curve tracing 2.pdf

1. Introduction For evaluating areas, volumes of revolution etc. we need to know the general nature of the given curve. In this chapter we shall learn the methods of tracing a curve in general and the properties of 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 symmetrical about any line with the help of the following rules :

( 1) The curve is symmetrical about the x axis if the equation of the curve -

remains unchanged when y is replaced by- y i.e. if the equation contains only even powers o f y.

(2) The curve is symmetrical about the y axis if the equation of the curve -

remains unchanged when x is replaced by- x i.e. if the equation contains o nly even powers of x.

(3) The curve is symmetrical in opposite quadrants if the equation of the curve remains unchanged when both x andy are repl aced by- x and y. -

(4) The curve is symmetrical about the line y = x if the equation of the curve remains unchanged when x andy are interchanged. 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 points of

intersection of the curve with the coordinate axes. Find also the equations of

the tangents at these points. IV. Asymptotes : Find out the asymptotes if any. V. Regions where no part of the curve lies : Find out the regions of the plane where no part of the curve lies. VI Find out dyldx : Find out dy/dx and the points where the tangents are parallel to the coordinate axes. •

Eng�Mrering Mathemat -

If

Tracing of Corwes

(7-2)

3. Common Curves You have already studied quite in detail straight line, circle, parabola, eJiipse and hyperbola, includi ng rectangular hyperbola. We shall briefly review these curves. Rectangular Coordinates

(a) Straight Line: General equation of straight line is of the form ax +

by + c = 0.

To plot a straight line we put y = 0 and find x. Also we put x = 0 and find y. We plot these two points

and join them to get the required line.

Some particular cases of straight Jines are shown below. (i) Lines parallel to the coordinate axes. y

y x=h

y=k X

0

X

0

Fig. (7.1)

Fig. (7.2)

(ii) Lines passing through origin y

y

y=mx mO X

X

Fig. (7 .3)

Fig. (7.4)

(iii) Lines making given intercepts on coordinate axes. y

X

X

Fig. (7 .5)

General equation of circle is :il g,- h) and radius=

2 + J2- c

Fig. (7 6) .

+

j + 2gx + 2/y + c = 0. Its centre is

Engineering Mathematics

-

II

Tracing of Corves

(7- 3)

(i) Circle with centre at origin and radius

a

and 2 + j2- c.

(ii) Circle with centre at (- g,- f) and radius 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. y -x

0

Fig. (7.9)

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 + y

Fig. (7.13)

l ± 2gx + c = 0, .-r? + i ± 2fy + c =' y

,_Fig. (7.14)

JTat:mg

v- ..,,

or �.;orves

(vi) Circle touching both axes. Its equation is of the form :Xl + l ±2ax :;t 2ay+ a2 = 0. y

Fig. (7.15)

l- 4x + 4y + 4 = 0 2 2 Sol. : We write the equation as (x - 2) + (y + 2) 2 = 2 Ex. 1 : Draw the circle :Xl

Its centre is

+

(2, -2) and radius = 2.

It is shown on the right.

Polar Coordinates Fig. (7.16)

(a) Straight line If We put X =rCOS 8, y = r Sin

8 in the equation Of the Straight line

ax+by+c=O we get,

ar cos e + br sin e + c = 0

i.e.

a

cos

e

+

b sin e

(i) A line parallel to the initial line. If a line is parallal to the initial line (or the

+

.:. = 0

r

x- ax is) at a ditance p from

it then, from the figure we see that its equation is rsin e =±p (It

p 0

is clear thatrsin e = p, if the line is e =- p, if the line is

above the initial line andrsin

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 the x-

axis) at a ditance p from it then from the figure we

see thatrcos6=±p

X

(It is clear thatrcos e = p, if the line is to the right of the pole and rcos a =- p, if the line is to the left of the

0

Fig. (7.18)

pole.) . 0

p

Fig. (7.19)

X

(iii) A line through the pole (or the

origin)

