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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 — 4.12 = 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 geepose. 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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