region in the 1st quadrant enclosed by:
2y x= , 4=y , 0=x
12) Evaluate ∫∫R
dA )(y sin 3 , R: the region bounded by x=y , 2=y & 0=x
13) dydx e x1
0
1xy2∫ ∫
y
References 1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
Polar Coordinates and Graphs
Polar Coordinate system
Each point P can be assigned polar Coordinates (r, θ) where:
1) r is the distance from the pole (origin) 0 to the point P. r is positive if
measured from the pole along the terminal side of θ and negative if
measured along the terminal side extended through the pole.
2) θ is the angle from the Initial ray to (op). The angle θ is positive if the
rotation is counterclockwise and negative if the rotation is clockwise.
Review in trigonometric functions:
functions odd
cot)cot(tan)tan(csc)csc(sin)sin(
−=−−=−−=−−=−
θθθθθθθθ
functionseven sec)sec(cos)cos(
=−=−
θθθθ
1cotcsc1tansec1cossin
22
22
22
=−
=−
=+
θθ
θθ
θθ
xxxx
yxyxyx
xxxxyxyxxxxxxxxyx
2
22
tan1 tan22 tan y if
tan tan1tantan)tan(
sin cos )cos(2 y if sinsin cos cosy) cos( cos sin 2)sin(2 y ify sin cosy cos sin)sin(
−=⇒=
±=
−=⇒=±=
=⇒=+=+
mm
m
22cos1sin
22cos1cos
2
2
xx
xx
−=
+=
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
Converting from polar to rectangular form and vice versa We have the following relationship between rectangular Coordinates
(Cartesian) ),( yx and polar Coordinates ),( θr : 222 ryx =+
xy
xy
sryry
rxrx
1tan or tan
in or sin
cos or cos
−==
==
==
θθ
θθ
θθ
θd d rrdxdydydx
dA ⇒
=
Cartesian Coordinates
)(xfy =
Polar Coordinates
)( θfr =
Graphing polar equations
Sketch
i) symmetric about x-axis if replacing θ by )( θ− does not change the function.
ii) Symmetric about y-axis if replacing θ by )( θπ − does not change the function.
iii) Symmetric about the origin if replacing r by )( r− does not change the function.
iv)
2
0
M
π
πθ =
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
Ex.1: Converting an equation from Cartesian form to polar form
0422 =−+ yyx Since 222 ryx =+ and θin sry =
θθ
θθθ
sin 4ronly keep and 0r discaredcan we 0,4sin-r ofgraph in the included is pole thebecause pole. theis 0r ofgraph the
sin4or 00)sin4( 0sin4 04
2
22
====
===−=−
=−+⇒
rrrr
rryyx
Ex 2: Converting an equation from polar form to Cartesian form
θcos 3 −=r
θcos 3 2 rr −= Multiply both sides by r
03 3
22
22
=++⇒
−=+⇒
xyxxyx
Ex 3: Converting an equation from polar form to Cartesian form
r cos(θ-π/3)=3 r(cosθ cos(π/3)+ sinθ sin(π/3))=3
63323
21
3sin23cos
21
=+⇒=+
=+
yxyx
rr θθ
Ex 4: Converting an equation from polar form to Cartesian form r=4cosθ
xyxrr 4cos4 222 =+⇒= θ
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
Some important curves
}
}
}
cos
rose Leafed 4 cos , sin
rose Leafed 3 cos , sin
cordioid ) sin (1 , ) sin -(1 ) cos(1 , ) cos-(1
cos , sin ,
22 θ
θθ
θθ
θθθθ
θθ
2
22
33
ar
arar
arar
arararar
circleararar
=
==
==
+==+==
===
Standard Polar Graphs
1) Circle a) a=r
2=r
M
2 2
2 4
2 0
=⇒=
=⇒=
=⇒=
r
r
r
πθ
πθ
θ
b) θsin ar = i) replace θ by -θ θθ sin )(-sin arar −=⇒=∴ Not symmetric about x-axis ii ) replace θ by π-θ θθπ sin )-(sin arar =⇒=∴ symmetric about y-axis iii ) Not symmetric about the origin.