If a line passes through the pole and makes an

Engineering Mathematics

ang l e

-

II

Tracing of Corves

(7- 5)

a with the initalline then every point on the line has coordnates (r, a), 0= a

where r is positive or negative. Hence the equati on of a line is

In particular the equation of a line making e = 1t/4.

an

•

angle of 45° is

(b) Circle If we put X= r cos e. y = r s in e in the equation 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 2 If We put X = r COS () y = r sin 0 in the equation (a, 0) and radius a i.e. in

of the circle with center

X

Fig. (7.20)

(x- a) 2 + y 2 = a 2 i.e. in x 2 + y 2- 2ax = 0 we get r 2 -

2a r cos e = 0:. r = 2a cos e

This is the equation of the circle with cen ter (a, 0) and radius a in polar coordniates.

X

If we put

X

= r cos e, y = r sin 0 in the

equation of the circle with center (0, a) and

Fig. (7.21) x

2 +

radius a i.e. in

(y- a) 2 =a 2 i.e.

we get, r 2 -

y

2ar sin e

x =

2 + y 2-

i.e. r =

2ay = 0

2a sin e

This is the equation of the circle with center (0, a)

and radius a in polar

y

coordniates

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 touching both the coordniate axes in the first quadrant X

Fig. (7.23)

(x- a)2 + (y- a)2 = a2 2 i.e. in x + l- '2ax- 2ay + a2 = 0

i.e. in

we get the equation r 2 - 2ar. (cos

Ex.

e +sin 0) +a 2 =0 in polar coordniates.

: Find the equation of the cricle in polar coordinates having centre

and radius the line y

3. Also find the polar coordinates of its point of intersection with x.

=

Sol. : The equation of the circle with centre (3, 0) and radius

(3, 0)

3 is (x- 3) 2 + y 2 = 3 2

.,

--

----- ----------:---

WI

�va-.aa

+y2-6x=O

i.e. x2

Putting X = r COS we get

••..•••11

,. - ..,,

r

2-

8, y = r Sin a

6r cos a= 0

i.e.

r

= 6 cos 8

The equation of the line is y = x. Putting X equation in r

sin

= r COS 0, y = r sin 8, we get the polar coordinates as

a = r cos 8 i.e. tan a = 1

Putting 8 Hence

=

1t/4 in

r=

:.

Fig. (7.24}

a= 1t /4

6 cos a , we get r = 6 cos

1t /4 =

the polar coordinates of the point of intersection

6

·

=

are

(3 J2

3 J2 .1t /4)

(iii) Parabola : General equation of parabola is

y = ax 2 + bx + c

ay 2 + by + c. By completing the square on x or on y and by shifting the equation can be written in standard form as y 2 = 4ax or x 2 = 4 by. x=

or

origin the

d x =

x

a

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

=

a (x- h)

4

(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-

y

h)2 = 4a (y- k)

(iv) the axis parallel to the y-axis, opening downwards is

(x- h)2

=-

4a

(y- k)

Ex. 1: Sketch the curve r

X

+ 4x- 4y + 8 = 0

Sol.: The equation can be written as

r- 4y + 4 =- 4x- 4

i.e.

(y- 2) 2

Fig. (7.27) =-

4 (x

+ 1)

Engineering Mathematics

Putting y

-

2

=

II

Y, x + 1

Its equation is Y Its vertex is at

-

2

=

Tracing of Corves

(7- 7) X,

=

- 4X.

(-1, 2) and it opens on the left.

Ex. 2 : Sketch the curve 2 x + 4x- 4y +

16

0.

=

Sol. : The equation can be written as (x + 2)

2

=

4

(y- 3).

Putting x + 2 = X, y 3 . 2 Its equation is X = 4 Y. -

Its vertex is (-2,

=

Y,

Fig. (7.28)

3) and it opens upwards.

