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# Note for Applied Mathematics-1 - M-1 By Bandana Panda

• Applied Mathematics-1 - M-1
• Note
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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.

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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

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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 xy 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  

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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 1n 1    x n  2 n  2    ......  x1    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 1n 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)