2a
6
a 2
0 0
π
π
θ r
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
c) θ cos ar = i) replace θ by -θ θθ cos )(- cos arar =⇒=∴ symmetric about x-axis
2a
3
0 2
a 0
π
π
θ r
2) Cardioids a) ) cos(1 θ+= ar Symmetric about x-axis
2a
32
23a
3
0
a 2
2a 0
π
ππ
π
θ r
Rapid polar sketching Ex: Sketch ) cos(1 4 θ+=r
θ varies from
Cos θ varies from
4 cos θ varies from
) cos(1 4 θ+=r varies from
0 to π/2 1 to 0 4 to 0 8 to 4 π/2 to π 0 to -1 0 to -4 4 to 0 π to 3π/2
-1 to 0 -4 to 0 0 to 4
3π/2 to 2π 0 to 1 0 to 4 4 to 8
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
b) ) sin (1 θ+= ar H.W
) sin -(1 ) cos-(1
θθ
arar
==
EX.: Find the area of the region enclosed by the cardioids
) cos-(1 θar =
2
3)2sin41
21sin2(
))2cos1(21cos21()coscos21(
)cos1(2
2
2
0
00
2
0
2
cos1
00
0
cos1
0
2r
πθθθθ
θθθθθθ
θθθ
θ
π
ππ
π
θ
π
π θ
=++−=
++−=+−=
−==
=
∫∫
∫∫
∫ ∫−
−
dd
dd
rdrdA
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
Problems
1) Converting equations from Cartesian form to polar form
a) 0622 =−+ xyx
b) 22 5 xyy −=
c) xy 42 =
d) 12 =xy
2) Converting an equation from polar form to Cartesian form
a) 0sin 2 =+ θr
b) 1sin 4cos3( −=− θθr
c) 4 =r
d) 4
πθ =
3) a) sketch ) sin (1 5 θ+=r
b) sketch 2 cos 8 θ=r
4) change the Cartesian integral into an equivalent polar integral. Then
evaluate the polar integral
a) ∫ ∫−
−1
1
1
0
2
x
dydx
b) ∫ ∫−
−
−−
1
1
1
1
2
2
x
x
dydx
c) ∫ ∫−
+1
0
1
0
22
2
)( y
dxdyyx
d) ∫ ∫6
0 0
y
dydxx
5) Use polar coordinate ∫ ∫−
++
a xa
0 02/322
22
)yx(1dydx
Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:
6) Find the area of the region R that lies inside the cardioid ) cos(1 θ+=r
and outside the circle 1=r .
7) Find the area of the region R that lies inside the cardioid
) cos(12 θ+=r and outside the circle 3=r .
8) Find the area of the region R that lies inside the circle ) (sin 4 θ=r and
outside the circle 2=r .
9) Find the area of the region R cut from the first quadrant by the cardioid
) sin (1 θ+=r .
10) Find the area of the region common to the ) cos(1 θ+=r and
) cos(1 θ−=r .
References: 1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
1
Polar Coordinates and Graphs 3) Leaves rose
θn cos ar = , θnsin ar =
}
}
cos
rose Leafed 4 cos , sin
rose Leafed 3 cos , sin
22 θ
θθ
θθ
2
22
33
ar
arar
arar
=
==
==
1- Number of leaves:
θn cos ar = , θnsin ar =
a) If n even → No. of leaves =2n b) If n odd→ No. of leaves =n 2- The major axes for the first leaf :
a) 001 cos =⇒=⇒= θθθ nn b)
nn
221sin πθπθθ =⇒=⇒=n
3- Limit for the first leaf (begin and end)
a)
−=⇒
−=
=⇒=⇒=
nn
nn
22
220 cos πθπθ
πθπθθn
b)
=⇒=
=⇒=⇒=
nn
nπθπθ
θθθ
000sin n
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
2
4- Complete drawing the other leaves: (360/No. of leaves) from major
axes. Ex.1: θ2 cos =r
1- No. of leaves = 4 2- 00212cos =⇒=⇒= θθθ
3-
−=⇒
−=
=⇒=⇒=
422
422
02cos πθπθ
πθπθθ
4- 24
360 π= , this mean every
2π leaf repeated.