(iv) Ellipse The equation of ellipse in standard form is

x2 y2 -;z+t;2=

1

x2 y2 (a> b), -;z+ b2

=I

(ab

Fig. (7.30)

Fig. (7.29)

If the centre of the ellipse is at (h, k) the equation of the ellipse becomes

h) 2 a2

(x-

+

(y- k) 2 b2

_

-1

If 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 Equation of hyperbola in standard form is

Mathematics

-

II

Tracing of Corves

(7- 8) y

{0, be)

y

(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�-k

Fig. (7.33)

Fig. (7.34)

If the lines y = + x and y = -x are the asymptotes, the equations of the 2 2 2= 2 = a , y -x a •

rectangular hyperbola are given by 2-

-l

y

X

X

'

x2-y2=a2

Fig. (7.35)

'

y2 -x2

=

a2

Fig. (7.36)

Polar equation of the rectangular hyperbola x2

by putting X = r cos e, y = r sin e. The equation is 2 2 2 2 i.e. r (cos e- sin 8) = a2 i.e. r2 cos 2 e = a Similarly the polar equation of

l- x2 = a2 is ,:z. cos 2 e = - a2

?-

-

l

2 cos

2 a is obtained 2 2 - r sin e = a2

=

e

the other rectangular hyperbola i.e. of

Engineering Mathematics

-

II

Tracing of Corves

(7· 9)

Parametric Equatioas Ellipse The parametric equations of the ellipse are X =a cos

e. y = b sin e.

Hyperbola The parametric equations of the hyperbola are X= a sec

e, y = b tan e or

X= a

COS/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 =

Curves of the form ym

ct and y = cIt. =

xm where m and n are positive intigers.

It is interesting to note the shape of the curves whose equations can expressed in the above form.

S ome of the curves are known to us and

be

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.

(2)

y

2

Fig. (7.38)

=x .

The staight line through

The parabola through the origin

the origin

and opening y

Fig. (7.39)

up y

Fig. (7.40)

Mat.tMma -II

Tracing of Corves

(7·10)

(3) y=xl. Cubical parabola

(4)

? =x.

parabola through the origian and opening on the right

The

X

X

Fig. (7.42)

Fig. (7.41) (5)

i=�.

(6)

Pair of two lines passing through the ori gin

i=�. Semi-cubical parabola

y

X

X

Fig. (7.44)

Fig. (7.43) (7)

y3

=x.

(8)

i =x2. Semi-cubical parabola

Cubical parabola

4. Some Well-known Curves Ex 1 : Cissoid of Diocles :Trace the curve i (2a- x) = � 2at3 2at2 or x = -2 , y = -2 (S.U. 1986, 90) l+t l+t Sol. : (i) The curve is symmetrical about the :as oN :II x- axis. : )( (ii) The curve passes through the origin and the tangents at the ori gin are j = 0 i.e. the x-axis (2a, O) x is a double tangent at the origin. (iii) Since

3

l=

-x

as

x � 2a, y � 0

2a,

i is negative. Hence, x > 2a.

the curve does not exist when

(Remark :By putting X= rcos a, y = r sine, show that the polar equation

of Cissoid is r = 2a tan 0 sin 0).

Englneertng Mathematics II

Tracing of Corves

(7- 11)

-

Ex. 2: Trace the curve a2� = y (2a- y) 3

(S.U. 1981,86,90, 2006)

Sol.: (i) The curve is symmetrical about they-axis. (ii) It passes through the origin and the x-axis is a tangent at origin. (iii) It meets the y-axis at (0,0) and (0, 2a).

Further d y l dx. is zero at (0, 2a); the tangent at this point is parallel to the x-axis. 3 2 2 (iv) Since x = y (2a - y) I a when y 2a 2 and y < 0, x is negative and the curve does not

>

Fig. (7.46)

exist for y > 2a and y < 0. y

Ex. 3 : Astroid or Four Cusped Hypocycloid : Trace the curve 213 + 213 1

=

(ylb)

(xla)

3

or X= a cos 3e, y = b sin e

(S.U.1985, 2003)

S ol. : (i) The curve is symmetrical about both axes.

