1. Introduction
For evaluating areas, volumes
of
revolution etc. we need to know the general nature of the given curve. In thisc
hapter
we shall learn the methods of tracing a curve in general and the propertiesof
some standard curves commonly met in engineering problems.2. Procedure for tracing curves given· in Cartesian equations
I.
Symetry : Find out whether the curve is symmetric
al about any line with the help of the following rules :( 1)
The curve is symmetrical about the x -axis if the equatio
n of the curve remains unchanged when y is replacedby-
y i.e. if the equation contains only even powers of y.(2)
The curve is symmetrical about they-axis if the equation of the curve remains unchanged when x is replaced by-xi.e.
if the equation contains only even powers of x.(3)
The curve is symmetrical in opposite quadrants if the equation of the curve remains unchanged whenboth
x andy arer
epl
aced by-
x and -y.(4) The
curve is symmetrical about the line y = x if the equationof
the curve remains unchanged when x andyare i
nterch
anged
.II.
Origin :
Find out whether the origin
lies on the curve.If it
does, find the equations of the tangents at the origin by
equating to zero the lowest degre terms.III. Intersection
with the cordinate
axes:
Find out the pointsof
intersection of the curve with thecoordinate
axes. Find also the equations ofthe
tangents at these points.IV.
Asymptotes:
Find out theasymp
totes if any.V.
Regions
where
no
part of the curve lies:
Find out the regions of thep
lane where nopart
of the curve lies.VI
•Find
out dyldx:
Find out dy/dx and the points where the tangentsare
parallel to the coordinate axes.Eng�Mrering
Mathemat - If(7-2)
Tracing
of Corwes3. Common Curves
You
have alreadystudied
quite in detail straight line, circle, parabola,eJiipse
and hyperbola,
incl
udi
ngrectangular
hyperbo
la. We shall briefly reviewthese
curves.Rectangular Coordinates
(a) Straight Line:
General equation of straight line is
of
theform
ax +by
+ c =0.
To plot a straight line we put y =
0
and find x.A
lso
we put x =0
and findy. We plot these two points
and
join themto
get the required line. Some particular cases of straight Jines are shown below. (i) Lines parallel tothe coordinate
axes.y
x=h
0 X
Fig.
(7.1)
(ii) Lines passing through origin y y=mx m>O X y y=k 0 X y
Fig. (7.2)
y=mx m<O X Fig.(7 .3)
Fig.(7 .4)
(iii) Lines making given intercepts on coordinate axes. y
X X
Fig.
(7 .5)
Fig.(7
.6)
General equation of
circle is:il
+j
+2gx
+2/y
+ c =0.
Its centre isEngineering Mathematics
-
II(7- 3)
Tracing of Corves(i) Circle with centre at origin
and radius a and(ii) Circle with centre at (- g,-
f)
and radius 2 +j2-
c. are shown below.y
Fig. (7.7)
Fig. (7.8)
(iii) Circle with centre on the
x-axis and passing through origin.
Its equation is of the form x2 +
i
±2ax
=0.
Fig. (7.9)
y
-x
0
Fig. (7.10)
(iv) Circle with centre on they-axis and passing through origin.
Its equation is of the form x2 +
i
±2by
=0.
y
0
Fig. (7.11)
Fig. (7.12)
(v) Circle with centre on the axes but not passing through origin.
Its equation is of the form x2 +
l
±2gx
+ c =0,
.-r?
+i
±2fy
+ c ='y
y
v- ..,, JTat:mg or �.;orves
(vi)
Circletouching
b
o
th axes.Its
equation is of the form:Xl
+l
±2ax :;t
2ay+ a2
= 0.y
Fig.
(7.15)
Ex. 1
: Draw the circle:Xl
+l- 4x
+4y + 4
= 0 Sol. : We write the equation as(x
- 2)2
+(y
+2)
2 =22
Its centre
i
s(2, -2)
and radius =2.
It is shown on the right.
Polar Coordinates
(a) Straight line Fig.
(7.16)
If We put X = r COS 8, y = r Sin
8
in the equation Of the Straight lineax+by+c=O
we get,
ar
cose
+br
sine
+ c = 0 i.e. a cose
+b
sine
+.:.
= 0r (i)
A
line parallel to the initial line.If a line is parallal to the initial line (or the
x-
ax is) at a ditance p from it then, from the figure we see that its equation is p r sine
= ± p (Itis
clear that r sine
= p, if the line is above the initial line and r sine
=- p, if the line is 0 Fig.(7.17)
below the initial line.)(ii)
A
line perpendicular to the initial line. If a line is perpendienlar to the initial line (or thex-axis) at a ditance p from it then from the figure we
see thatrcos6=±p
(It
is clear that r cose
= p, if the lineis
to the right 0 of the pole and r cosa
=-
p, if the line is to the left of the pole.)X
Fig.
(7.18)
. 0 p
Fig. (7.19)
X (iii)
A
line through the pole (or theorigin)
Engineering Mathematics
-
II(7- 5)
Tracing of Corvesa
n
gl
ea
with the initalline then every point on theline
has coordnates(r, a),
where r is positive or negative. Hence the e
q
ua
ti
on of a line is0 =
aIn particular the equation of
a linemaking
an • angle of45°
ise
=1t/4.
(b) Circle
If
we putX= r
cose.
y= r
sin e
in the equ
ati
on of the circle with center at the origi
n and radius a i.e.in
x 2 + y 2 =a 2 we
get,
r 2=
a 2If
We put X =r
COS()
y= r
sin 0 in the equationof the circle with center
(a,
0) and radius a i.e.in
X
Fig. (7.20)
(x- a)
2 + y 2=a
2 i.e.in
x 2 + y2- 2ax =
0X
we get r 2
- 2a r
cose =
0:.r = 2a
cose
This is the equation of the circle with cen
ter(a,
0)and radius a in polar coordniates.
If we put X
= r
cose,
y= r
sin0 in
theFig. (7.21)
equation of the circle with center (0, a) and radius a i.e. in
x 2 +
(y-
a)2 =a
2 i.e. x 2 + y 2-2ay =
0 we get, r 2- 2ar
sine
= i.e. r =2a sin e
This is the equation of the circle with center (0, a)
y and radius a in polar
coordniates
y
X
Fig. (7.22)
If We put X= r COS 0, y = r sin 0 in the equation
of the cricle with center (a, a) and t
o
uching
both the coordniate axes in the first quadrantX i.e. in
(x-
a)2 +(y-
a)2=
a2Fig. (7.23)
i.e. inx2
+l-
'2ax-
2ay + a2= 0
we get the equation r 2- 2ar. (cos
e
+sin 0) +a 2= 0
in polar coordniates.Ex.
