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 30 13 30 Discussion Location MS 5117 T and 5138 R Website http www math ucla edu jedyang 31a 1 10w SMC Jan 6 Mar 11 M R 09 00 15 00 MS 3974 T 12 00 13 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 segment joining 1 2 and 5 4 1 Solution Slope of segment is m1 4 2 5 1 2 Slope of perpendicular bisector is 2 4 m2 1 m1 2 Mid point is 1 5 2 2 So the equation can be written as y 3 2 x 3 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 4m 3 hence m 43 So the equation can be written as y 34 x 3 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 y the slope m is the change of y over the change of x Symbolically m x or y m x This concept will be useful later when we deal with differentials dy m dx Since the change in x is x 3 1 2 we get that the change in y is y y 4 2 2 4 hence y 8 This method seems longer but conceptually it is easier to do in one s head and will lead to intution for calculus later 3 4 Exercise 1 4 55 Use the addition formulae for sine and cosine to prove tan a b cot a b tan a tan b 1 tan a tan b cot a cot b 1 cot b cot a 1 2 Proof Recall that sin a b cos a b sin a cos b cos a sin b cos a cos b sin a sin b 3 4 Now tan a b sin a b cos a b sin a cos b cos a sin b cos a cos b sin a sin b sin b sin a cos a cos b sin a sin b 1 cos a cos b tan a tan b 1 tan a tan b 5 6 7 8 where we get from 6 to 7 by dividing top and bottom by cos a cos b x The case for cotangent is completely analogous Remember cot x cos sin x and that sin b sin b and cos b cos b Work out the details and convince yourself 3 5 Exercise 1 4 56 Let be the angle between the line y mx b and the x axis Prove that m tan Proof This is trivial 3 6 Exercise 1 4 57 Let L1 and L2 be the lines of slope m1 and m2 respectively 2 m1 1 Show that the angle between L1 and L2 satisfies cot m m2 m1 Proof This is immediate by using Exercises 55 and 56 3 7 Exercise 1 4 58 Perpendicular Lines Use Exercise 57 to prove that two lines with nonzero slopes m1 and m2 are perpendicular if and only if m2 1 m1 Proof What is cot 2 Math 31A Yang 3 3 8 Exercise 1 4 59 Apply the double angle formula to prove p a cos 8 21 2 2 q p b cos 16 21 2 2 2 and of cos 2 n for all n Guess the values of cos 32 Proof Recall cos2 t 1 cos 2t For the general case let a0 0 and define in2 ductively an 2 an 1 We claim that for n 1 we have cos 2 n 12 an 1 The base case is trivial By induction assume cos 2 n 12 an 1 By the half angle q q 12 1 21 an 1 14 2 an 1 12 an formula we get cos 2n 1 4 Basic Limits 4 1 Basic Limit Laws Assume that limx c f x and limx c g x exist Then a Sum Law lim f x g x lim f x lim g x x c x c x c b Constant Multiple Law For any number k R lim kf x k lim f x x c x c c Product Law lim f x g x lim f x lim g x x c x c x c d Quotient Law If limx c g x 6 0 then lim x c f x limx c f x g x limx c g x 4 2 Exercise 2 3 22 Evaluate the limit limz 1 z 1 z z 1 Solution Recall that limz 1 z 1 and limz 1 1 1 By the Quotient Law 1 1 z 1 1 z limz 1 z 1 limz 1 z 1 lim limz 1 z 1 1 By the Sum Law limz 1 z limz 1 z 1 1 2 By the Sum Law limz 1 z 1 2 So by the Quotient Law 1 1 z z 2 z 1 z limz 1 z z 1 lim limz 1 z 1 2 1 4 3 Exercise 2 3 29 Can the Quotient Law be applied to evaluate limx 0 sin x x Solution The Quotient Law requires the limit of the denominator namely limx 0 x to exist and be nonzero This is not the case so we cannot apply directly 4 4 Exercise 2 3 30 Show that the Product Law cannot be used to evaluate limx 2 x 2 tan x Solution The Product Law requires the limit of each factor to exist However limx 2 tan x does not exist 4 5 Exercise 2 3 31 Give an example where limx 0 f x g x exists but neither limx 0 f x nor limx 0 g x exists Math 31A Yang 4 Solution Let f x be any function defined on a neighborhood of 0 but not necessarily at 0 such that limx 0 f x does not exist e g f x 1 x Let g x f x Then of course limx 0 g x also does not exist otherwise by the Constant Multiple Law limx 0 f x also exists But notice f x g x is identicaly zero in a …