(ii) The curve cuts the x-axis in (± a, 0)

and they- axis in (0, ±b)

(iii) Neither x can be greater than

Fig. (7 .47)

a

can be greater than b.

Ex. 4: Witch of Agnesi: Trace the curve xl = a2 (.rz- x) (S.U. 1980, 91, 92, 98, 2003, 04) y Sol.: (i) The curve is symmetrical about the x-axis. (ii) The curve passes through the point

(111 ) s·mce . . .

,v

2

= a

2

(a--x) X

'

negative. Also x cannot be greatc:.· than

(iv) As x � 0, y�

oo,

0)

.

x cannot be a.

they-axis is an

a sym ptote .

Ex. 5: Trace the curve l = (x- a) (x- b) (x- c) whete a, b, c are positive. Sol. : We consider the following cases : (a) Case I :

a

O

Fig. (7.48)

nor y

Engineering Mathematics

-

(7- 12)

II

Tracing of Corves

l is negative l >O

when b < x< c, whenx>c,

Hence, there is no curve to the left-of

x =a

and also between

x =band

x=c.

(4) If x> c and increases then l also increases y

x

Fig. (7.49) (b) Case II: a= b

a, l is negative., Hence there

no curve beyond x = a,

Engineering Mathematics

II

-

Tracing of Corves

(7- 15)

Fig. (7.59)

Ex. 5: Trace the curvee i

=

(x- a) (x- b)2

Sol.: We have already dicussed the above curve in Ex.

L.ase III on page 7.12.

�

replace y by

The shap of the curve will remain unchanged if y i.e. the shape of the curve of the equation

cl

=

(x -

a)

(x -

same.

(a) Trace the curve: (S.U. 2004,06)

9al

=

b)2

.,Jc . y

y

will be the

x (x- 3a)2,

a> 0.

Sol.: Replacing c by 9a, a by zero and b by

3a,

we get the above curve. The curve is shown on the right.

Fig. (7.60)

y

i

(b) Trace the curve x=t

2

=x(l-�r

or

t3

3

Sol.: Replacing a by zero,

b by 113 and c

1,

we get the above curve. The curve is shown on the right.

:Fig. (7.61)

al x (x- a)2 Sol. : Putting a 0 and b a we get the above

(c) Trace the curve: =

=

y

=

curve. The curve is shown on the right. When

x

<

0, y is imaginary and there is

no curve for x < 0 When x -7

oo

or x -7- oo, y -7 oo.

Fig. (7.62)

Engineering Mathematics

-

(7-16)

II

Tracing of Corves

(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. y

X

When 0 < x < a, y is real and when x>a, is imaginary. Hence, there is no curve for

(a, 0)

x>a When x

--7

oo

or x

--7- oo, y --7

oo.

Fig. (7.63)

(e) Trace the curve: al=4.l- (a-x) (S.U.1984,95)

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

x=b, x=O.

(a, 0)

2

. s· 2 x(x-b) (tv) mce y =

, when x=a, a-x y is infinite. Hence x= a is an asymptote. (a) Trace the curve: l (4-x)=x (x-2)2

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

5.

l =Xi (2a-x)

6.

l (a2-.l-) = a3 x

(.l- + 4a2)= 8a3

(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

Tracing of Corves

(7- 17)

7. x2i=a2 ci -�) 8. al=x2(x-a)

(S.U.1985)

9.

x2=i (2-y) 10. al = x(a2 + x2) 11. 2l = (4+ �) 12. l (a-x)=� 14. a2� = l(a2 -/) 1 6. l(x2+ /) a2(y2-�) 18. x2 (� + l> = a2(y2 -x2) 19. i=x3-6�+ llx-6 20. l (x.Z + 4) = �+ 2x 22. x2(�-4a2)=l(�-a2)

