Lecture for Week 2 (Secs. 1.3 and 2.2–2.3)Functions and Limits1First let’s review what a functi on is. (SeeSec. 1 of “Review and Preview”.) The best wayto think of a function is as an imaginary ma-chine, or “black box”, that takes in any of variousobjects, l abeled x wheneve r there’s no reason tocall it something else, and p roce s ses it into a newobject, l abe led y = f(x).A function is not necessarily given by aformula.2Usually both x and y are numbers, and inthat case we can easily think of a function as be -ing the same thing as a graph. A curve, or anyset of points in the x–y plane, defines a f unction,provided that no verti cal lin e intersects the setmore than once.........................................................................................................................................................................................................................................................................................xy3Sec. 1.3, however, deals with fun ctions r(t)whose values are vectors, or points, not numbers.The graph of such a function still exists, butit lies in a more abstract space (with three dimen-sions in this case, one for t and two for r).A point is not quite the same t hi ng as a vec-tor. A point i s represented by a vector, r = hx, yi,when we choose an origin of coordinates in space.If you move the origin, the numbers x and y willchange, but the numerical components of true vec-tors, such as velocity and force, will not change.4Exercise 1.3.7Sketch t he curve represented by the parametricequationsx = 3 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π,and eliminate the p arameter to find the Carte-sian equation of the curve.Note, no actual vector notation here, althoughone could have written r(θ) = h3 cos θ, 2 sin θi.5Well, first I’d plot t he points forθ = 0,π4,π2, . . . , 2π.(The last point is the same as the first one.)Then I n otice thatx32+y22= cos2θ + sin2θ = 1,so the curve is recognized as an ellipse. (It’s th ewhole ellipse, since we ob s erved that the pathcloses.)6Fortunately my plotting software has an el-lipse command, so I di dn’t need to do the arith-metic for the first part..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................xy7Exercise 1.3.27Find (a) a vector equation, (b) parametric equa-tions, and (c) a Cartes ian equation for the li nepassing through the points (4, −1) and (−2, 5).(The exercise in the book doesn’t ask for aCartesian equation, but the next group of exercisesdoes, so I’l l do it here.)8First let’s find the vector pointing from thefirst vec t or to the second:v ≡ h−2, 5i − h4, −1i = h−6, 6i.If we add any multiple of v to any point on th eline, the result is a point on the line (and you getall the points t hat way). So an answer to (a) isr(t) = h4, −1i + th−6, 6i = (4 − 6t)i + (− 1 + 6t)j.9r(t) = h4, −1i + th−6, 6i = (4 − 6t)i + (−1 + 6t)j.Notice that the question asked for “a” vector equa-tion, not “the” vector equation. There are manyother correct answers to (a) , corresponding todifferent starting points on the line or differentlengths and signs for v.For (b), just write t he components sepa-rately:x(t) = 4 − 6t, y(t) = 6t − 1.10To get a Cartesian equation, we need toeliminate t. In the present case that is easilydone by ad ding the two parametric equ ations:x = 4 − 6t, y = 6t − 1 ⇒x + y = 3.In more general situations, you would need tosolve one equation for t and substitute the resultinto the other equation.11Limit is the fundamental concept of calcu-lus. Everything else is defined in terms of it :• continuity• derivative• int egral• sum of infinite seriesIt took mathe maticians 200 years ( of cal cu-lus history) to arrive at a satisfactory definiti onof “limit”. Not surprisingly, the result is not easyfor beginne rs to absorb.12For that reason, Sec. 2.4, “The Precise Def-inition of a Lim it”, is not a required part of oursyllabus. That doesn’t mean that you are forbid-den to read it! But you might find it more mean-ingful if you come back to it after gaining someexperience with how limits are used and why theyare impor tant. Two natural places in the textbookfrom w hi ch to l oop back here are• after infinite sequences and series (Chap. 10);• when functions of several variables arise (Sec.12.2). (In that place Stewart simply states themultivariable generalization of the “precisedefinition” without fuss or apology. )13So, we have to make do with “intuitive”ideas of a limit. The graphical p roblems on p. 89are a good place to start. However, they don’tlend themselves to this projector pre sentation,,because I have no good way to reproduc e thegraphs. So we’ll look at them in real time....I’ll come back to infinite limits and verticalasymptotes later.14Let’s go on to Sec. 2.3. The key issue inthat section is t his: W hen you are presentedwith a formula de fining a functi on, such asf(x) =√x2+ 1x − 3,usually the limit of t he function at a point is justthe value of the fun ction at that point, but notalways.limx→1f(x) =√11+ 11 − 3= −√22= f(1).15f(x) =px2+ 1x − 3!But limx→3f(x) does not exist ; the function val-ues get arbi t rarily large near x = 3. ( Even worse,they are positive on one side, negative on theother.)16So, the big question is, when can you getaway with just sticking the number into the for-mula to find the limit?limx→af(x) = f ( a) ?Textbooks give you a list of “limit laws” thatstate conditi ons that guarantee that the limitcan be taken in the obvious way. Let’s turn thequestion around and try to identify “dangersigns” that label situations whe re the obviousway might go wrong.17In practice, the most common trouble is azero of the denominator. (Notice that in t helimit laws on p. 91, the last on e, concerning di-vision, is the only one that needs a caveat(“if limx→ag(x) 6= 0”).)Now two things can happen:1. The limit of the numerator as x → a
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