Ex.2: θ2sin =r
1- No. of leaves = 4 2-
42212sin πθπθθ =⇒=⇒=
3-
=⇒=
=⇒=⇒=
22
0020sin πθπθ
θθθ2
4- 24
360 π= , this mean every
2π leaf repeated.
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
3
Ex.3: θ3 cos =r
1- No. of leaves = 3 2- 00313cos =⇒=⇒= θθθ
3-
−=⇒
−=
=⇒=⇒=
623
623
03cos πθπθ
πθπθθ
4- 3
23
360 π= , this mean every
32π leaf repeated.
Ex.4: θ3sin =r
1- No. of leaves = 3 2-
62313sin πθπθθ =⇒=⇒=
3-
=⇒=
=⇒=⇒=
33
0030sin πθπθ
θθθ3
4- 3
23
360 π= , this mean every
32π leaf repeated.
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
4
Ex.5: θ4 cos =r
1- No. of leaves = 8 2- 00414cos =⇒=⇒= θθθ
3-
−=⇒
−=
=⇒=⇒=
824
824
04cos πθπθ
πθπθθ
4- 48
28
360 ππ== , this mean every
4π leaf repeated.
Ex.6: θ4sin =r
Ex.7: θ5 cos =r
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
5
Ex.8: θ5sin =r
1- No. of leaves = 5 2-
102515sin πθπθθ =⇒=⇒=
3-
=⇒=
=⇒=⇒=
55
0050sin πθπθ
θθθ5
4- 45
25
360 ππ== , this mean every
52π leaf repeated.
Ex.9: θ2 cos a 22 =r
1- No. of leaves = 2 2- 00212cos =⇒=⇒= θθθ
3-
−=⇒
−=
=⇒=⇒=
422
422
02cos πθπθ
πθπθθ
4- 2
22
360 π= , this mean every π leaf repeated.
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
6
Ex.10: θ2sin a 22 =r
1- No. of leaves = 2 2-
42212sin πθπθθ =⇒=⇒=
3-
=⇒=
=⇒=⇒=
22
0020sin πθπθ
θθθ2
4- ππ==
22
2360 , this mean every π leaf repeated.
Problems
11) Find the area of the region enclosed by the cardioid θ2 cos 42 =r .
12) Find the area of the region enclosed by the cardioid )3 (cos12 θ=r .
13) Find the area of the region in the first quadrant bounded by 1=r and
θ2 sin=r , 24πθπ
≤≤
References: 1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
7
Graphs of some problems: 6) Find the area of the region R that lies inside the cardioid ) cos(1 θ+=r
and outside the circle 1=r .
7) Find the area of the region R that lies inside the cardioid
) cos(12 θ+=r and outside the circle 3=r .
8) Find the area of the region R that lies inside the circle ) (sin 4 θ=r
and outside the circle 2=r .
4
3
2
4
2
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
8
9) Find the area of the region R cut from the first quadrant by the cardioid
) sin (1 θ+=r .
10) Find the area of the region common to the ) cos(1 θ+=r and ) cos(1 θ−=r .
11) Find the area of the region enclosed by the cardioid θ2 cos 42 =r .
12) Find the area of the region enclosed by the )3 (cos12 θ=r .
13) Find the area of the region in the first quadrant bounded by 1=r and
θ2 sin=r , 24πθπ
≤≤
Mathematics: Lecture 10 مدرس مساعد ازهار مالك Polar Integral:
9
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
1
Fourier Series: Are series of cosine and sine terms and arise in the important practical task of
representing general periodic functions.
