0107 - Math 31A 2010.01.07 MATH 31A DISCUSSION JED YANG 1....

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Unformatted text preview: Math 31A 2010.01.07 MATH 31A DISCUSSION JED YANG 1. Introduction Lecture 1 Instructor: Steve Butler. Location: HAINES 39. Sections 1A and 1B Email: mailto:jedyang@ucla.edu . Office: MS 6617A. Office hours: R 12:3013:30. Discussion Location: MS 5117 (T) and 5138 (R). Website: http://www.math.ucla.edu/ ~ jedyang/31a.1.10w/ . SMC: Jan. 6Mar. 11, MR 09:0015:00, MS 3974, T 12:0013:00. 2. Administration HW due Fridays in lecture, can turn in early to me, and I will hand back in section. Textbook: Rogawski, Single Variable Calculus, 2008. Confirm office hour. 3. Precalculus Review 3.1. Exercise 1.2.21. Find the equation of the perpendicular bisector of the seg- ment joining (1 , 2) and (5 , 4). Solution. Slope of segment is m 1 = 4- 2 5- 1 = 1 2 . Slope of perpendicular bisector is m 2 = 1 /m 1 = 2. Mid point is ( 1+5 2 , 2+4 2 ). So the equation can be written as y 3 = 2( x 3). square 3.2. Exercise 1.2.23. Find the equation of the line with x-intercept x = 4 and y-intercept y = 3. Solution. Equation of the line is y = mx + b , where b is the y-intercept, hence b = 3. The x-intercept x = 4 will yield y = 0 (by definition), so substituting, we may solve for m . We get 0 = 4 m + 3, hence m = 3 4 . So the equation can be written as y = 3 4 x + 3. square 3.3. Exercise 1.2.24. A line of slope m = 2 passes through (1 , 4). Find y such that (3 , y ) lies on the line. Math 31A Yang 2 Solution. One way is to write down an equation of the line in point-slope form: y = 2( x 1) + 4. Then we see clearly that if x = 3, then y = 8. Alternatively, the slope m is the change of y over the change of x . Symbolically,....
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0107 - Math 31A 2010.01.07 MATH 31A DISCUSSION JED YANG 1....

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