: Find the equation of the cricle in polar coordinates having centre(3,
0) and radius3.
Also find the polar coordinates of its point of intersection with the line y =x.
Sol.
: The equation of the circle with centre(3,
0) and., --- ---:--- -- ,. - ..,, ••..•••11 WI �va-.aa
i.e. x2
+y2-6x=O
Putting X = r COS
8,
y = r Sina
we g
et
r 2-6r
cosa=
0 i.e. r= 6
cos8
The equation of the line is
y=
x.Putting X
=
r COS0, y
= r si
n8,
we get theequation in
polar
coord
ina
tes
asr sin
a
= r cos8 i.e. tan a =
1 :.a=
1t /4Fig. (7.24}
Putting
8
=1t/4
in
r =6
cosa
, we get r= 6
cos 1t/4 = 6
· =3
J2Hence
the
polar coordinates of the point of intersection are(3
J2
.1t/4)
(iii)
Parabola :
General equation of parabola is
y
= ax2 +
bx+ c
or x=
ay 2 + by + c.
By completing the square on x or on
y
and by shifting the origin theequation can be written in standard form as
y 2
= 4ax
or x 2=
4by.
d
x = a x
Fig. (7.25)
Fig. (7.26)
The equation of the parabola with vertex at
(It,
k)
and(i)
the axis parallel to the x-axis, opening to the right is(y- k)2
= 4a
(x-
h)
(ii)
the axis parallel to the x-axis, opening to the left is(y
-k)2
=-
4a
(x-h)
(iii) the axis parallel to the y-axis, opening upwards is (x-
h)2
=4a
(y-
k)
(iv) the axis parallel to the y-axis, opening downwards is
2
(x-
h)
=-4a(y-
k)
Ex. 1:
Sketch the curver +
4x- 4y+
8 =
0Sol.:
The equation canbe
written asFig. (7.27)
r-
4y + 4 =-4x-4
i.e.(y- 2)
2 =-4 (x+
1)
y
Engineering Mathematics -II
Putting y -2 = Y, x + 1 = X,
Its equation is Y 2 =
-
4X.(7- 7)
Tracing of CorvesIts vertex is at
( -1,
2) and it opens on the left.Ex. 2
: Sketch the curvex2 + 4x- 4y +
16
=0.
Sol.
: The equation can be written as(x + 2)2 = 4
(y- 3).
Putting x + 2 =
X,
y -3
= Y,Its .equation is X 2 = 4
Y.
Its vertex is (-2,
3)
and it opens upwards.(iv) Ellipse
The equation of ellipse in standard form is
Fig.
(7.28)
x2
y2
x2
y2
-;z+t;2=
1(a> b), -;z+ b2
=I (a<b) y y a>bFig.
(7.29)
Fig.
(7.30)
If the centre of the ellipse is at (h, k) the equation of the ellipse becomes (x-
h)
2(y- k)
2 _a2
+b2
-1If a> b, the major axis is parallel to the x-axis and if
a<
b the major axis is parallel to the y-axis.(v) Hyperbola
Mathematics -II
(7- 8)
Tracing of Corvesy
y
{0, be)
(0,- be)
Fig.
(7.31)
Fig. (7.32)
(vi) Rectangular Hyperbola
If the coordinate axes are the asymptotes the equation of the rectangular hyperbola is xy =
k
or xy = -k.
l
xy= k xy�-kFig.
(7.33)
Fig. (7.34)
If the lines y = + x and y = -x are the asymptotes, the equations of the rectangular hyperbola are given by
2-
-l
= a2, y2 -x2 = a2•y X x2-y2=a2
Fig.
(7.35)
' ' y2 -x2 = a2
Fig. (7.36)
XPolar equation of the rectangular hyperbola x2 -
l
= a2 is obtainedby putting X = r cos
e,
y = r sine.
The equation is?-
cos2e
- r2 sin2e
= a2i.e.
r2 (cos2e-
sin2 8) = a2i.e.
r2 cos 2e
= a2Similarly the polar equation of
the
other rectangular hyperbolai.e. of
Engineering Mathematics -II
(7· 9)
Tracing of CorvesParametric Equatioas
Ellipse
The parametric equations of
the
elli
pse are X = a cose.
y =b
sine.
Hyperbola
The parametric equations of
the
hyperbola areX= a sec
e,
y =b
tane
or X= aCOS/t
1, y =b sinh
1.Parabola
The parametric equations of the
parabola l
=4ax
are x =a?
, y =2at.
The
parametric equations of the parabola x2 = 4ay are x =2at,
y =a?
.Rectangular Hyperbola
The parametric equations of the rectangular hyperbola xy = c2 are x = ct and y =
cIt.
Curves of
the form ym = xm where mand
n arepositive intigers.
It is interesting to note the shape of the curves whose equations can
be
expressed in the above form.
S
ome of the curves are known to us and some will be studied in the following pages. We give below the equations and the graphs of these curves.y y
Fig. (7.37)
(1)
y =x.The staight line through the origin y
Fig. (7.39)
Fig. (7.38)
2(2)
y =x .The parabola through the origin and opening
up
y
Mat.tMma -II
(3)
y=xl.
Cubical
parabola
Fig.
(7.41)
(5)
i=�.
XPair
of twol
ines passing through theor
ig
iny
(7·10)
Tracing of Corves
(4)
?
=x.The
parabo
la throughthe origian
and opening
onthe
right
Fig.
(7.42)
(6)
i=�.
Semi-cubical parabola X X XFig. (7.43)
(7)
y3=x.
Cubical parabolaFig. (7.44)
(8)
i =x2.
Semi-cubical parabola4. Some
Well-known Curves
Ex 1 : Cissoid of Diocles :Trace the curve
i
(2a- x)
=
�
:as oN :II
:
)(2at2
2at3
or x
=
--2 ,y =
--2(S.U. 1986, 90)
l+t
l+t
Sol. : (i) The curve is symmetrical about
the x-axis.
(ii)
The
curve passesthrou
gh
the origin and(2a, O)
xthe
tangentsat the
ori
g
i
nare
j
=0 i.e.
thex-axis
is a
double tangent atthe
origin. 3(iii) Since
l =
-x asx
�2a,
y � 0<\the line x =
2a
is an asymptote.(iv)
W
he
nx
>2a,
i
isnegative. Hence,
Fig.