(S.U. 1979, 2002) (S.U. 1981, 94)

X

=

24. x5

+

l=5a2x2l

13.

a2/=�(2a -x)

15.l=d 17.x2(x2 + l> = a2(x2-l)

1)(x-2)( x- 3)) 21. x5 + l-5a2�y = 0 23. x4+ y4 = 2a2xy (Hint: y2 = (x

25.

y= 1

x

( s.u. 2005)

+x2 27. (a-x) l=a2 x 29. a4l=x4(a2--�) -

26. x6 + l = a2x2l 28. l(x2 + a2)=x2(a2-�) 30. a4l+ b2 x4 = a2b2x2 31. l(a x)(x- b)=� (a, b >0 a> b) [Ans.: (1)

(2)

y

Similar to the previous Ex. 1 with a = 2

Fig.(7.65) (3)

(4)

y

y (0, 2a) 2a

X

X

Fig.(7.66)

Fig.(7.67)

a.JI�UIIIOIIOIIII� IWIUioiiiiOIIIIWiol

(5)

..

.,

\'- IUJ

1•

··- .. ···

(6)

y

(2a,

-·

--·

·--

y

) �·

:(-a,O)

x

:.

.. X .:x =a

Fig. (7.69)

Fig. (7.68)

(8) Compare with solved Ex. 5 Case II (page 7.12)

(7)

y

y

x

X

Fig. (7.71)

Fig. (7.70) (9) Similar to solved Ex. 2 (page 7.11) with (11) Similar to Ex. 10 above with

a=

a=

(1 0) 1.

y

2.

X

0

(12) Similar to solved Ex. 1 page (7.10) with a= a /2.

Fig. (7.72) (13)

(14)

y

x

o

Fig. (7.73)

Fig. (7.74)

Engineering Mathematics

(15)

-

II

Tracing of Carves

(7- 19) (16)

y

Y

(O,a)

Fig. (7.76)

Fig. (7.75) (18)

(17)

Fig. (7.78) (19) Similar to solved Ex. 5 Case I page (7.11)

Fig. (7.77)

Fig. (7.80)

Fig. (7.79) (23)

(22)

y

y

Fig. (7.82)

Fig. (7.81) (24)

y

(20)

(21)

�+,

(25)

y

X

X

Fig. (7.83)

Fig. (7.84)

- II

Engineering Mathematics

(26)

(7- 20)

Tracing of Corves

(27)

y

y

X

Fig. (7.85)

Fig. (7.86)

(31)

(28) Same as 17

(29) Similar to Ex. 17. (30) Similar to Ex. 17.

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 replacing r by - r i.e. if the powers of r are even then the curve is symmetrical about the pole. The pole is then called the centre of the curve.

II. III.

r for both positive and negtive values of e. e which give r =0 and r =oo .

Form the table of values of

Also find the values of 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§

IV.

Find out if the values of

greatest or least value of

6 (b) page (2.11))

r and e lie between certain limits i.e. find the

r so as to see if the curve lies within or without a

certain circle.

Ex. 1 : Cardioide r =a (l +cos e)

(S.U. 1991, 95)

(b) r =a (1 -cos e)

(S.U. 1986, 90)

(a)

(c)

Vr =Va cos (e/2)

y

6-n

X

Sol. : 1 (a) : (i) The curve is symmetrical about the initial line since its equation remains unchanged by replacing (ii) When

e by - e.

e =0, r = 2a and when e =7t, r =0.

r

=a ( 1

Fig. (7.88)

+ cos

6)

Engineering Mathematics

Also when

(7- 21)

II

8 = rt/2,

r = a

Tracing of Corves

and when

r=a.

