curve tracing

04/20/23 Prepared by Dr. C. Mitra 2

Different forms of CURVES• CARTESIAN FORM:

• POLAR FORM:

• PARAMETRIC FORM:

• PEDAL FORM:

• INTRINSIC FORM:

( , ) 0f r

( , ) 0f x y

( )

( )

x t

y t

( , ) 0f r p

( , ) 0f s


SOAPSymmetry:• About X axis : Every where in the equation

powers of y are even.• About Y axis : Every where in the equation

powers of x are even.• In opposite quadrants or with respect to origin :

On replacing x by –x and y by -y in the equation of the curve, no change in the equation.

• About the line y = x : Replace x by y and y by x no change in the equation.


SOAPOrigin:

• If the independent constant term is absent in the equation of the curve then it passes through the origin. i.e. if x=0 and y=0 satisfies the equation of the curve then it passes through the origin.

• If the curve passes through the origin, then the Tangents to the curve at origin can be obtain by equating to zero lowest degree term of the equation.


Asymptote:

• A tangent to the curve at infinity is called its asymptote.

• Asymptote can be classified into two category

– Parallel to X-axis or parallel to Y-axis.– Oblique Asymptote: Neither Parallel to X-

axis nor parallel to Y-axis.

SOAP


1. Asymptote parallel to Y-axis

• Asymptote Parallel to Y-axis can be obtained by equating to zero the coefficient of highest degree term in y.

2. Asymptote parallel to X-axis

• Asymptote Parallel to X-axis can be obtained by equating to zero the coefficient of highest degree term in x.


3. Oblique Asymptote:

• If the Equation of the curve is in the implicit form i.e. f(x, y) = 0, then it may have oblique asymptote.

• Procedure: Let y = mx + c be the asymptote.

• Solving f(x, mx+c) = 0 and equating the co-efficients of highest and second highest power of x to 0 we get required values of m and c.


POINTS• Find X-Intercept Or Y-Intercept: • X-Intercept can be obtained by putting y = 0 in

the equation of curve. • Y-Intercept can be obtained by putting x = 0 in

the equation of curve.

• For the Curves ,

x

y

fdy

dx f

( , ) 0f x y

SOAP


• If =0, then tangent will be parallel to X-axis.

• If = , then tangent will be parallel to Y-axis.

• If > 0, in the Interval a < x < b, then y is increasing function in a < x < b.

• If < 0, in the Interval a < x < b, then y is decreasing function in a < x < b.

dy

dx


Region of Absence:

• If y2<0, for x < a (x=a is an X-intercept), then there is no curve to the left of the line x=a.

• If y2<0, for a < x < b (a and b are X-intercepts) then there is no curve between the lines x = a and x = b.

• If y2<0, for x > b ( x = b is an X-intercept) then there is no curve to the right of the line x=b.


2 2 ( )xy a a x


2 2 3(2 )a x y a y


3 3 3x y axy


2 ( 1)( 2)x y y y


2 2 2 2 2 2( ) ( )y a x x a x


2

2

( 1)

( 1)

xy

x


2 2 2 2 2( )x y a y x


Symmetry:: • About the initial line: Replace by no

change in the equation.• About Pole: Replace r by -r no change in the

equation.• About the line perpendicular to the initial line (Y

axis): Replace by and r by - r. OR replace by no change in the equation.


Pole :: If for certain value of , r=0 the curve will pass through the pole.

• Tangents at pole: Equate r=0 and solve for , lines =constant are tangents at the pole.

• Prepare table of values of r and .


• Angle between tangent and radius vector:: be angle between tangent and radius vector then ,

• tangent and radius vector coincides and

• tangent and radius vector are perpendicular.

tan

drrd

0

2


Region of absence:• If for < <, r2 <0, then no curve lie

between = , = .• |sin | 1, |cos | 1 .So for the curves r =

acos n , r = asin n ,r a. So, no curve lies outside the circle of radius r.

• In most of the polar equations, only functions sin , cos occurs and so values of between 0 to 2 should be considered. The remaining values of , gives no new branch of the curve.


Polar Curves Type I

cos / sin ( 1, 1 / 2, 1 / 2,2)n n n nr a n r a n n

Sine curve can be obtained from corresponding cosine by rotating the plane through an angle /2n.


1 1 cosr a 1 1 sinr a


n = – ½ Parabola

21 cos

ar

2

1 sina

r


cos Circle touching axis at originr a Y

sin Circle touching axis at originr a X


n = 2 Bernoulli’s Lemniscates

2 2 cos2r a 2 2 sin2r a


n = 1/2 Cardiode

1/2 1/2 cos , i.e. (1 cos )2 2

ar a r

(1 cos )2a

r

(1 sin )2a

r

(1 sin )2a

r


Type II Rose Curves

cos / sinr a n r a n

Type III

cosr a b

Type IV Spirals

, Equiangular Spiralmr ae


cos 2r a

sin 2r a


sin 5r a

cos5r a


(1 cos )r a


cos ,r a b a b


Symmetry :: • About X axis – Replace t by –t , x remain

unchanged and y changes the sign then curve is symmetric about X- axis.

• About Y axis - Replace t by –t , x changes the sign and y remain unchanged then curve is symmetric about Y- axis.

• Note that :: For trigonometric equation replacing t by , y remains unchanged and x changes the sign curve will be symmetric about Y- axis.

• About Origin - Replace t by –t both x and y changes the sign curve is symmetric about origin.


• Points of Intersections ::– If for some value of t, both x and y are

zero curve passes through origin.– Find X and Y intercepts.

• Find asymptotes if any.

• Find region of absence.


• Tangents

Find /

/

dy dy dt

dx dx dt

Form the tableof values , , ,dy

t x ydx


3 3cos , sinx a y a


( sin ), (1 cos )x a y a


WITCHES OF AGNESI

The curve is obtained by drawing a line OB from the origin through the circle of radius a and center (0,a), then picking the point with the y coordinate of the intersection with the circle and the x coordinate of the intersection of the extension of line OB with the line y=2a .


CYCLOID


EPICYCLOID


CARDIOID


PASCAL’S LIMACON


PASCAL’S LIMACON

Let P be a point and C be a circle whose center is not P. Then the envelope of those circles whose center lies on C and that pass through P is a limaçon.


HYPOCYCLOID


ASTROID

curve tracing

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