(7 .45)
thecurve does
not exist whenx
>2a.
(Remark
:By putting X= rcosa, y
= r sine,show
that the
po
lar equationEnglneertng Mathematics
-II(7- 11)
Tracing of CorvesEx. 2:
Trace the curvea
2�
=
y3(2a-
y)(S.U. 1981,86,90, 2006)
Sol.: (i)
The
curve is symmetrical about they-axis. (ii) It passes throughthe
origin and the x-axis is a tangent at origin.(iii) It meets the y-axis at
(0,0)
and(0, 2a).
Further d yl
dx. is zero at(0,
2a);
the tangent at this point is parallel to the x-axis.(iv) Since x2
=
y3(2a
- y)I a2
when y>
2a
and y <0,
x2 is negative and the curve does notexist for y >
2a
and y <0.
Fig.(7.46)
y
Ex. 3
: Astroid or Four CuspedHypocycloid : Trace the curve
(xla)
213 +(ylb)
213=
1or
X= ac
os3e, y = b
sin3e (S.U.1985, 2003)
S
ol. : (i)
The curve is symmetrical about both axes.(ii)
The curve cuts the x-axis in(± a, 0)
andthey-
axis in(0, ±b)
Fig.
(7 .47)
(iii) Neither x can be greater than anor
ycan be greater than
b.
Ex. 4:
Witch of Agnesi: Trace the curve xl = a2 (.rz-x)
(S.U. 1980, 91, 92, 98, 2003, 04)
ySol.
: (i)The curve
is symmetricalabout
the x-axis. (ii)The curve passes through the point
0)
.. . .
) s·
2 2(a--x)
(111
mce
, v = ax
cannot beX '
negative. Also
x
cannot begreatc:.·
than a.(iv)
Asx � 0, y�
oo, they-axis is ana
symp
tote.Ex. 5:
Tracethe curve l
=(x- a)
(x-b)
(x- c)
whete a,b, c
are positive.
Sol.
: We considerthe following
cases : (a) CaseI
: a< b < c
(1) It is symmetrical about
the x-axis.(2) It meets the
x-axisi
n(a, 0), (b,O)
and(c, 0).
(3)
Whe
n x< a, l is negative.
whena < x < b,
l>O
Engineering Mathematics
-
II(7- 12)
Tracing of Corveswhen
b <
x<
c,l
is negative whenx>c,l >O
Hence, there is no curve to the left-of x =a and also between
x
=bandx=c.
(4) If
x > c and increases thenl
also increasesy
x
Fig. (7.49)
(b) Case II:
a=b
<
cy The �quation now becomes
2
2
y
=(x-
a)
(
x-c)
( 1)
As before the curve is symmetrical aboutx the x-axis.
(2)
It meets the x-axis in points(a, 0)
and(c,
0).
Fig. (7.50)
(3)
(a
, 0)
is an isolated point because ifx
<a
(i.e.
x<c), l
is negative and ifa<
x <c, l
is negative.( 4) If�
> c and increasesy2
also increases.(c) Case III :
a
<b
= cy The equation now becomes
y2
=(x-a)(x-b)2
x
(1)
The cmve is symmetrical about thex-axis.
(2)
It meets the x-axis in points(a, 0)
and(b, 0).
Fig. (7 .51)
(3) If
x<a, l
is negative.( 4)
If x>
b
and increases,l.
also increases. y(d) Case IV :
a
=b
= c(S.U. 2003)
The equation now becomes
l
= (x-a)3
(a,
x(1)
As before the curve is symmetrical about thex
- axis.Engineng
Mathematics-
II(7·13)
(3) If
x
<a,j
is negative
(4) If
x >a
and
increases,
i
also
increases.
Ex. 6
:Trace
the
curve
x112 + y112 = a112 YSoL
:
(i) The
curveis symmetrical about the line
y=x.(ii)
If
y =x we get
2x112
= a112:. 4x=a i.e.
x=a/4. The liney=xcuts the
curve
is (a/4, a/4).
(iii)
The
x-axis
is
a tangent at (a,
0) and t
he
y-
axis is a tangent
at
(0, a).
5. Some
Common Lops
Ex. 1
:Strophold
Trace the curvee
j
(a
+ x)
=:?
(b -x)
Sol.
: (i) Curve
is
symmetrical about the x-axis.
(ii) The curve passes through the origin and the equations of the tangents
at the origin
areo
b
tai
ne
d
by equating to zero the lowe
s
t
deg
re terms.II >< Y
� .+/
:
.
a,Z=bx2
:. y=±
·x
��--� , __;./
(b, 0)
(iii) The curve
meets
the x-axis in
(0, 0) and (b, 0)
2(
•)
S'
2 X(b-x)
when
tvmce
y
=(a+x)
x =- a,
yis infinite.
Hence
x =-a is
anasymptote.
Fig. (7.54)
(v) When
x >band
when
x<-a,
l
is
negative. Hence, the curve
does not exist when
x >
b and
x <-a.The c
urve
is
shown
in
the figure.
(a) Trace the
curve :1'
(a
+x)
=:?
(3a
- x)
(S.U. 1980, 86, 91, 20)
When b
is replared by 3a we
get the above
curve
which is shown on
the right.
y
FJa. (7.55)
Engineering Mathematics -II
(7- 14)
Tracing of Corvesy
'
(b) Trace the
:l (a - x)
= x2 (a +x)
If we replace
b by a and
xby
-- xwe get
the above curve which is shown on the
left.X
(c) Trace the curve
: y(a+ x) =
(a-x)
(-a, 0)
Sol.
required curve.
:
If we replace b by a in Ex. ( 1) we get the
(d)
Trace the curve
: x(x2
+ l) =a (x2 -Jl)
(S.U.
1980, 85, 90)
Fig. (7.56)
The equation can the written as
x3 +xl=d-al
:. xi +
al
= ax2- x3 :.i (a+ x) =
:x?
(a x)
This is the same as the first equation where b
=a.
Ex.
2:Trace the curve
.\y2+ (x + a)2 (x + 2a)
=0
Sot: (i) The curve is symmetrical about the x-axis.
j
(ii)
Thecurve cuts the x--axis in (
-a, 0),
0)
(-2a O)
... ,. " 2 •
(m)Smcey�=-(x+a) (x+2a)lx, xcannot
be
positive.
xcannot be less than- 2a.
(iv) They- axis is an asymptote.
Fig. {7.57)
Ex. 3: ·Folium ofDecartes. Trace the curve
x3
+ y 3 = 3axy... X
or x=--3at y=---3at2
l+t3, l+t3
(S.U. 2003)
Sol.
