Lecture for Week 10 (Secs. 4.5–8)Derivative Miscellany III1Growth and decay problemsI already discu ss ed the theory of these prob-lems two weeks ago, so let’s just do an example.2Exercise 4.5.3A culture starts with 500 bacteria, and after 3hours there are 8000.(a) Find the formula for th e number after thours.(b) Find the number after 4 hours.(c) When will the population reach 30,000?3The basic assumption is that the numberof new bacteria is proportional to the numberalready there (parents). SoB(t) = B(0)ektfor some constant k. So acc ording to the data,8000 = 500e3k,or 3k = ln805= ln 16, or k =13ln 16.B(t) = 500e13t ln 16= 500(16)t/3.4ThenB(4) = 500(16)4/3= (calculator output).For the last part,30, 000 = 500(16)t/3⇒t3= log163005⇒ t = 3 log1660= 3ln 60ln 16.5Inverse trigonometric functionsThere are two aspects of inverse trig func -tions that need to be s tudied:• the defini ti ons (especially branch choices);• their derivatives.The most important i nverse trig functionsare sin−1and tan−1.6Both of the problems we encountered for thesquare root function also appear for the inversesine.1. sin θ is always between −1 and 1, so there isno way to define sin−1z if |z| > 1 (unless wego into c omplex numbers, which we won’t).2. For −1 ≤ z ≤ 1 there is more than oneθ with z as its sine. (In fact, the re are in-finitely many.) So, we have to choose a prin-7cipal value (or branch) of the inverse func -tion. The standard choice is to pick θ sothat −π2≤ θ ≤π2.Please refer to the book for the graphs (p. 276and p. 278 in this case). Recall that to get thegraph of an inverse function, you can plot the orig-inal function on a transparent sheet and fl ip it overso that the horizontal and vert i cal axes are inter-changed.8Thus s i n(sin−1z) = z always, butsin−1(sin θ) = θ is fal se if θ is not in the principalinterval.Recall also that sin−1z does not mean(sin z)−1(that is, 1/ sin z), althou gh sin2z doesmean (sin z)2. This notational inconsis tency isunfortunate, but we ’ re stuck with it. (Let’s noteven ask what si n−2z means.) Another notationfor the inverse is arcsi n z.9The inverse tan gent is easier (see graphsp. 279), because it is d efined for all z and all thebranches look the same (have positive slope).But there are still infinit ely many branches, andthe standard choice is −π2< θ <π2. (Why is it“<” here but “≤” for the inverse sine?)This tan−1z is a very nice func ti on. It in-creases smooth l y between h oriz ontal asymptotesat θ = −π2and θ = +π2.10The usual technique for di ffere ntiating animplicit or inverse function yiel ds the formulasddxsin−1x =1√1 − x2,ddxtan−1x =11 + x2.These are ordinary algebraic func ti on s ! (All traceof trig seems to have di sappeared.) One reasoninverse trig fun ctions are important is that theyhelp prov i de the antiderivatives of certain alge-braic function s.11Exercise 4.6.51Find the derivative of g(x) = sin−1(3x + 1) andstate the domains of g and g′.ExerciseFind an antiderivative off(x) =3√4 − 4x2−10x2+ 1.12g(x) = sin−1(3x + 1).g′(x) =3p1 − (3x + 1)2(which c ou l d be si mpli fied). For g to be definedwe need |3x + 1| ≤ 1.Case 1: 3x + 1 ≥ 0. Then3x + 1 ≤ 1 ⇒ x ≤ 0,3x + 1 ≥ 0 ⇒ x ≥ −13.13Case 2: 3x + 1 < 0. Then−3x − 1 ≤ 1 ⇒ x ≥ −23,3x + 1 < 0 ⇒ x < −13.So the domain of g consists of the two intervals−13≤ x ≤ 0 and −23≤ x < −13,which fit together to give−23≤ x ≤ 0.14For g′to be de fined we also need 3x + 1 6= 0,hence the interval shrinks to −23< x ≤ 0. (Seethe vertical tangents at the ends of the graph,Fig. 4 on p. 278.)Alternative solution of the inequality:|3x + 1| ≤ 1 ⇐⇒x +13≤13.This clearly describes the numbers whose dis-tance from −13is at most13— namely, the interval−23, 0 .15F′(x) =3√4 − 4x2−10x2+ 1;what i s F ?F′(x) =321√1 − x2−10x2+ 1,so the obvious choice isF (x) =32sin−1x − 10 tan−1x.16Soon we will reach the proof that the only otherantiderivatives are eq ual to thi s one plu s a con -stant.(What if the two numbers inside the squareroot were not the same? Look forward to the ex-citement of Chapter 8 in Mat h. 152!)17Hyperbolic functionsThis topic is not in the syllab us for Math.151 at TAMU. To see why it s hould be, read mypaper in College Math. Journal 36 (2005) 381–387. It also explains why I d on’t talk about cot,csc, sec−1, etc.18Indeterminate forms (l’Hospital’s rule)(The name is pronounced “Loap-it-ALL”(more or less) and sometimes spelled “l’Hˆopital”.)In my opinion, the two most importantthings to learn about l’ Hosp i t al’ s rule are• when not to use it• what it teaches us about limits of exp andln at infini ty.19Suppose we want to calculate the limit off(x)g(x)as x → a (a may be ∞), and suppose thatboth f(x) and g(x) approach 0 in that limit, orboth app roach ∞. L’Hospital’s rule state s th atthat limit is the same as the limit off′(x)g′(x)(whichmay be easier to calculate).Please don’t confuse this formula with the“limit law” for a quotient, or with the formula forthe derivative of a quotient. They are three differ-ent things!20Here is an example of the correct use of therule:limx→0sin(5x)7x=“00”= limx→05 cos(5x)7=57.However, you didn’t really ne ed the rule to dothis problem, did you? You already kn ow thatsin(5x) ≈ 5x when x ≈ 0 (or c an appeal tolimu→0sin uu= 1).21After studying Taylor series (Chapter 10)you will know many other situations where thebehav i or of the functions f and g ne ar a is obvi-ous, so l’Hospital is unnecessary. Many studentsoveruse l’Hospit al’ s rule, relying on it as a “b l ackbox” when they would learn much more (andsolve the problems equal ly fast) by just taking aclose look at, and comparing, the behavior of thenumerator and denominator as x → a.22Here is an example where using the rule isabsolutely wrong: We know that limx→0+cos xx=+∞, because the numerator approaches 1 whil ethe denominator approaches 0. If you incor-rectly applied l’Hospital’s rule, you would getlimx→0+− sin x1= 0. This fraction did not satisfy thehypotheses of l’Hospital’s rule, because th e limitof the numerator, cos x, is not 0.23Finally, consid er limx→0(x ln x). This is an in-determinate form of the type “ 0 × ∞”. To applyl’Hospital’s rule we must rewrite i t as a quotient.First try: limx→0+x(ln x)−1is an
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