lecture10compl-09[1] - Math 18.02(Spring 2009 Lecture 10...

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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 10 Second derivative test. Global max/min: Boundaries and infinity. Level curves and lever surfaces February 26 Last time: Max/Min problems. Least squares approximation. Today: Second derivative test. Global max/ min: Boundaries and infinity. Level curves and level surfaces. Reading Material: From Simmos 19.1, 19.7. From the Lecture Notes SD. 2 Second derivatives Given a function z = f(:L', y) we denoted the first partial derivatives of f as _5f _3f fw—é—fl: and fyqa—y. Then it is simple to define the second partial derivatives: 6(%)_82_f f” = 2‘“ derivative w.r.t. :r; = 89: — 3:62 fmy = derivative W.r.t. y of derivative w.r.t- :3 = (fmly 5 _ 6‘ (5i) _ 62f _ 8y _ 83; rise a 9i 2 fyy : 2nd derivative w.r.t. y = (6y) 3 f 53; = 53;? Remark. Theie is a very surprising theorem: if all second derivatives are continuous then fay = fox: and this in fact is going to be the case for all functions we will be dealing with in this class! 3 211d derivative test Last time we learned that for a function z = f (a, y) local max/min occur on critical pints, that is on points (:30, yo) such that { fm(~’3oay0) = 0 fyll‘oryo) = 0 We also learned that not all critical points are either a local max or min, take for example the point (0, 0) critical point for the function 2 = 3:2 -— y2 which is a saddle point. (0,0,0) 4.5 >>< Question: Is there a way to recognize if a critical point is a max a min or a saddle? The answer to this question is a “partial” YES. We first have to define the Discriminant D: Definition 1. Given a function z = f(a:,y) we define D = fmwfyy _ (fmy)2 fan: fay fym fyy note fly = fym. We have the following result: Theorem 1. Assume that (30,310) is a critical point for the function 2: = f(a:,y). Let D be the discriminant above evaluated at ($0,310). Then I. if D < 0 —> the critical point is a saddle 2. ifD > 0 (note fm = 0 impossible) (a) and fan: > 0 —> the critical point is a min 3. if D > O (a) and fm < 0 —> the critical point is a max 4. if]? = 0 no info Exercise 1. Consider first the function f (3;, y) = 2:2 — y2. We already know that (0,0) is a critical point. We have: fa, = 2:6, fy = —2y and hence fwm =2 for = _2 fym=0 which implies D = —4 < 0 and (0,0) is a saddle point: A 2- 7 Sim/0M (012% ¢ 9 K On the other hand if f(rs, y) = 1:2 + y2 7 > :5 d X Wit/1 POM/«f we still get the same critical point (0,0), but in this case fmm=2 foo: D=2(2)—02=4, so mamormin but fm=2>050 min. We finally consider the function f(:c,y) = .124 + y4 > R» s X AIM/M POI” It is easy to check that again (0, 0) is a critical point, but now 3‘” = 1232, fyy = 123,12 and fmy = 0 and at the point (0,0) 19:0, so there is no conclusion that we can make, We need to make a sort of "manual” inspection: since for any point P = (e, 6) # (0,0) near zero we have f(e,5) = 54 +64 > 0 = f(0,0) it follows that (0,0) is a local min point. Remark. A proof of the Second Derivative Test is rather involved. Bat there are two fundamental steps in it: the quadratic approximation of a function near a point (otherwise called the Taylor polynomial of degree 2) and the Second Derivative Test for quadratic functions. These two facts are stated in the following lemmas: Lemma 2. [Quadratic Approximation of a Function of two variables] Given a function z = f (cc, y) and a point P0 = (530,110) we have the following second order appromimation for f near P0: f($1y) N flifloiyo) + fz($01?