: (i) The curve is symmetrical about the line
x = y .(ii) The curve passes through the origin.
(iii) The curve intersects the line x =
yin
(0, 0) and
[
•]
(iv) The line
x+
y+a = 0 is an asymptote.
(v)
If
x<
0 andy< 0, l.h.s. is negative while
r.h.s. is positive. Hence, no prut
of the curve is in the
third quadrant.
Ex. 4 :
Trace the curve a2l =
x2(a
2-x2)Sol. : (i) The curve is symmetrical about both the axes.
y �
-1.-3a]
'2
X
Fig. (7.58)
(ii) The points
(0, 0) (a, 0) and(- a, 0) lie on the curve.
(iii)
If
x >a, l is negative., Hence there no curve beyond x = a,
x=-a.
Engineering Mathematics -II
(7- 15)
Fig. (7.59)
Ex. 5:
Tracet
he curveei
=(x-
a)
(x-
b)2
Tracing of Corves
Sol.:
We have already dicussed the above curve in Ex. L.ase III on page 7.12. The shap�
of the curve will remain unchangedif
y replace y by.,Jc
. y i.e. the shape of the curve of the equationcl
=(x-
a)
(x -
b)2
will be thesame.
(a) Trace the curve:
9al
=x (x-
3a)2, a>
0.(S.U. 2004,06)
Sol. :
Replacing c by9a, a
by zero andb
by3a,
we get the above curve. The curve is shown on the right.
y
(b) Trace the curve
2 t3
x=t
3
yFig. (7.60)
i
=x(l-�r
orSol.:
Replacinga
by zero,b
by 113
andc
1, we get the above curve. The curve is shown on the right.:Fig. (7.61)
(c) Trace the curve:
al
=x (x-
a)2
Sol. :
Puttinga
= 0 andb
=a
we get the abovecurve. The curve is shown on the right. When x <
0,
y is imaginary and there isno curve for x < 0
When x -7 oo or x -7-oo, y -7 oo.
y
Engineering Mathematics
-
II(7-16)
(d) Trace the curve:
3al =
.l-
(a-x)
(S.U.1980,85,89,97)
Sol. : Putting
c= 3a, a =
0 and
b = a
we get the
above equation. The curve is shown on the right.
When
0
<x
<a,
yis real and when
x > a,
y
is imaginary. Hence, there is no curve for
x>a
When
x
--7 ooor
x
--7-
oo, y --7 oo.(e) Trace the curve:
al = 4.l- (a-x)
(S.U.1984, 95)
Tracing of Corves
X
(a, 0)
Fig. (7.63)
Sol. : Multiplying by
4
and then putting
a = a/12
in (d) we get the above curve.
Hence, it is similar to it.
(f)
Trace the curve :
al = x2 (a-x)
(S.U.1985)
Sol. : Putting
a = a/3
in (d) we get the above curve.
Ex. 1 :Trace the following curve
l (a-x )= x (x-b)2, a> b
Sol.
:(i) The curve is symmetrical about the
x-axis.
(ii) The curve passes through the origin.
(iii) The curve intersects the x-axis at
x=b, x=O.
2
(.
tv mce
)
s·
y2 x(x-b)
=
,
w en
h
x =a,
a-x
y
is infinite. Hence
x = a
is an asymptote.
(a) Trace the curve:
l (4-x) =x (x-2)2
X
(a, 0)
Fig. (7.64)
Sol.
:Putting
a = 4
and
b = 2
in the above equation, we get, the required
curve.
Exercise -I
Trace the following curves
1.
al = x (a2
- .l-)
2.
2l =x (4-.l-)
3.
a2l =.l-
(x-
a) (2a
-x)
4.
y(.l-
+4a2) = 8a3
5.
l =Xi (2a -x)
6.
l (a2 -.l-) = a3
x
(S.U.1994,97,2002,2005)
(S.U. 1978, 81, 88)
(S.U. 1979, 84, 2006)
(S.U. 1980, 96, 2006)
(S.U. 1983, 85, 93)
(S.U. 1984, 99, 2004)
Engineering Mathematics -II
7.
x2i =
a2
ci
-�)
8.
al=x2(x-a)
9.
x2
=
i (2
-
y)
10.
al =
x (a2
+x2)
11.
2l =
X( 4
+
�)
12. l
(a-x)=�
14.
a2�
=
l (a2
-/)
1 6.
l
(x2
+/)
=a2 (y2 -
�)
(7- 17)
18.
x2
(�
+l>
=
a2 (y2
-x2)
19.
i=x3-6�+ llx-6
20. l
(x.Z
+
4) = �
+
2x
22.
x2 (� -4a2) = l (� -a2)
24.
x5
+l = 5a2x2l
26. x6
+l
=a2x2l
28.
l (x2
+a2) = x2 (a2 -�)
30.
a4l
+ b2
x4
=
a2b2x2
Tracing of Corves(S.U.1985)
(S.U. 1979, 2002)
(S.U. 1981, 94)
13.a2/ =� (2a
-x)
15.l=d
17.
x2(x2
+l>
=
a2 (x2 -l)
(Hint:
y2
=
(x 1) (x-2) ( x-
3))
21.
x5 +l-5a2�y
=
0
23.
x4
+
y4
=2a2xy
x25.
y = -1 2( s.u. 2005)
+x
27.
(a-x) l = a2 x
29.
a4l
= x4 (a2-
-�)
31. l
(a x)(x-
b)=� (a, b >0 a>
b)
[Ans.:
(1)
(3)
y
y
Fig. (7.65)
2a
Fig.
(7.66)
X(2)
Similar to the previous
Ex.
1 with
a
=2
(4)
y
(0, 2a)
Fig. (7.67)
Xa.JI�UIIIOIIOIIII� IWIUioiiiiOIIIIWiol
..
., 1•(5)
y(2a,
xFig. (7.68)
\'- IUJ ··- .. ···:.
-· --··--(6)
y)
:(-a,O)
�·
Fig. (7.69)
. . .
X :x=a
(7)
(8) Compare with solved Ex. 5 Case
II(page 7.12)
y XFig. (7.70)
y xFig. (7.71)
(9) Similar to solved
(1 0)
yEx. 2 (page 7.11) with
a=1.
(11) Similar to
Ex. 10 above with
a=2.
(12) Similar to solved
Ex. 1 page (7.10)
with
a= a/2.
