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- Applied Mathematics-1 - M-1
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Disclaimer: The lecture notes have been prepared by referring to many books and notes prepared by the teachers. This document does not claim any originality and cannot be used as a substitute for prescribed textbooks. The information presented here is merely a collection of materials by the author of the subject. This is just an additional tool for the teaching-learning process. The teachers, who teach in the class room, generally prepare lecture notes to give direction to the class. These notes are just a digital format of the same. These notes do not claim to be original and cannot be taken as a text book. These notes have been prepared to help the students of BPUT in their preparation for the examination. This is going to give them a broad idea about the curriculum.

3 Module -1 ASYMPTOTES DEFINITION A straight line is said to be an Asymptote of an infinite branch of a curve if as point P recedes to infinity along the branch the perpendicular distance of P the straight line tends to zero. Y Example: xy = 1 We can write either y 1 1 or x x y P when x 0 , y or y 0, x O M METHODS FOR FINDING ASYMPTOTES X Ist Method Let y = mx + c be an equation of a straight line which is not parallel to y-axis. If p = PM be the perpendicular distance of any point P(x,y) on the infinite branch of a given curve from the line y = mx + c, we have p y mx c Y 1 m2 p 0 as x lim p lim x y mx c x 0 1 m2 1 1 m2 P lim y mx c x lim y mx c 0 x lim y mx c 0 x c lim y mx x O M X

y c m x x Taking the limit as x , we get Also y = mx + c or lim x y c lim m m 0 x x x Therefore m xlim y y mx c and lim x x Working Rule y 1. Divide the equation by the highest power ‘x’ and write the equation in the form x 2. Taking the limit as x , it implies m xlim y x 3. Write the equation in the form (y – mx) and taking the limit x y mx c 4. Put lim x 5. Put the values of m and c in the equation y = mx + c to get the required asymptotes. Example: x 3 y 3 3axy 0 Divide term x3 i.e. highest power, we get 3 y y 1 1 3a 0 x x x Taking x , we get 2 1 m3 0 1 m 1 m m 0 m 1 Other values are imaginary, Next, we have to calculate c lim y mx , c lim y x x x 3 3 So, we write x y 3axy 0 xy 3axy x xy y 2 lim x y lim 3axy x xy y 2 x x 2 2 y 3a 3a 1 x c lim a 2 2 x 1 y y 1 1 1 x x

The required asymptote is y = mx + c Putting the value of m and c y = –x –a, y+x+a=0 2nd Method To determine the asymptotes of the general algebric equation of nth degree. Let the equation of the curve be An + An–1 + An–2 + ....+A2+A1+A0 = 0 ......(i) where Ar is a homogeneous expression of degree r, in x and y. y r Here A r x r x y y where r is a polynomial in of degree ‘r’. The above equation can written as x x y y y y y x n n x n 1n 1 x n 2 n 2 ...... x1 0 0 x x x x x .......(ii) Dividing equation (ii) by xn, we get 1 y 1 y y 1 y y 1 n n 1 2 n 2 .... n 1 1 n 0 0 x x x x x x x x x Taking limit x , we get .....(iii) y 1 1 1 n m =0 Sincelim m and lim 0, lim 2 0.....lim n 0 x x x x x x x x Again we have y = mx + c c y m x x Putting the value of y in equation (ii), we get x c c x n n m x n 1n 1 m ....... 0 x x Expanding each term by Taylor’s Theorem, we have c 1 c2 c x n n m n m m ...... x n 1 n 1 m n 1 m ..... 2 x 2x x x n 2 n 2 m ...... 0 .....(5)

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