}0){$ — 330) + fy($oayo)(y — so) + glam-(930, yollm — $0)2 + éfyy(30:90)(y — yo)2 + fym($0:yo)(y - yo)($ — $0)- Lemma 3. [Second Derivative Test in a Special Case] Consider the quadratic function 1 w(.v, y) = we + 5(14332 + 23:31; + CyZ). Clearly O = (0,0) is a critical point. Then AC — 32 > O,A > 0 or C > 0 =:- (0,0) is a minimum point AC —B2 > [LA < 0 or C < 0 => (0,0) is a mamimnm point AC — B2 < 0 => (0,0) is a saddle point. For the proof of this lemma see SD in Lecture Notes. Remark. Notice that in Lemma 5’ A = wm(0,0),B : wmy(0,0), C = wyy(0,0) and hence D = A0 —Bz. To prove the Second Derivative Test for arbitrary functions one first nses Lemma 2 to approeimate the function f(rc,y) near the critical point by a quadratic function. Then one uses Lemma 3. 4 Global max/ min: Boundaries and Infinity In this section we pass from considering local max/ min to global ones. Let’s start with the following: Exercise 2. Consider the function flay) = 3:2 + :1;2 defined on the domain W = flaw/43:2 +3;2 5 1}. A i Compute global mas: and min of f on this domain. Solution: We first observe that if the global max and min occur inside W then it must be also a local max and min. We know from the previous example that there is no local max for f, but there is a local min at (O, O). This is also the global minimum since for any other point P = (mmyg) one has flmoayol > 0- But what about the global max? For this we have to look at what happens to f at the boundary (perimeter) of W, also denoted with EWA We have to evaluate f on this boundary and decide if f has a max on it. Since the function and the domain W are symmetric with respect to the .r—axis we will only check the values of f on the upper boundary of W, let’s denote it 641/17: .L It is easy to parametrize 8+W by Z M— F(x) 2 all + 1 — 43325. The value of f on B+W is given by fa, \/1 — 49:2 = m2 + 1 — 4.1-2 = 1 — 39:2 = [1(3), for a: e [—1/2, 1/2]. New Man) is a function of one variable and it is not difficult to show that on its domain its max occurs at :I: = O, which translate into the fact that f has two max on the boundary of W, more precisely at 13+ = (0,1) and P. = (01 —1)‘ We also have that f(P+l = f(P—) = 1, and this is the global max of f on W. Question: What is the global max of f when we allow it to be defined on the whole plane? Since there is no boundary we can evaluate f on points P = (at, 3;) “going to infinity“, that is such that lm| -> 00 or |y| —> 00. Clearly on these points the value of f = 3:2 + y2 will tend to infinity as well, so the function does not admit a global max. 5 Level curves and level surfaces You already have seen an example of level curves: any topographic map is made by drawing several level curves relative to the 3D landscape that it represents. In fact if Ma, y) is the function that represents the altitude at point P = (:L', y), then the graph of this function is the landscape and the topographic map is made by representing the (level ) curves that are obtained by considering all the points at a certain altitude. We have the following definition: Definition 2. Given a two amiable function 2 = f (33,14), for any scalar C we call the level curve ”yo off the set 'YC = {Carl/flay) = C}- These level curves are represented on the nay—plane and their collection is called contour map: Exercise 3. Consider the function f (313,31) 2 y — :32. Draw the contour map of this function. bvxz: C, We have a similar definition for functions of three variables: Definition 3. Given a three variable function it) = f(:1:,y,z), for any scalar C we call the level surface Sc of f the set SC={(£E,T,Z)/f($,y,2)=0}- These level surfaces are represented on the {eye—space and their collection is also called contour map. Exercise 4. Consider the function f(.’£,'l ,z) = 4.132 + y2 + 922. Draw the contour map of this function. 4 XZ‘FJP‘l 7&2: C lg? OCCO(C: 30a Lula 3W Coma/«trout > ”&%35‘\oz gee-pose. Study Guide 1. Think about the following questions: In Why in Theorem I I said that ifD > 0 than it must be f” 7% D? ...
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