(14)
Fig. (7.73)
(13)
0 XFig. (7.72)
yo
xFig. (7.74)
Engineering Mathematics -II
(15)
yFig. (7.75)
(17)
Fig. (7.77)
(21)
Fig. (7.79)
(22)
Fig. (7.81)
(24)
y �+, XFig. (7.83)
(7- 19)
Tracing of Carves(16)
Y(O,a)
Fig. (7.76)
(18)
Fig. (7.78)
(19) Similar
tosolved
Ex.
5 Case
Ipage (7.11)
(20)
y(23)
(25)
Fig. (7.80)
yFig. (7.82)
yFig. (7.84)
XEngineering Mathematics
-
II(7- 20)
Tracing of Corves(26)
yFig. (7.85)
(28)
Same as 17(29)
Similar to Ex.17.
(30)
Similar to Ex. 17.(27)
(31)
y XFig. (7.86)
x=a
(\
Fig. (7.87)
6. Procedure for tracing curves given in polar equations :
I.
Find out the symmetry using the following rules.( 1) If on replacing
e
by- e
the equation of the curve remains unchanged then the curve is symmetrical about the initial line.(2)
If on replacingr
by -r i.e. if the powers of r are even then the curveis symmetrical about the pole. The pole is then called the centre of the curve.
II.
Form the table of values ofr
for both positive and negtive values ofe.
Also find the values ofe
which giver = 0
andr = oo
.III.
Find tan«1>.
Also find the points where it is zero or infinity. Find the points at which the tangent coincides with the initial line or is perpendicular to it. (Refer to the explanation and the fig. in§6
(b) page (2.11))IV.
Find out if the values ofr
ande
lie between certain limits i.e. find the greatest or least value ofr
so as to see if the curve lies within or without a certain circle.Ex. 1 : Cardioide
(a)
r =a (l
+cose) (S.U. 1991, 95)
(b) r =a (1
-cose) (S.U. 1986, 90)
(c)Vr = Va
cos(e/2)
Sol. : 1 (a)
: (i) The curve is symmetrical about the initial line since its equation remains unchanged by replacinge
by- e.
y
6-n
X
r
= a
(
1 + cos6)
Fig. (7.88) (ii) When
e = 0, r = 2a
and whene =
7t,r = 0.
Engineering Mathematics
-
II(7- 21)
Also when
8 =
rt/2, r = a and when8=3rc/2,
r=a.
·
rd 8
a(1
+cos8)
Now tan$=
(Tr"
= -a sin8
:. tan «l» =-cot
[
+
: . $ = + Hence,'I' = 8 +
«l» gives1t
8
1t38
'1'=8+2 +2 =2 +T
Tracing of Corves
When
8
=0, 'I'
= i.e. the tangent at this point is perpendicular to theinitial
line. When8 =
1t,
'I' = 2rt
i.e. the tangent at this point coincides withinitial
line.(iv) The following table gives some values of rand
e.
1 (b)
: (i) The curve is symmetrical about the initial line.Y (ii) When
8
=0
,
r =0;
8
=1t/2,
r=a
whene =
1t, r=2a;
8 = 31t/2,
r =a...
d 8
a (1-
cos8)
(111)
tan $ = r(Tr"
= a sine
r = a(
1-
cos6)
8
8
Fig. (7.89)
=tan2
:
.
«l»= 2
Hence, 'I'=
8
+ «l» gives'I'= 8 +
=When
8
=0, 'I'= 0
i.e. the tangent at this point coincides with the initial line whene
= 1t,'I'
=31t/2
i.e. the tangent at this point is perpendicular to the initial line.
(iv) The following table gives some values of rand
8.
1 (c)
: Squaring¥r
=Vi cos ()
we get r =a cos2(
)
=(1
+cos8).
This is the cardioide similar to the cardioide shown in
1
(a) witha/2
in place of a.Engineering Mathematics -II
(7- 22)
Ex:. 2: Pascal's Limacon
r =a+ b cos 0(a) Case
1 : a > b(i) The curve is symmetrical about the initial line.
(ii) Since a > b
,
r is always(iii) The following table gives some values of rand e. r
0
a +b(b) Case II
: a = b n/3 a+b/2 n/2 aWhen a = b we get r = a (l +
cos
0), the cardioideshown
in theex.
1 (i)(c) Case III
: a < b1992)
(i) The curve
is s
ymmetri
cal aboutthe
initial line.
(ii) r =
0 when
a + b cos 0 =0
. I
( !.!:_)
z.e. e = cos -b .
Tracing of Corves XFig. (7.90)
2n/3 1t a--b/2
a-b y a+b xFig. (7.91)
(iii) When 0 =
0,
r =a + b and is maximum.(iv) When e = cos -1
(
-alb), r =0 i.e.
th
e curve passes throughthe
origin.
(v) When cos
-I[-%]<
0 < n, r is negative and at 0 = n, r =a-- b.L•
: It may be noted that to plot a point
(
- r, 0), we3
/
/
rotate the radius vector through 0 and r in[,'
o opposite direction in this position. For example, to getthe point(-
3, 45°),
we rotate the radius vector from OXthrough 45° into the
position OL. We have tomeasure
3 along OL a distance- 3 i.e. we have to measure a distance
F
(7 92)
3 not along OL, but in opposite direction. Producing LOP
Ig.
•toP so that OP = 3 units we get the required
point
P. (vi) The following givessome
of rand 0n/2 a
cos -1 (-alb) cos -1 (-a/b) < e < 1t
0
negative1t a-b Note that we get a loop inside another toop as shown
in
the figure.Engineering Mathematics -II
(7- 23)
Tracing of CorvesEx.
:Trace the curve
r =2
+ 3cos8
Y(S.U. 2007)
Sol. : Following the above line we see that the curve r = 2 + 3cos
e
can beshown as on the right. y
5
Fig. (7 .94)
(d) Pascal's Limacon
is also given byr =
a
+b
sin8
where the y-axis becomes theinitial line.
For example, the curve r =
2
-
3sin8
isshown in the neighbouring figure.
Ex. 3
:Bernoulli's Lemniscate
,2
=a2
cos28
or(x2
+l> 2
=a2
(�
-l)
(S.U.1995,97,98,2003)
9 = 37tl4 y 9 = 7tl4Fig. (7.95)
8
0
n/6
,2
a2
a2/2
ra
aN2
Sol.:
(i) Since on changing r by-r the equationremains unchanged, the curve is symmetrical about the initial line.
(ii) Since on changing
8
by -8
theequation remains unchanged the curve is symmetrical about the pole.
