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by: Kavon Feest


Kavon Feest

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Class Notes
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This 4 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 016B at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 65 views. For similar materials see /class/226603/math-016b-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15
Math 16B 7 F05 Supplementary Notes 2 SecondDerivative Test To understand what is behind the second derivative test for functions of two variables we shall start by looking at the simplest nontrivial example that of a polynomial of degree 2 First the test will be stated Let the function fy have a critical point at ab The second derivative test involves the function 2 62f 62f 62f D 7 7 7 my My and it applies when Dfab 31 0 Note that if Dfab gt 0 then T ab and 37 ab must have the same sign The test distinguishes three cases l lfffa b gt 0 and ab gt 0 equivalently a b gt 0 then a b is a relative minimum 0 ll lfffab gt 0 and a b lt 0 equivalently 3a b lt 0 then a b is a relative maximum 0 III lf Dfab lt 0 then ab is a saddle point of f neither a relative maximum nor a relative minimum Now we look at the simple example fy 0amp2 QBxquyyz where 0467 are constants not all 0 This quadratic polynomial has a critical point at 00 where it takes the value 0 We have if 7 if 7 3x2 7 a7 3y 7 Y7 3ny DAM 4CW 7 62 The second derivative test can be derived for this function f by means of elementary algebra To illustrate the case a gt 0 will be discussed the other cases are similar except for the case a y 0 31 6 which is simpler in fact trivial Assuming a gt 0 we use the method of completing the square Namely we look at a22 xy the rst two terms in the expression for f and determine what expression of the form 6y2 we can add to it to produce a perfect square After a little thought one discovers that 6 6204 works 267 62 B 2 azz 2 y Eyz xEx g We can thus rewrite f as 2 7 2 lt1 my m Now lets examine what happens in case Df0 0 4a y 7 62 is positive negative or zero i If Df00 gt 0 then the coef cient multiplying y2 in 1 is positive and f is everywhere positive except at 00 which is thus a relative minimum in agreement with the test ii If Df 07 0 lt 0 then the coef cient multiplying y2 in 1 is negative On the y axis the function f is positive except at 007 but on the line x y 0 it is negative except at 07 0 The critical point 07 0 is a saddle point again7 in agreement with the test iii If Df07 0 0 then the yZ term in 1 drops out The function f is nonnegative and vanishes everywhere on the line xEm y 07 every point of which is thus a relative minimum This case is not covered by the test The preceding discussion illustrates the fact that for the function fy 0amp2 2 y yyz one can derive the second derivative test by purely algebraic means It will now be indicated how the general case can be deduced from this special one by an approximation argument We return to the general function fzy The local linear approximation of f near a point 17 b was discussed in Supplementary Notes 1 The approximation is expressed by 6 6 lt2 mm M b was 7 a View 7 b The expression 2 is shorthand for a more precise statement that indicates how the error in the approximation behaves as my approaches Lab The approximation 2 is called a rst order approximation because it is an approximation of f by a rst degree polynomial There are analogous7 more accurate7 higher order approximations The second order approximation7 which approximates f near 17 b by a second degree polynomial7 reads N 5f 5f lt3 mm flta b a a was a 6y abgtlty b 162 2 92f 162 2 fi hbx a m hbx My b ify ml W b 7 and there is an associated error estimate Now suppose cab is a critical point of f7 and let 162 1 92f 162 04 g wvl 5 6z6ya7 7 Y gig2W Then 3 reduces to 4 f967y fa7b M96 i 602 2595 i 0M i b 71 Z02 The second degree polynomial on the right side here is a trivially modi ed version of the one discussed earlier it has a critical point at a7b instead of at 07 07 and it may not vanish at the critical point7 but those changes are merely super cial One can verify the second derivative test for the polynomial on the right side of 4 by purely algebraic means Finally7 using the error estimate that accompanies 47 one can show that7 as long as Dfab 4a y 7 62 is not zero7 the nature of 17 b as a critical point of f is controlled by the nature of cab as a critical point of the polynomial on the right side of This is how the second derivative test is established Math 16B 7 F05 Supplementary Notes 4 The Derivatives of sint and cost One can derive the formulas d d 1 ampsin t cos t Ecos t isint starting from the relations sint lt2 15 T 1 t i 1 3 lim LS 0 taO t Note that 2 and 3 just say that 1 holds at the origin since cos 0 1 and sinO 0 Taking 2 and 3 temporarily for granted let7s derive By de nition of the derivative d i t h 7 i t Wm 53 We use the addition formula sint h sintcos h cos tsin h to rewrite this as 071 t cos171 tsinh dt s1n ihlir s1n 7h cos h By 2 and 3 the limit on the right side equals cos t which establishes the rst formula in The second formula in 1 can be deduced from the rst one by means of the identities 1 7T 1 7T costs1nltt7gt s1nt7coslttigt 2 2 and the chain rule We have 003 ew Coslttgtlttgt 7139 r cos ltt ismt So to establish 1 it only remains to establish 2 and Once 2 is known 3 follows easily In fact cost71 i cost71cost1 cos2t71 t tcos t 1 T tcost 1 isinzt sint 1 7s1nt 7 7 tcost 1 It cost 1 As It tends to 0 the rst factor on the right side tends to 0 since sin 0 0 and the last factor tends to since cosO 1 By 2 the middle factor tends to 1 so the product tends to 01 0 which gives The relation 2 is thus the basic one We7ll derive it using some simple geometry Let t be a sint small positive angle Since T is an even function of t it suf ces to establish 2 as It tends to 0 through positive values From the point A cos tsin t on the unit circle we construct the tangent line to the circle7 and we let B denote the point where the tangent line intersects the s axis see Figure 41 The origin will be denoted by O The distance of the point A from the s axis is sin t which is less than the length of the arc of the unit circle subtended by the angle t The preceding arc has length It by the de nition of radians7 so we have the inequality sint lt t which we can write as lt 1 To obtain a lower bound for 51 we consider the right triangle OAB The side adjacent to the angle t has length 17 so the side opposite the angle t has length tan t The area of the triangle is therefore 1tan t The triangle contains the sector of the unit circle cut off by the angle It so its area is larger than the area of the sector The area of the sector equals the area of the whole circle7 which is 7T times 2 the ratio of t to the length of the full circle The area of the sector is thus giVing us the inequality gt g which we can rewrite as S tit gt cost We now have the pair of inequalities sint cost lt T lt1 Since lirntnocos t 17 the relation 2 follows


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