Periodic functions: A function )(xf is called periodic if it is defined for all real x and if there is
some positive No. T such that
)()( xfTxf =+
The No. T is called a period of )(xf .
Fourier said If )()( xfTxf =+ , T: periodic No. Then
∑∞
=
++=
1
0 2sin 2cos 2
)(n
nn xTnbx
Tnaaxf ππ
Where nn baa & , 0 are Fourier coefficients and
dxxfT
aB
A
)( 20 ∫=
dxxTnxf
Ta
B
An 2cos )( 2 π
∫=
dxxTnxf
Tb
B
An 2sin )( 2 π
∫=
BxA pp Notes:
No.integer , ,....) 2 , 1 , 0 ( , 0 sin nnn ±±==π
==
= ... , 4 , 2 , 0 1
... , 5 , 3 , 1 1- os
nn
nc π
....) , 3 , 2 , 1 , 0 ( allfor 1 2 os ±±±== n ,nnc π
sin )(-sin cos )(- cos
odd xxeven xx
==
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
2
EX.: Write Fourier series for π2 0 , )( pp xxxf =
ππ 2 0 - 2 T ==⇒ First we find nn baa & , 0
[ ] πππππ
ππ
2 0421
21
22
)( 2
22
0
22
0
0
=−=⋅==
=
∫
∫
xdxx
dxxfT
aB
A
dby , cos 1
2
2cos 22
2cos )( 2
2
0
2
0
vudxnxx
dxxnx
dxxTnxf
Ta
B
An
∫
∫
∫
=
=
=
π
π
π
ππ
π
π
[ ] 0 )11(12 cos1 cos 11
sin 1sin 1 1
22
2
02
2
0
2
0
=−==⋅=
−⋅= ∫
ππ
ππ
ππ
ππ
n0 cos-n
nnx
n
dx nxn
nxn
x
0a =∴ n
xnxdxxnx
dxxnxb
B
A
B
An
sin dv , u , sin 1
2
2sin 22
===
=
∫
∫
π
ππ
π
n
nxnn
xnxn
nxn
x
2b
sin1)012(1 1
d cos1 ) cos1( 1
n
2
02
2
0
2
0
−=
+−⋅
−=
−−
−⋅= ∫
π
ππ
ππ
π
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
3
)xx x
)xx x
)xx x
a ,nxxf n
L
L
L
+++−=
+−
+−
+−
+=
++++=
==⇒
−=
−=
−=
∑∞
=
3sin 312sin
21sin ( 2
3sin 322sin
22sin
12(
3sin b2sin bsin (b
0sin b 22)(
32b ,
22b ,
12b
321
1nn
321
π
π
ππ
Fourier even & odd functions 1) If )(xf is even , then
∫
∫
⋅=
⋅=
=
B
B
dxxTnxf
T
dxxfT
0n
00
n
2cos)( 22a iii)
)( 22a ii)
0b i)
π
2) If )(xf is odd , then
2sin)( 22b ii)
0aa i)
0n
n0
∫⋅=
==B
dxxTnxf
Tπ
Def.: A function )(xf is even if )()( xfxf =− for all x . For example, 2)( xxf = . A function )(xf is odd if )()( xfxf −=− for all x . For example, 3)( xxf = . Notes: - If )(xf symmetric about y-axis ⇒ even.
txfxfxfxxfxxf constan==== )( , )()( , cos)( , )( 2 - If )(xf symmetric about origin ⇒ odd.
sin)( xxf = EX.:
Write Fourier series for −
= 2 0 , 1 0 2- , 1
)(pp
pp
xx
xf
4 (-2) - 2 T ==⇒
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
4
i) From sketch ⇒ symmetric about origin ⇒ odd.