(iii) Considering only positive values of
r we get the following table.
n/4
1t/4<8<31t/4
3n/4
1t
0
negative0
a2
0
imaginary0
a
(iv) Consider the equation
(�
+l)2
=a2(x2 -l).
If we put fromx2 -l=
0
i.e. y = ±x
i.e.8
=n/4
and8
= 3n/4
, we see that8
=1t/4
and8
=3n/4
are tangents at the origin.7. Curves of the form
r =
a
sinn e
or r = a cosn e
Sol.
: ( 1) Since sinn8
or cosn8
cannot be greater than one in both the casesr cannot be greater than a. Hence, the curve wholly lies within the circle of
Engineering Mathematics -II
(7- 24)
Tracing of Corves(2)
To find the curve r= a
sinnO
putr = 0.
ThennO = 0,
1t, 2n, ...
1t 21t
:. O=O, n, n
Draw these lines.
If
n is odd there aren
loops in alternate divisions and if n is even there are2n
loops one in each division.(3)
To trace the curve r= a
cosn
0
putr = 0
Then,1t 1t 31t 51t
nO=-
2 '2, 2
'
2 ,
....1t
1t
31t
51t
:. 0=-
' 2n ' 2n ' 2n
Draw these lines. If
n
is odd there aren
in alternate divisions and ifn
is even there are 2n loops one in each division.Ex. 1
:Three-leaved Rose
(i) r= a
sin30
(S.U.1996, 97, 99, 2004)
(ii)
r =a
cos30.
(S.U.1999)
9 =2n/3
--- --- 9 = n/3Sol.
: (i) The curve consists of three loopslying within the circle
r = a.
' '' Put
r= 0.
:. sin3 0
= 0
:. 30
= 0,
1t ' 21t ,
.... e ' ' ' ' ' '1t 1t 21t
41t 51t
:.0=0·
3
·
3
·
3
·1t·
3
·
3
Fig. (7.96)
Draw these lines and place equal loops in alternate divisions.
9 =
n/2
(ii) The curve consists of three loops9 =
5nt6/
' ' ' ' ' ' 9 =3n/2
Fig. (7.97)
Ex. 2
: Four leaved Roselying within the circle
r = a.
Putr = 0.
:. cos
30
= 0
:. 30
=
_2: 2:
2'2'2'2···
31t 51t
0
_
n n n sn 7n 3n
--6'6'2'6'6'2
Draw these lines and place equal loops in alternate divisions.
(i)
r= a
sin20
(S.U.1978, 1980)
(ii)r= a
cos20
(S.U.1997, 2003)
Sol.
: (i) The curve consists of four(2
x2)
loops lying within the circler =a
. Putr = 0,
sin20
= 0
Engineering Mathematics - II () = n/2 8 = Sn/4 ',___ __.-' () = -n/4 Fig. (7.99)
(7- 25)
:. 20 = 0, n, 2n ....n
3n
:. 0 = 0,2
' 1t , 2-Tracing of CorvesDraw these lines and place equal loops in all divisions.
(ii) The curve consists of four (2 x 2) loops lying within the circle
r
= a.
Put r =0.
:. cos 20 =
0
1t
n 3n
5n
:. 20 =-
2
·2
·T
·2
• ····n
1t3n 5n
:. 0=-
4 , 4
,4 ' 4
' 0000Draw these lines and place equal loops in all divisions.
Ex. 3
: Spiral of Archimedes r = a 0Sol. : Let us consider first only positive values of 0. When 0 = 0, r = 0, so the curve passes through
the pole. As 0 increases
f
, 1t ,3
27t , 2n .. and so on we see that r increases and the pointtracing the curve goes away from the pole as shown
in the neighbouring figure. When 0 � oo, r � oo. Fig. (7.100)
Fig. (7.101) 0�=.
r�O.
If we consider negative values of 0 also we get a curve shown by dotted line. (Fig.
7.101)
Ex.
4: Reciprocal Spiralr
0 =a (S.U. 1985)Sol. : Let us first consider only positive values of
e.
From the values considered below we see that as 0 increases r decreases. This means as 0 increases the point tracing the curve goes nearer and nearer to thepole and when
3n
1t0 = ... 2n ,
2,
n ,2 ....
_!I_ ..! 2a !I_1E._
Engineering Mathematics -II
-;- X
Fig.
(7- 26)
Tracing of CorvesFurther from the above values we note that as
e
-)0,
r � oo . If we consider negative valuesof 0 also we get a curve shown by dotted line. It may
be
noted that the line y =a
is anasymptote.
Ex.
5
: Equiangular Spiral r= ae
be(S.U.
1985)
Sol. : When 0
= 0
, r= a.
As 0 increases r alsoincreases. When 0 -� oo , r � oo •
When 0 -> - 9 r ->
0
dO
Further, tan
<I> =
rdr
r r 1:. tan
<I>= a bee
=t;
\Fig.
(7.104)
Thus,
<I> i.e.
the between the tangent and the radius vector is always constant. Hence, the name equiangular. Ex.6
: r= a
(sec 0 + cos 0)(S.U. 1979,
2004)
Sol. : The equaition can be written as
r2
0)
=a
[r
cos 0 r 2 =a (r sece
+ r cos 0) 0(2a,
.
X Fig.(7.105)
•2
2
[x2+y2
)
.. x +y +x :. J2 (x-a) =
x2
(2a
x)(i)
The curve is symetrical about x-axis. (ii) When x =0,
y =0. Further
i
+2:l-
=0
are the tangents at the origin.The tangents at
the
origin are imaginary. The origin is an isolated point. (iii) x =a
is an asymptote.Exercise II
1.
(a) r= 2 (1
+cos 0), (b) r= 1
+cos 0 (S.U.1978, 80, 93, 94, 97)
2. r
=
[I)
(1
+ cos 0) (S.U.1989)
Engineering Mathematics
-
114. r =
1 +2
cos8
5.
,.z
=
4 cos 28
6. r (
l + cos8)
=2a
7. r (1
-cos8) = 2a
9. r=
3 cos28
10. r =a
(1 + sin8)
{7- 27)
8.
r= 2a
cos8
Tracing of Corves(S.U. 1980)
(S.U. 1980, 99)
(S.U. 1981)
(S.U. 1988)
(S.U. 1981, 2003)
11.
?
cos28 = a2
orr2
=a2
sec28
12. r2 = a2
sin28
13. r =a
(1 -sin8)
14. r = 2a
tan8
sin8
[Ans. :
(S.U. 2005, 06)
(S.U. 2003)
1.
