curve tracing
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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
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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.
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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.
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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
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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.
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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.
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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
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• 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
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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.
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2 2 ( )xy a a x
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2 2 3(2 )a x y a y
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3 3 3x y axy
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2 ( 1)( 2)x y y y
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2 2 2 2 2 2( ) ( )y a x x a x
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2
2
( 1)
( 1)
xy
x
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2 2 2 2 2( )x y a y x
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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.
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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 .
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• 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
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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.
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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.
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1 1 cosr a 1 1 sinr a
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n = – ½ Parabola
21 cos
ar
2
1 sina
r
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cos Circle touching axis at originr a Y
sin Circle touching axis at originr a X
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n = 2 Bernoulli’s Lemniscates
2 2 cos2r a 2 2 sin2r a
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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
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Type II Rose Curves
cos / sinr a n r a n
Type III
cosr a b
Type IV Spirals
, Equiangular Spiralmr ae
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cos 2r a
sin 2r a
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sin 5r a
cos5r a
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(1 cos )r a
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cos ,r a b a b
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cos ,r a b a b
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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.
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• 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.
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• Tangents
Find /
/
dy dy dt
dx dx dt
Form the tableof values , , ,dy
t x ydx
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3 3cos , sinx a y a
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( sin ), (1 cos )x a y a
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( sin ), (1 cos )x a y a
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( sin ), (1 cos )x a y a
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( sin ), (1 cos )x a y a
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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 .
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CYCLOID
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EPICYCLOID
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CARDIOID
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PASCAL’S LIMACON
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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.
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HYPOCYCLOID
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ASTROID
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