8=3rc/2, rd 8

Now tan$= (Tr" ·

-

a =

(1

+cos

-a

sin 8

[

:. tan «l» =-cot : . $=

Hence,

+

8) +

'I' = 8 + «l» gives

1t 1t 38 8 '1'=8+2 +2 =2 +T

8

When

initial initial

=

0, 'I'

line. When

=

8=

1t

line.

i.e. the tangent at this point is perpendicular to the

, 'I' = 2rt

i.e. the tangent at this point coincides with

e.

(iv) The following table gives some values of rand

1 (b)

:

(i) The curve is symmetrical about the initial line. (ii) When

Y

,

8 =0

r

8 =1t/2, e = 1t,

when

=0; r=a

r=2a;

8 = 31t/2,

r = a

(

1

-

... (111) tan $ cos

6)

=tan

Fig. (7.89) Hence, 'I'=8 + «l» gives

=

d8 r (Tr"

'I'= 8 +

r =a

=

8 2

a (1- cos 8) a sin e 8 : «l»= 2

.

=

8 =0, 'I'= 0 i.e. the tangent at this point coincides with the initial e =1t, 'I' = 31t/2 i.e. the tangent at this point is perpendicular to the

When line when initial lin e

.

(iv) The following table gives some values of rand

1 (c)

: Squaring

¥r

=Vi cos (

)

we get

r =a

cos

2

(

)

8.

This is the cardioide similar to the cardioide shown in place of a.

(1

=

1

+cos

(a) with

8).

a/2

in

Engineering Mathematics

-

Ex:. 2: Pascal's Limacon (a) Case 1

:

II

r

Tracing of Corves

(7- 22)

=a+ b cos 0

a>b

(i) The curve is symmetrical about the

X

initial line. (ii) Since a

>

b ,

r

is always

(iii) The following table gives some

Fig. (7.90)

va l ues of rand e.

r

(b) Case II :

a

0

n/3

a +b

a+b/2

n/2

2n/3

1t

a

a-- b/2

a-b

=b y

When a = b we get r= a (l +cos 0), the cardioide

shown in the ex. 1 (i)

(c) Case III : a < b

1992)

a+b

(i) The curve is symmetri cal about the

x

initial line. (ii) r= 0 . z.e. e

when a + b cos 0 = 0 =

cos

(iii) When 0 = 0,

I r

( !.!:_b ) .

Fig. (7.91)

-

=a+b and is maximum.

(iv) When e = cos -1

( -alb),

r

= 0 i.e. the curve passes through the

r

is n egative and at 0 = n,

r

It may be noted that to plot a point

(

origin.

[,'

(v) When cos

L• :

3

/

o

/

< n,

=a-- b. -

r,

0), we

rotate the radius vector through 0 and

r in

opposite direction in this position. For example, to get

3, 45°), we rotate the radius vector from OX through 45° into the position OL. We have to measure

the point(-

3 P

-I[-%]< 0

along OL a distance- 3 i.e. we have to measure a distance

FIg. (7 92) •

3 not along OL, but in opposite direction. Producing LO toP so that OP = 3 units we get the required

(vi) The following

gives

some

n/2

cos -1 (-alb)

a

0

point P.

of rand 0 cos

-

1 (-a/b) < e < 1t

negative

Note that we get a loop inside another toop as shown

1t

a-b

in the figure.

Engineering Mathematics

-

II

Tracing of Corves

(7- 23)

Ex. : Trace the curve r = 2 + 3cos 8 (S.U. 2007)

Y

Sol. : Following the above line we see that the curve

r =

2 + 3cos

e can be

shown as on the right. y

(d) Pascal's Limacon is also given by r = a + b sin 8 where the y-axis becomes the initial line. For example, the curve

r =

2

- 3sin 8 is

shown in the neighbouring figure. 5

Fig. (7 .94)

Ex. 3 : Bernoulli's Lemniscate ,2

=

a2 cos 28 or (x2 + l> 2 a2 (� -l) =

(S.U.1995,97,98,2003) 9

=

y

37tl4

9

=

Sol.: (i) Since on changing r by- r the equation

7tl4

remains unchanged, the curve is symmetrical about the initial line. (ii) Since on changing 8 by

-

8 the

equation remains unc hanged the curve is symmetrical about the pole. (iii) Considering only positive values of

Fig. (7.95) r

we get the following table.