Similar to solved Ex.l (1) Fig.(7.88).
3.
Similar to solved Ex.2
(a) Fig.(7.90).
5.
Similar to solved Ex.3.
Fig.(7 .95).
2.
Similar to solved Ex. 1(1).4.
Similar to solved Ex. 2 (c) Fig. (7.91)6.
Parabola opening on the left.Fig. (7.106)
8.
CircleFig. (7.108)
7.
Parabola opening on the right.Fig. (7.107)
9.
Cricle similar to the previousone with a=
3/2
10.
y
Engineering Mathematics
-
II(7- 28)
Tracing of Corves11.
13.
y12.
Fig. (7.110)
yFig. (7.111)
y
14.
See remark underX
Fig. (7 .112)
Ex.
1
page (7.1 0)
See Fig. (7.45).
8. Some More Curves :
(1) Cycloids
Because of its graceful form and beautiful properties, mathematicians call this curve the Helen of Geometry. (After all, mathematicians are not dry as they are sometimes called !) When a circle rolls, on a straight line without sliding any fixed point on its circumference traces a cycloid.
The curve traced by a fixed point on the cricumference of a cricle, which rolls without sliding on the cricumference of another fixed cricle is called an epicycloid or hypocycloid.
If the rolling circle is outside the fixed cricle the curve is called the
epicycloid
and if the rolling cricle is inside the fixed cricle, the curve is called thehypocycloid.
We have seen one particular hypocycloid in Ex. (3) page7.11.
If the radius of the rolling circle is equal to the radius of the fixed cricle, the epicycloid is called cardioide because of its heart like shape (Cardio
=
Heart, eidos =shape). We have studied this curve Ex.
1
on page7.20.
The cyloid is generally given in one of the following forms.
(a)x=a(t +sint),y=a(l +cost) (b)x=a(t-sint),y=a(l +cost)
(c) x =a (t +
sint), y =a (1
- ccist)
(S.U. 2004, 07)
Engineering Mathematics
-
II(7- 29)
(d)
x �""a(t --sin t),
y =a (1 -cost)dx dy .
Sol.
:(a) dt =a (1
+cost),dt =-a sm t dy dy!dt --asint
(t)
:. dx = dx/dt =a( 1 +cost)
=-tan
2
Some of
the values of x, y and dyldxare
-n -n/2 0
-an
--a ( 1t/2 +1)
()
0 a 2a 00 1 0 Tracing of Corves(S.U. 1987)
n/2 n a(1t/2+1) an a 0 - 1 -oo
From the above table we see that at t =
-
n and at t = n, the tangents areparallel to they-axis. These points are called cusps.
Further
att
= 0,the
tangentis parallel
to
the x-axis.y t = 0
--�:
2aFig. (7.113)
dx dv
(b)
-=a (l- cost) --=-a sin tdt ' dt
dy dyldt --asint
[t)
: · dx =
d;·d
d t = a ( 1-
cost)
=
- cot 2
Some of
the values of x, y, dyldx are0 n/2 n X 0 a (7tl2 -1) an y
2a
a 0 dyldx -00
-1 0 31t/2 2n a (3n/2 +1) 2an a 2a 1 00From the above table we see that at t = 0 and t = 2n
the
tangents are parallel to they-axis. These are cusps. Further att
= n, y = 0 and dy/dx = 0 i.e.the x-axis is the tangent.
t=O an: an:
---
+
Engineering Mathematics -II
(7-30)
Tracing of Corvesdx
(c)
=a(1
+cost),
=a sin
tdy
dyldt
asint
_[t]
. - = ---- = -- tan--.. dx
dxldt
1+cost)
2Some of
the values of x, y,dyldx
aret
-n:
-n:/2 ---0
n:/2
1t.n/2
+1)
0 a(n:/2+1 an
dyldx
_a
0 a
--1 0
1
y t = 1t : xFig. (7.115)
dx
dy
dt-
=a
(1
-cost)
, dt
= asin
t. !!r_
=dyldt
=_E._� =
cot(1.)
· ·dx
dxldt
a
(
1 -cost) 2Some
of the values of x,y,
dyldx aret 0 rrJ2
2a
002n:
X0 a (rrJ2-
1)
n
an a (31tl2 +
1)
2arr.
y
dy/dx 0
00 ya
1 t = n ' ' ' : 2a an : an2a
0
a -- 1 = 0 t:: 27tFig. (7.116)
(2) Catenary
0
-00(S.U. 2003)
If a heavy uniform string is allowed to hang freely under gravity, it hangs in the form a curve called catenary. Its equation can be shown to be
Engineering Mathematics
-
II(7- 31)
Tracing of Corves Its shape is as shown in the figurey
\
Fig. (7.117)
Its lowest point is at a dirtance c from the -cause,
when x =
0,
y = c cosh0
=c.Further, we see that y
=
cc
osh;
= c2
Hence, if x is replaced by - x the equation remains the same. Hence, the
curve is symmetrical about they-axis
.
d
yexlc
_ e -xlcSm
c
e=
is zero. when x = c, the tangent atx
= c isd.:r 2 parallel to the x-axis.:
Sol.:
(3)
Tractrixx = a cos t +
t
a log tan 2( t]
, y = a sint
dx
.
a -=-asmt +dt
22tan (f)
sec2(f)·-}
tan 2[ f]
a = -a sin t + 2 sin( 1-)
cos( f]
= -a sint
+--(!.-
=--(!.-
(1
- sin 2t)
sm t sm tdy
a cos 2t
sint
dt
=acost
. dy
=dyldt =a cost.
sin t
=tan t
· ·dx
dxldt
a cos
2t
dy
Some of the
valuesoft, x,
y,dx are ,
Engineering Mathematics - II
(7- 32)
Tracing of Corvest -n -n/2
0
n/2 1tX 00
0
000
00y
0
-a0
a0
dy/dx
0
-000
000
From the above data we see that as t �- n the point on the curve is (oo,
0)
and as t �0
the point is(- oo,0).
