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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 11
Summary of Global inax/ min: Boundaries and inﬁnity. Chain
Rule. Differential. Vector Fields February 27 Last time: Second derivative test. Global max/min: Boundaries and inﬁnity. Level curves and
level surfaces. Today: Summary of Global max/min: Boundaries and inﬁnity. Chain Rule. Differential. Vector
Fields. Reading Material: From Simmos 19.4 and 19.6. 2 Summary of Global 111ax/ min: Boundaries and inﬁnity Last time we considered the following exercise: Exercise 1. Consider the function f(:r,y) = 2:2 + y2 deﬁned on the domain
W I {(1‘,g)/’1:r2 + y2 51}. Compute global mum and min of f on this domain. I would like to summarize the method we used to ﬁnd the absolute {or global) max/min in this
example by keeping the situation pretty general: Assume a function e = (rt, 3;) is given and with it a domain W with a boundary 8er which can be
represented as a closed curve not intersecting itself. The absolute (or global) max/min are obtained
in the following way: c Find all critical points strictly inside W. Analyze them using the second derivative test and
decide which ones are local max and which ones are local min. Assume that P1, ...,P” are local
min and Q1, ..., Q,” are local max. 0 Now it is time to analyze what goes on on the bundary. Find parametric equations {or it
F05) =.1:(t)i +y(t)j', For t in [cub].
0 Evaluate the function f(:1;,y) on the boundary by deﬁning Mt) = flTltlsllll Now this is a function of one variable and you can study it to ﬁnd global max and min in the
interval [n,b]. Suppose that the global max is at $1 and the global min at t2. This means that A”! = f(I'(£;),y{t1)) : global max of f on 6147 and
TH = f(.i'(t2), y{t2)) = global min of f on 3W I we are now ready to summarize: — the global max of f on W is the largest value among f(Q1),...,f(Qm) and M.  the global min of f on ‘W is the smallest value among
f<Pl):1f(Pn) and in. Remark. If the domain is not enclosed in a closed curve, meaning there are some parts of it going
to 00, then the problem is more triehy. You can proceed as above but you still need to understand if in that part of the domain that goes to inﬁnity the function can become inﬁnitely large in the positive
or negative sense. Exercise 2. Consider the function f(a:,y) = .132 + y2. Find, if it eaists, the global mare and min of
f over the domain W = {(n We 2 y} Since the only: critical pointwis (0, 0) there are no critical points in the region where a: > y. On the
boundary of W', denoted ("ﬂ/V, that is where a: = y, we certainly have the minimum at (0,0). To check for a possible maa‘ we deﬁne the function. h(a‘) = f(:e,a:), the restriction. off on 3W5 Then
we observe that hfa‘) : 23:2 has not may; since lim h(a:) : +00.
:cHoo To conclude on Pl” the function f (:3, y) 2 3:2 + y2 has minimum value 0 and no mazimum value. 3 Chain Rule Consider the two one variable functions y = flu) and a = a(a:). Since it chages with a: and y chages
with u then it makes sense to ask for the change of y with respect. to 3;. More precisely we want @_ﬁ
a'atwdas' Two methods: 1. Compose functions: Let [1(a) = f(a($)). Then we Compute a _ a
da‘ — dre.