0aa n0 ==⇒
1)- (cos2- 2
cos 2-
d 2
sin 1
d 4
2sin 1 242b
2
0
0
0n
ππ
ππ
π
π
nn
xnn
xxn
xxn
B
B
==
⋅=
⋅⋅=
∫
∫
) cos1(2b n ππ
nn
−=∴ …. (1)
To find 1b , put n = 1 in eq. (1)
ππ
πππ
πππ
54))11(
52b
0))11(52b ,
34))11(
32b
0))11(22b , 4))1(1(2b
5
43
21
=+=
=−==+=
=−==−−=∴
.....) x2
1 x2
1 x2
4
..... x2
4 0 x2
4 0 x2
4
x2
nxf
+++=
+++++=
=⇒ ∑∞
=
ππππ
ππ
ππ
ππ
π
5sin 5
3sin 3
(sin
5sin 5
3sin 3
sin
sin b )( 1n
n
Notes:
y)] cos(y) cos([21 cos cos
cos 2cosy) cos(y) cos(
add sinsin cos cosy) cos( sinsin cos cosy) cos(
−++=
=−++
+=−−=+
xxyx
yxxx
yxyxxyxyxx
We can obtain yx sinsin by subtraction.
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
5
)sin()sin(21ysin cos
ysin cos2)sin()sin(
nsubtractio y sin cosy cos sin)sin(y sin cosy cos sin)sin(
yxyxx
xyxyx
xxyxxxyx
−++=
=−++
−=−+=+
EX.:
Write Fourier series for
=
23
2 , 0
2
2- , os
)(ππ
ππ
pp
pp
x
xxcxf
πππ 2 2
2
3 T =+=⇒
)(xf is even in 0b 2
2
-n =⇒
ππpp x
This is true if and only if the other interval = 0
ππππ
ππ
2 ]01[2 sin 2 cos 222 )( 22a
0
2
0
2
00 =−==⋅=⋅= ∫∫
B
xdxxdxxfT
....(2) 0)22
(sin1
10sin)22
(sin1
11
)(sin 1
1)(sin 1
11
])( cos )[cos(21 2
....(1) n cos cos 222
2
2ncos )( 22a
2
0
2
0
2
0
2
0n
−−
−+
−+
+=
−
−++
+=
−++=
⋅=
⋅=
∫
∫
∫
πππππ
π
π
π
π
π
π
π
π
nn
nn
nxxn
nxxn
dxnxxnxx
dxxx
dxxxfT
To find 1a , put n = 1 in eq. (1)
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
6
21 ))00(0
2(1
2sin211 )2 cos(1
212 cos2a
2
0
2
0
2
01
=+−+=
+=+⋅=⋅=∴ ∫∫
ππ
πππ
πππ
xxdxxdxx
in eq.(2)
ππ
πππ
32 )1
31(1
}2
sin{}2
3sin{ 311a2
=+−
=
−
−=
0 }}0{21-0{
411a3 =
=
π
ππ
πππ
152 )
31
51(1
}23sin{
31}
25sin{
511a4
−=−=
−
−=
+−++=
+=⇒ ∑∞
=
......xxx1
xnxf
4 cos 15
22 cos 32 cos
21
cos a 2a )(
0nn
0
πππ
Problems: Write the Fourier series for the following functions:
1)
−=
2 , 0 ,
)(ππ
πpp
pp
xaxa
xf
2) ππ - , )( pp xxxf =
3)
−+
= 0 ,
0 - , )(
ππππ
pp
pp
xxxx
xf
4) ππ - , sin)( pp xxxf =
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
7
5) −
= 2 0 , 1
0 2 , 0)(
pp
pp
xx
xf
6) −−
= 1 0 , 20 1 , 1
)(pp
pp
xxx
xf
7)
−
=
23
2 , 0
2
2 ,
)(ππ
ππ
pp
pp
x
xkxf
8)
−
=
23
2 , -
2
2 ,
)(πππ
ππ
pp
pp
xx
xxxf
9) ππ pp xxxf - , )( = . 10) ππ pp xxxf - , )( 3= .
