CURVE TRACING
Different forms of CURVES CARTESIAN FORM: POLAR FORM: PARAMETRIC FORM: PEDAL FORM: INTRINSIC FORM: 1/18/2019 Prepared by Dr. C. Mitra
CARTESIAN CURVES 1/18/2019 Prepared by Dr. C. Mitra
SOAP Symmetry: 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. 1/18/2019 Prepared by Dr. C. Mitra
SOAP Origin: 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. 1/18/2019 Prepared by Dr. C. Mitra
SOAP 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. 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
Procedure: Let y = mx + c be the asymptote. 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. 1/18/2019 Prepared by Dr. C. Mitra
SOAP 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 , 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
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POLAR CURVES 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
Prepare table of values of r and . 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 . 1/18/2019 Prepared by Dr. C. Mitra
tangent and radius vector coincides 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. 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
Polar Curves Type I Sine curve can be obtained from corresponding cosine by rotating the plane through an angle /2n. 1/18/2019 Prepared by Dr. C. Mitra 24
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n = – ½ Parabola 1/18/2019 Prepared by Dr. C. Mitra
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n = 2 Bernoulli’s Lemniscates 1/18/2019 Prepared by Dr. C. Mitra
n = 1/2 Cardiode 1/18/2019 Prepared by Dr. C. Mitra
Type II Rose Curves Type III Type IV Spirals 1/18/2019 Prepared by Dr. C. Mitra
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PARAMETRIC CURVES 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
Tangents Find 1/18/2019 Prepared by Dr. C. Mitra
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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 . 1/18/2019 Prepared by Dr. C. Mitra
CYCLOID 1/18/2019 Prepared by Dr. C. Mitra
EPICYCLOID 1/18/2019 Prepared by Dr. C. Mitra
CARDIOID 1/18/2019 Prepared by Dr. C. Mitra
PASCAL’S LIMACON 1/18/2019 Prepared by Dr. C. Mitra
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. 1/18/2019 Prepared by Dr. C. Mitra
HYPOCYCLOID 1/18/2019 Prepared by Dr. C. Mitra
ASTROID 1/18/2019 Prepared by Dr. C. Mitra
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THANK YOU 1/18/2019 Prepared by Dr. C. Mitra