2. Chain Rule: d df cl
1’ _ _ _“
da _ du d1" Often (2) is really the only option we have, like for example when f(u) = sin u and u(a:) 2 2:2. [\3 What happens if we consider multivariable functions? Let us take the functions u. : u(t) z = “1"”) v = v(t) If we then change it clearly u(t) and DOE) change, but also f (u, 1;) does. So it makes sense to compute
iii—ft. We in fact have: Chain Rule:
of: _ Lil—j _ 8f du 8f do aane+na dz) (W) (“i (W) (”i
(“it (an) 6“ (“(Inlmﬂtﬂ) dt (to) 5"“ (uoumonn 0” (in) Exercise 3. Consider 2 : e‘t‘a‘yn and :c 2t and y = t — 1. We want % at t = 1. Solution: I’Ve ﬁrst observe that 3(1) = 1 and y(1) = U, then by the Chain Rule we have
dz 82 d3: 6: dy a aa+ﬁa
2_ 2 _ s_‘ 2
(—2me‘9‘ 1" )(1'0)(1)+(—2ye m y ){1,o)(0)
\__\,_—, \—V———/
1 0
: —2€—1 : —23 Note that one can also start by calculating the composition of the various functions and then take
derivative with, respect to 15, but it is may messier.’ More general chain rule: Let us now consider 5 2 ﬂow) it = ”(139) v=v(w,y)
Then we have
_2 _ e one an
81' _ an 61' an 33:
2 _ d_ eiee
8y _ an By an 8y 4 Application of Chain Rule Before trying to understand where the chair rule comes from let’s see how for example the chain
rule can be used to prove the Product Rule for functions of one variable:
Consider two functions f (3.) and g(:L'), then you know that s d e
(Ci?)=ég+d—:f. (4.1) Assuming the Chain Rule, one can prove (4.1) in the following way: deﬁne Mn, 1)) = in: and u = f(.'L')
and 'U : g(rr). Then by Chain Rule My) _ dh _ ah dn on at _‘ [if . d9 m df ‘dg
div __ do: _ an (11' + 31) d3; _ Udm + ”a; —9dm +1 do.” 5 Idea of the proof of Chain Rule We recall that if a function 2 : f(:r,y) is “nice” in a neighborhood of a point (mmyu), then the
values of f(rc,y) near (mmyu) can be approximated by the tangent plane at (13“,!qu That is for
A2 = fiTn + Amayﬂ 4" Ail} — f(~'vu,yu) we write AZ % fi(mnwulﬁr + Men, yolAy (51) Now suppose that .1 : raft) and y = y(t) and that {EU = $(tg) and ya = Mt”) . Then if At is the
increment in t near to, if we divide {5.1) by it we obtain A: A2; A3; Kt W fm("t'o.yu)E + fyll'u‘ 310le Ii" now we let At —a 0 we obtain 13 d .
(fl—t) (in) = f2:[fﬂﬂayll)d—:(t(il + fi,(:1;g,,yu):—1:(1sn) and this should be compared to (3.1). Now that we recalled the approximation formula we can deﬁne the differential of a function: Deﬁnition 1 (Differential). Given a function 2 : (3,1,) such that approximation. by tangent pienee
holds, then we deﬁne the (total) differential off a: 82
CL, — $61.1. 'i' ady. 6 Vector Fields In 2D 3. vector ﬁeld is a vector valued function 13(33, y} = f(:c, y)i+g(a:, 31)}, where f(:c, y) and g[:r:, y)
are two scalar functions representing the coordinates. Exercise 4. Represent the vector ﬁeld I? = mi + y} and G" = —yi + 3:} In 3D a vector ﬁeld is a vector valued function FIT, y‘ 3) : f(.2:, y, z)l+g(.1', y, :ﬁ—‘rhm‘, y, 3N9, where f(:c,z ,3),g(2;,1,.,) and h(1‘,y,3} are three scalar functions representing the coordinates. Exercise 5. Represent the vector" ﬁeld F‘(:L‘,y, z) = :Li + yj + 21;. A2} 51244: ”3": “Khanz"? Wm Deﬁnition 2 (Gradient). Given a scalar function f{a:,y, 2) we can deﬁne a very special vector ﬁeld
linked to f called gradient: Won 1 ,zJ = no, w)? + no, y. 2)} + mm, y, szc
The symbol 37 formally can be deﬁned as
6 = 361: + 531, + La.
Exercise 6. Consider the function f(:r:,y,.z) z .1' + 3y. Compute the gradient off. Solution: We have ﬁz:($.y,z)=1,fy($,y,z)= z and f:(rc,y,z) = y, hence ﬁf(a;,y,z) :i+23+ylc. U! Study Guide 1. Answer the following questions: 0 Do you have an idea of how to prove the more 98118qu chain rule in page 3? I Can you prove, by using the chain rule, that given. two ﬂirtatious ﬁr) and. g(:c) the derivative
of their sum is the sum of their derivatives? ...
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