11) −
= 0 , 1
0 - , 1 )(
ππ
pp
pp
xx
xf
12)
−
=
23
2 , 0
2
2 ,
)(ππ
ππ
pp
pp
x
xxxf
13) Find the Fourier series of the function )(xf which is assumed to have the period π 2 . 14) Find the Fourier series of the function )(xf
π/2 π -π π/2
1
k
π -π
Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:
8
15) Find the Fourier series of the function )(xf References: 1- Advanced Engineering Mathematics (Erwin Kreyszic)- 8th Edition. 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
k
-k π 2π
Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:
1
Partial Differentiations
0z)y,f(x,or ),( == yxfZ
==∂∂
==∂∂
yy
xx
fZyZ
fZxZ
1st partial derivatives
2
2
2
2
2
2
=∂∂
∂
=∂∂
∂
==∂∂
==∂∂
xy
yx
yyyy
xxxx
Zyx
Z
Zxy
Z
fZyZ
fZxZ
2nd partial derivatives
yxxy ZZ = Ex.1
If yxZ = , find , yZ
xZ
∂∂
∂∂
functionpower constant x , ln , constant y y y1-y ⇒⋅⋅=∂∂
=∂∂ dyxx
yZx
xZ
Ex.2
If xyZ 1tan−= , show that xyZ=yxZ
22 yxy
+−
=−
⋅+
=xy
xy
Zx
2
2
1
1
(1) )()(
2.)1)((222
22
222
22
Lyxxy
yxyyyxZ yx +
−=
++−+
=
Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:
2
22 yxx+
=⋅+
=x
xy
Z y1
1
1
2
2
(2) )()(
2.)1)((222
22
222
22
Lyxxy
yxxxyxZxy +
−=
+++
=
(1) & (2) are equal
Properties: 1) If y),g( , )( xf == υυω
rulechain
or
∂∂
⋅∂∂
=∂∂
∂∂
⋅∂∂
∂∂
⋅∂∂
=∂∂
yy
xf
xx
υυωω
υυ
υυωω
2) If s)h(r,y , s)g(r, , ),( === xyxfω \
rulechain
∂∂
⋅∂∂
+∂∂
⋅∂∂
=∂∂
∂∂
⋅∂∂
+∂∂
⋅∂∂
=∂∂
sy
ysx
xs
r
ry
yrx
xr
ωωω
ωωω
3) Total differential If ,.....),,( zyxf=ω
...
...
+++=
+++=
dzdydxdordzfdyfdxfd
zyx
zyx
ωωωω
ω
Ex.1
If 222),,( zyxxyzzyxf +++==ω , Find dxdω
By property (3)
Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:
3
dxdzzxy
dxdyyxz
dxdxxyz
dxd
dzzxydyyxzdxxyzddzdydxd zyx
)2()2()2(
)2()2()2(
+++++=
+++++=
++=
ωω
ωωωω
Ex.2
If (1) ..... )()( ctxgctxf −++=ω , Show that 2
22
2
2
xc
t ∂∂
=∂∂ ωω
There are two methods to solve this Ex. First method: Let , sctxrctx =−=+ Eq.(1) becomes (1) ..... )()( sgrf +=ω
)()()(
)()(
csgcrfts
ssg
tr
rrf
t−⋅′+⋅′=
∂∂
⋅∂
∂+
∂∂
⋅∂
∂=
∂∂ω
( ) ( )
( ) ( ) (2) ...... )()(
)()()(
)()(
22
2
2
2
sgrfct
csgccrfc
ts
ssgc
tr
rrfc
t
′′+′′=∂∂
∴
−⋅′′−⋅′′=
∂∂
⋅∂
∂−
∂∂
⋅∂
∂=
∂∂
ω
ω
1)(1)(
)()(
⋅′+⋅′=∂∂
⋅∂
∂+
∂∂
⋅∂
∂=
∂∂
sgrfxs
ssg
xr
rrf
xω
11)(11)(2
2
⋅⋅′′+⋅⋅′′=∂∂ rgrf
xω
In eq.(2)
2
22
2
2
xc
t ∂∂
=∂∂ ωω
Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:
4
Second method: t) مباشرة بالنسبة لـ 1نشتق معادلة (
[ ] (2) .....
