1224 Lecture 1

1224 Lecture 1 - MODULE 1 1. REVIEW OF SOME BASIC ALGEBRA...

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MODULE 1 1. REVIEW OF SOME BASIC ALGEBRA (1) Solving Equations You should be able to solve for x : a + 1 b = c a d + e x + c and get x = e ( ba +1) b ( c a ) d ( ba c Common mistakes and strategies: . a b + c 6 = a b + a c , but a + b c = a c + b c [You can’t separate a demominator, but you can divide a single denominator into each numerator term] . 1 y = a + 1 a + b implies y = 1 a + 1 a + b = a + b a ( a + b )+1 = a + b a 2 + ab +1 [The Frst step is by inverting both sides, being careful to invert each side as a whole; the second step is by multiplying numerator and denominator by ( a + b )] Note that combining di±erent algebraic fractions can usually only be done by multiplying numerator and denominator of each fraction in such a way as to create a common denominator: 1 a + 1 a + b + 1 c = 1 a · ( a + b ) c ( a + b ) c + 1 a + b · ac ac + 1 c · a ( a + b ) a ( a + b ) = ( a + b ) c + ac + a ( a + b ) a ( a + b ) c = 2 ac + bc + a 2 + ab a 2 c + abc Similarly, a fraction in the denominator can be simpliFed by multiplying entire numerator and entire de- nominator by common denominator of the fractions in the denominator: a + c 1 a + 1 a + b = a + c 1 a + 1 a + b · a ( a + b ) a ( a + b ) = a ( a + b ) 1 a · a ( a + b )+ 1 a + b · a ( a + b ) = a 2 + ab ( a + b a = a 2 + ab 2 a + b (2) Quadratic equations The standard equations studied in high school algebra in two dimensions involving quadratic powers of the variables are the following. In these a constant may be written as a 2 or b 2 to indicate it is positive, since squares of real numbers are always positive. Constants written as a or b may be either positive or negative. . a 2 x 2 + a 2 y 2 = c 2 (circle) . a 2 x 2 + b 2 y 2 = c 2 (ellipse) . ½ a 2 x 2 b 2 y 2 = c 2 b 2 y 2 a 2 x 2 = c 2 (hyperbola) . ½ y = ax 2 + bx + c x = ay 2 + by + c (parabola) . xy = c (hyperbola)
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The scientist or engineer should be able to name and sketch these equations on sight. This is most efficiently done by knowing the possible shapes and then checking the easy cases x =0and y = 0. For example, consider the ellipse 9 x 2 + y 2 = 9. An ellipse is always “egg-shaped” but is it elongated along the x -axis or the y -axis? Consider x = 0. Then y 2 =9so y = ± 3, and the ellipse goes through the points ± 3on the y -axis. Consider y = 0. Then 9 x 2 =9 ,so x = ± 1 and the ellipse goes through the points ± 1onthe x -axis. Thus the ellipse is elongated along the y -axis. Its graph is given in plot A below. Note that x =0 corresponds to points on the y -axis and y = 0 corresponds to points on the x -axis. This is an observation which sometimes confuses students. Consider 6 x 2 y 2 = 1. It is clearly a hyperbola, but along which axis is it oriented? Consider x = 0. Then y 2 = 1 which is never true for any real number, so x = 0 can never satisfy the equation. This already tells you that no part of the y -axis is on the curve; ie., the hyperbola is oriented along the y -axis. Consider y =0. Then 6 x 2 = 1, so the hyperbola goes through the points ± 1 6 on the x -axis. The graph is given in plot B below. The graph of 6 y 2 x 2 = 1 may be sketched by similar reasoning, and is given in plot C below.
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1224 Lecture 1 - MODULE 1 1. REVIEW OF SOME BASIC ALGEBRA...

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