9. Some Solids
t = n/2
X
Fig. (7.118)
You have studied to some extent plane and straight line
in
three dimensions. We shall here get acquainted with some more three dimensional solids such as sphere, cylinder, cone, paraboloid etc. z1. Plane
: The general equation of a plane is linear of the form a'x + b'y + c'z + d' =0
which can be written as a' b' c' -x+-y+-z=1 d' -d'-d'
y=O X x=O i.e. ax + by + cz = 1Fig. (7.119)
ySimplest planes are yz plane whose equation is x =
0;
the zx plane whoseequation
is y =0;
the xy plane whose equation is z =0.
z y
!+2$,+!=1
a a a y X XFig. (7 .120)
Fig. (7.121)
The planes parallel to the coordinate planes are
x =±a, y = ±b,
z =±c.The plane ax+ by+
cz = 1cuts
off intercepts 1/a, 1/b, 1/c on thecoordinate
axes. Still more common form of the plane is
�+l'.+�=l a b c
Engineering Mathematics -II
(7- 33)
Tracing of CorvesThis plane cuts off intercepts a, b, con the coordinate axes.
2. Cylinders
: An equationinvolving
only two variables represents acylinder in three dimensional geometry. Thus,
f
(x, y)
=O,.f (y, z)
= 0, y (z,x)
= 0 representcylinders in three dimensions.
(a) Right Circular Cylinders :
x2
+y2
=a2
is a right circular cylinder, whosegenerator is parallel to the z-axis. Similarly,
y2
+z2:: !J2
is a cylinder whose generators are paralel to the x-axis. z2 +x2
= c2 is a cylinder x/
whose generators are parallel to they-axis.
z
..
X X
Fig. (7.122)
(7.123)
Fig
.(7.124)
(b) Parabolic Cylinders
: The equationy2
= 4ax represents a paraola intwo dimensions. But in three dimensions it represents a parabolic cylinder.
The equations z2 =
4by,
x2 = 4cz also represent parabolic cylinders asshown the following figures.
Fig. (7.125)
Fig. (7.126)
3. Sphere
: The equation of sphue in standard form i.e. with center at the origin and radiusa
is x2 + y2 +z2
= a2.4. Cone
: The right circular cone given byx2
+y2
= z2 is shown in thefollowing
figure7.129.
The other
two cones
y2 +z2
= x2 and x2+
z2
=y2
are shown in thefollowing
figures7.130
and7.31.
Engineering Mathematics -II
(7- 34)
Tracing of Corves X 0y
Fig. (7.129)
z
y2+z2=x2
XFig. (7.130)
5.
Ellipsoid
:The equation of ellipsoid in standard form is x2l
z2 -+-+-= 1 a2 b2 c2 zFig. (7.131)
zThe sections of the ellipsoid by planes parallel to the coordinate planes are ellipses.
6.
Hyperboloid
: The equations ofhyperboloid of one sheet and two sheet are
Fig. (7.132)
respectivelyx2
l
z2 x2l
z2-+-+-== 1 and -+-+-=-1.
a2 b2 c2 a2 b2 c2
The hyperboloid of one sheet is shown on the right. Sections of the hyperboloid of one sheet parallel to the xy-planes are ellipses and sections parallel to
y
yz-plane or zx-plane are hyperbolas. x
z The hyperboloid of two
sheets
is
shown on the left.Fig. (7 .133)
Fig. (7.134)
Sections of this hyperboloid by planes paralles to the xy-plane i.e. z ==
k,
I
k
I
> c are real ellipses.Y
Sections of this hyperboloid by planes parallel to the zx-plane i.e. y =k
are zhyperbolas.
7. Paraboloids
: We shall discuss here only one elliptic paraboloid given byx2
l
2z-+-==-a2 b2 c X
Engineering Mathematics -II
(7- 35)
Tracing of CorvesSections of this paraboloid parallel to the .xy-planes are ellipse. Sections parallel to the yz-plane or the zx-plane are parabolas.
Sketch the following figures. 2 2 2
1.
y=x ,x +y =2 2 2 2 2 2 X y2.
X + y = b ,2
+2
1 a b 73.
y- = X + 6, X = y4.
y = x2, y = 1, X = 1, X = 25.
y=x2, y=x+2,x=-1,x=26
• x + y = 2 2 ax, ax = by 7. x2 +i=by,ax=by 2 2 2 28.
X - y = 1, X - y = 2, X)' = 4, xy = 29.
xy = 2 - y10.
y = x2 -3x, y = 2x11.
i =-4 (x-1),l
=-2 (x-2)12.
a2 x2 = 4l (a2-l)
or x =a sin 2t, y =a sin t13.
r (cos 8 +sin 8) = 2a sin 8 cos 8 or (x + y) (x2 +l)
= 2axy14
. x + y = a xy or 4 4 2 2 r 2 = 2a2 sin8 cos84 4 sin 8 +cos 8 . 2 8 7 8
15
• x + y = a x y or 6 6 2 2 2 r 2 a 2 sm · cos-6 6 sin 8+ cos 816.
x4-2.tya2+ a2l =0
or?
= 2a tan 8 sec2 8 - a2 tan 2 8 sec2 817.
y=x2 -6x+3, y=2x-918.
y = 3x2 - x-3, y =- 2x2 + 4x + 719.
l
= 4x, y = 2x- 420.
y = 4x -x2, y = x21.
A cylindrical hole in a sphere.22.
A cone and paraboloid given b y z 2 = x2 +land z = x2 +l
23.
A paraboloid� +l
= az, and the c ylinder� +l
= a22 2
24.
The paraboloids z 4 -x2 - Y4 , z = 3x2 + y4Engineering Mathematics -II
(7- 36)
[Ans. :
(1)
(3)
(5)
(7)
( 6, 0)
y x2+y2=2Fig. (7.136)
yFig. (7 .138)
yFig. (7.140)
yFig. (7.142)
X X X Tracing of Corves(2)
Y x2+y2=b2 XFig. (7 .137)
(4)
y=1 x=1 x=2Fig. (7.139)
(6)
y XFig. (7.141)
(8)
y XFig. (7.143)
Engineering Mathematics -II
(7- 37)
Tracing of Corves(9)
y(10)
XFig. (7.145)
Fig. (7 .144)
(12)
(11)
X(2, 0)
XFig. (7.146)
Fig. (7.147)
(13)
y(14)
y X XFig. (7.148)
Fig. (7.149)
(15)
y(16)
y X XFig. (7.150)
Fig. (7.151)
Engineering Mathematics II
{7- 38)
(6, 3)
:Fig.
(7.152)
(19)
y(2, 7)
(-1, 1)
Fig.
(7.154)
(21)
XFig.
(7 .156)
(23)
yFig.
(7.158)
(18)
(20)
(22)
(24)
(-1, 1)
Tracing of Corves yFig. (7
.153)
Fig. (7
.136)
Yt
:Fig. (7 .157)
Fig. (7
.159)
Engineering Mathematics -II
(25)
yFig.