)()()()(
)()()(
22
2
2
2
gfct
ccctxgccctxft
cctxgcctxft
′′+′′=∂∂
∴
−⋅−⋅−′′+⋅⋅+′′=∂∂
−⋅−′+⋅+′=∂∂
ω
ω
ω
[ ] (3) .....
11)(11)(
1)(1)(
2
2
2
2
gfx
ctxgctxfx
ctxgctxfx
′′+′′=∂∂
∴
⋅⋅−′′+⋅⋅+′′=∂∂
⋅−′+⋅+′=∂∂
ω
ω
ω
From (2) & (3)
2
22
2
2
xc
t ∂∂
=∂∂ ωω
Ex.3
If
=
xyfxz n , Show that nz
yzy
xzx =
∂∂
+∂∂
(1) ..... )( )(
)( )(
)( )()(
1
12
12
xyfxn
xyfyx
xzx
xyfxn
xyfyxxx
xzx
xyfxn
xy
xyfx
xz
nn
nn
nn
⋅+′⋅−=∂∂
⋅
⋅+′⋅⋅−⋅=
∂∂
⋅
⋅+−
⋅′⋅=∂∂
−
−−
−
(2) ..... )(
0)1()(
1
xyfyx
yzy
xxyfx
yz
n
n
′⋅=∂∂
⋅
+⋅′⋅=∂∂
−
Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:
5
From (1) & (2)
nzxyfnx
yzy
xzx n
=
=∂∂
⋅+∂∂
⋅
)(
Ex.4
Express r∂
∂ω and s∂
∂ω in terms of s & r if 22 zyx ++=ω ,
rrsrx 2z , s lny , 2 =+==
srzr
szr
s
rz
zry
yrx
xr18441222211 +=++=⋅+⋅+⋅=
∂∂
⋅∂∂
+∂∂
⋅∂∂
+∂∂
⋅∂∂
=∂∂ ωωωω
ssrz
ssr
sz
zsy
ysx
xs202121 22
−=⋅+⋅+
−⋅=
∂∂
⋅∂∂
+∂∂
⋅∂∂
+∂∂
⋅∂∂
=∂∂ ωωωω
Problems:
Find zf
yf
xf
∂∂
∂∂
∂∂ , ,
1) xyzzyxf 1-sin ),,( =
2) 22 yxcos2y)-x(2),,(+
=zyxf
3) Find υω
∂∂ when
2-2uy , 12 , if 0 , 0 2 υυωυ +=+−=+=== uxxyxu
4) If
+
= 22 yxxyfω , show that 0=
∂∂
+∂∂
yy
xx ωω
5) If ),( yxf=ω , and θθ siny , cos rrx == , show that
2222
2 )(1)( yx ffyrx
+=∂∂
+∂∂ ωω
6) If xzyxzyxf
dd find , z & 0),,( +==
Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:
6
7) Find the directional derivative of xyxyxf 1tan),( −= at (1,1) in the
direction of jiArrr
−= 2
8) In which direction is the directional derivative of 22
22
),(yxyxyxf
+−
=
9) The D.D. of ),( yxf at )2,1(0p in the direction towards )3,2(1p is
22 and the D.D. at )2,1(0p towards )0,1(2p is -3 , find D.D. at
0p towards the origin. References:
1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)
Mathematics: Lecture 13 مدرس مساعد ازهار مالك Partial Differentiations:
1
Partial Differentiations The Gradient & Directional Derivative
0z)y,(x, =f
rsunit vecto & , , fyf
xf kjik
zji
rrrrrrr
∂∂
+∂∂
+∂∂
=υ
The Directional Derivative of z)y,(x,f at ),,(p 0000 zyx in the direction of 321 kajaiaA
rrrr++=
gradient & AAu , uD.D. υυ
rr
rrrr
=⋅=⇒
Ex.1: Find D.D. of zxyf −−= 23x at (1, 1, 0) in the direction of
632 kjiArrrr
+−= Sol.: First we find υr
1z
)1)(1(2 2y
)1()1(33x
22
-
-2
2
=∂∂
=−=−=∂∂
=−=−=∂∂
f
xyf
yxf
22
yx
kji
kzfjfif
rrrr
rrrr
−−+=∴
∂∂
+∂∂
+∂∂
=∴
υ
υ
74
49 6-64
1 , 3694
632) 22(
AA
, uD.D.