8

0

n/6

n/4

1t/4

c

are real ellipses.

Sections of this hyperboloid by planes parallel to the zx-plane i.e. y =

k are

z

hyperbolas.

7. Paraboloids : We

Fig. (7.134)

shall discuss here only one

elliptic paraboloid given by 2 x 2z -+-==2 a b2 c

l

X

Fig. (7 .135)

Engineering Mathematics

-

II

Tracing of Corves

(7- 35)

Sections 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

7

+y

2

2 =b ,

y- =X + 6,

3.

X

2

X

2

y

2+ 2 a

1

b

=y

2 y=x , y=1,X =1,X =2

4.

2 y=x , y=x+2,x=-1,x=2

5. 6

2

X

2.

2 2 x + y = ax, ax =by

•

2 x +i=by,ax=by

7.

X

8.

2

2 2 2 - y =1,X - y =2,X)' =4,xy=2

xy= 2 - y

9.

10. y=x2 -3x, y=2x

11. i=-4 (x-1), l =-2 (x-2) 12. a2 x2 =4l (a2 13.

r

-l)

or x=asin 2t, y=asin t

2 (cos 8 +sin 8) =2asin 8 cos 8 or (x+ y) (x +

2 14. x4 + y4 =2a xy or

r

2

2 15 x6 + y6 =a2x y or •

2

r

= 2

2

l) =2axy

2a sin8 cos8

sin 4 8+cos 4 8 a

2

·

78 . 2 8 cossm

6 6 sin 8+cos 8

16. x4-2.tya2+a2l = 0 or ? =2atan 8 sec2 8 - a 2 tan 2 8 sec2 8 17. y=x2 -6x+3, y=2x-9

18. y 3x2 - x-3, y=- 2x2 +4x+ 7 =

19.

l =4x,

y=2x- 4

21.

A cylindrical hole in a sphere.

20. y=4x-x2, y=x 22. 23.

2 2 A cone and paraboloid given b y z =x +land z =x2 + A paraboloid� +

l =az,and the c ylinder� +l =a2

24. The paraboloids z 25. y=4x(1 -x)

2 2 4 -x - Y , 4

z

=

3x2 +

y2 4

l

Engineering Mathematics

[Ans. : (1)

-

II

Tracing of Corves

(7- 36)

(2)

y

Y

x2+y2=b2

X

Fig. (7.137)

x2+y2=2

Fig. (7.136) (4)

y

(3)

y=1 X

( 6, 0)

x=1 x=2

Fig. (7.139) Fig. (7.138) (5)

y

(6)

y

X X

Fig. (7.140) Fig. (7.141) (7)

y

(8)

y

X X

Fig. (7.142) Fig. (7.143)

Engineering Mathematics

(9)

Tracing of Corves

(7- 37)

II

-

(10)

y

X

Fig. (7.145)

Fig. (7.144)

(12)

(11)

(2, 0)

X X

Fig. (7.146)

(13)

Fig. (7.147) (14)

y

y

X

X

Fig. (7.148) Fig. (7.149) (15)

(16)

y

y

X

X

Fig. (7.150)

Fig. (7.151)

Engineering Mathematics

II

Tracing of Corves

{7- 38) (18)

y

(6, 3) (-1, 1)

:Fig. (7.152)

Fig. (7 .153)

y

(19)

(20)

(2, 7)

(-1, 1)

Fig. (7.154) (22)

(21)

Fig. (7 .136)

Yt

X

Fig. (7 .156) (23)

y

Fig. (7.158)

:Fig. (7 .157) (24)

Fig. (7 .159)

Engineering Mathematics

(25)

-

II

(7- 39)

Tracing of Corves

y

Fig.

(7.160) •••