=+
=
=⋅+++−+
⋅−−+=
⋅=
⋅=⇒
iikjikjirr
rrrrrr
r
rr
rr
υ
υ
Mathematics: Lecture 13 مدرس مساعد ازهار مالك Partial Differentiations:
2
Maxima, Minima & Saddle point y)(x,f will have M , m, S according to: 1) 0 , 0 == yx ff to find the suggested point (a,b).
2) 0)( 2yy fxyxx fff −⋅
Then (a,b) is M or m according to xxf negative or positive. 3) 0)( 2
yy pxyxx fff −⋅
Then (a,b) is a saddle point. 4) 0)( 2
yy =−⋅ xyxx fff Ex.1: Locate M,m & S (if any) 422yx 22 −+++−= yxxyf
...(2) 022...(1) 0 2y -2x
22 2y-2x
=++−=+
++−=+=
yx
yxff
y
x
multi (1) by 2 + (2)
(-2,-2) , -2y , -2 x 063x ==⇒=+⇒
1 , 2 , 2 yy −=== xyxx fff
0 31)2)(2()( 2yy f=−=−⋅ xyxx fff
Since m is (-2,-2) 0 ⇒fxxf Ex.2: Locate M,m & S (if any) axyf 3yx 33 −+=
Mathematics: Lecture 13 مدرس مساعد ازهار مالك Partial Differentiations:
3
...(2) 033...(1) 03ay -3x
33
3ay-3x
2
2
2
2
=−
=
−=
=
axy
axyff
y
x
From (1) a
2xy =⇒
In (2) 0x 2
4
=−⇒ axa
0) x(0
33
34
=−
=−⇒
axxax
ayyaxx
==∴==⇒
, 0 , 0
a)(a, & )0,0( ⇒
afyfxf xyxx 3 , 6 , 6 yy −===
1) afffat xyxx 3 , 0 , 0 (0,0) yy −===⇒
0 9)( 22yy pafff xyxx −=−⋅
point saddle a is (0,0)
2) afafafat xyxx 3 , 6 , 6 a)(a, yy −===⇒
0 a 27 a 9-a 36
9)6)(6()(222
22yy
f==
−=−⋅ aaafff xyxx
i) if m is a)(a, 0 0 ⇒⇒ ff xxfa ii) if M is a)(a, 0 0 ⇒⇒ pp xxfa Problems:
Find zf
yf
xf
∂∂
∂∂
∂∂ , ,
1) xyzzyxf 1-sin ),,( =
2) 22 yxcos2y)-x(2),,(+
=zyxf
Mathematics: Lecture 13 مدرس مساعد ازهار مالك Partial Differentiations:
4
3) Find υω
∂∂ when
2-2uy , 12 , if 0 , 0 2 υυωυ +=+−=+=== uxxyxu
4) If
+
= 22 yxxyfω , show that 0=
∂∂
+∂∂
yy
xx ωω
5) If ),( yxf=ω , and θθ siny , cos rrx == , show that
2222
2 )(1)( yx ffyrx
+=∂∂
+∂∂ ωω
6) If xzyxzyxf
dd find , z & 0),,( +==
7) Find the directional derivative of xyxyxf 1tan),( −= at (1,1) in the
direction of jiArrr
−= 2
8) In which direction is the directional derivative of 22
22
),(yxyxyxf
+−
=
9) The D.D. of ),( yxf at )2,1(0p in the direction towards )3,2(1p is
22 and the D.D. at )2,1(0p towards )0,1(2p is -3 , find D.D. at
0p towards the origin. References:
1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)