First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009

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1 Mth 3B Review Steve 18 Mrch 2009

2 About the finl Fridy Mrch 20, 3pm-6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No Riemnn sums

3 Integrtion The ide behind integrtion is to slice wht we re interested in finding (i.e., re) into little pieces tht we cn mnge. We then dd up ll of the little pieces nd get the result. If we do this for the function f (x) where we slice between x = nd x = b we get the following interprettion. b [ ] re bove f (x) dx = x-xis [ re below x-xis ]. Things to wtch out for In cses where our re is mde of tringles, squres nd/or circles it is usully better to find the integrl by finding the pproprite re.

4 Properties of integrls b b f (x) dx = 0. f (x) dx = f (x) dx = c b f (x) dx. b f (x) dx + f (x) dx. c

5 Differentition nd integrtion re closely connected. Fundmentl theorem of clculus I If f is continuous on [, b] nd F (x) = x f (u) du then F (x) = f (x). Fundmentl theorem of clculus II If f is continuous on [, b] then b f (x) dx = F(b) F(), where F(x) is ny nti-derivtive of f (x).

6 Appliction of FTC I: Leibniz s rule Combining the fundmentl theorem of clculus with the chin rule we hve ( d h(x) ) f (u) du = f ( h(x) ) h (x) f ( g(x) ) g (x). dx g(x) Exmple Find lim x 0 sin x x e u du x 4.

7 Appliction of FTC II: Esier integrls If we know n nti-derivtive of f (x) we cn now find integrls. Functions tht we know ntiderivtives for include: x k, e x, sin x, cos x, sec 2 x, sec x tn x, x 2 Things to try when looking for ntiderivtive Expnding polynomils. Using trigonometric identities. Breking into sum of simpler pieces.

8 Applictions of integrtion: Are When finding the re between curves y = f (x) nd y = g(x), if f (x) g(x) between x = nd x = b then the re is given by Are = b ( f (x) g(x) ) dx. Things to wtch out for If no bounds re given then we need to solve for intersection, i.e., f (x) = g(x). If the curves intersect or severl curves bound the region then brek the problem into smller pieces. Sometimes it is esier to do the integrl with respect to y.

9 Applictions of integrtion: Volume We find volume by dding up the re of cross sections. The only cross sections tht we looked t (i.e., could be tested on) re circles, formed by solids of revolution. So if f (x) g(x) 0 nd we revolve the region between these curves nd x = nd x = b round the x-xis, the resulting volume is b ( Volume = π (f (x)) 2 (g(x)) 2) dx. Things to wtch out for Don t forget to squre the functions nd be creful when simplifying. Volume (s with re) should never be negtive.

10 Applictions of integrtion: Averge vlue The verge vlue, f vg of the function on the intervl x b is the vlue so tht the rectngle with width (b ) nd height f vg hs the sme re s under the curve, i.e., Averge vlue = f vg = b f (x) dx b Things to wtch out for Averge vlue cn be negtive (it cn even be 0). Men vlue theorem tells us tht for continuous function on [, b] there is some c between nd b so tht f (c) = f vg.

11 Substitution rule Rules for integrtion come from rules for differentition. The most importnt rule for integrtion is the substitution rule which comes from the chin rule. f ( g(x) ) g (x) dx = f (u) du where u = g(x). For definite integrls we hve b f ( g(x) ) g (x) dx = g(b) g() f (u) du Things to wtch out for Look for functions inside of functions. If you don t see substitution to mke, try rewriting the integrl in different form.

12 Integrtion by prts The product rule for integrtion becomes integrtion by prts. u dv = uv v du or b u dv = uv b b v du. This technique is often used when there is product of two functions; lso when there is some function which is hrd to integrte but esy to differentite (i.e., ln x or rctn x). Things to wtch out for Alwys try substitution before integrtion by prts. Sometimes must be done severl times. When doing definite integrl sometimes esier to do the indefinite integrl to find nti-derivtive nd then t the end use FTC to evlute.

13 Prtil frctions A rtionl function is polynomil over polynomil if we hve rtionl function then we cn use the method of prtil frctions. First check degrees. If the degree on top is greter thn or equl to tht on the bottom do long division. Next fctor the denomintor nd use prtil frction techniques to brek it into smll chunks. Integrte ech chunk. Things to wtch out for If it is not polynomil over polynomil don t use prtil frctions. The denomintor will lwys fctor s liner nd irreducible qudrtics.

14 Prtil frctions the tricky step ( ) (x + b) k ( ) = C 1 x + b + C 2 (x + b) C k (x + b) k + ( ) (x 2 + bx + c) l ( ) = D 1x + E 1 D l x + E l x bx + c (x 2 + bx + c) l + To solve for coefficients we first cler the denomintors then choose nice vlues for x; or group coefficients nd set them equl. (Note: the number of coefficients tht need to be solved for is the sme s the degree of the denomintor.)

15 Improper integrls An improper integrl is one tht involves infinity in some wy; either bounds involving / or verticl symptotes (such s division by 0 or ln 0). The wy to del with improper integrl is to pproximte it by n integrl which is not improper nd then tke limit, i.e., b Things to wtch out for f (x) dx = lim t b t f (x) dx. Wtch out for verticl symptotes ( sneky wy to hide n improper integrl). It might need to be broken into severl prts. Sometimes better to first do integrl s n indefinite integrl L Hospitl s rule is useful for evluting limits.

16 Tylor polynomils We cn pproximte functions by polynomils. To find the best polynomils round x = we use informtion bout derivtives, we cll these Tylor polynomils. P n (x) = f ()+f ()(x )+ f () 2 (x )2 + + f (n) () (x ) n. n! We cn use these polynomils to pproximte the function for points ner x =. Things to wtch out for Mke sure tht the coefficients re numbers, i.e., don t forget to evlute. Given function we cn find the Tylor series, but we lso cn tke Tylor series nd sy something bout the function.

17 Differentil equtions Differentil eqution reltes how dependent vrible is chnging (i.e., the derivtive) in terms of the dependent vrible nd the independent vrible. These re very importnt in physics, biology, chemistry, economics, nd getting good grde in Mth 3B. We restrict ourselves to solving seprble differentil equtions, Method to solve: dy dx = f (x)g(y) with y(x 0) = y 0. 1 Seprte, put ll y terms on one side, x on the other. 2 Integrte, integrte ech side. 3 Simplify, solve for the constnt nd (perhps) solve for dependent vrible.

18 (more) Differentil equtions Things to wtch out for We cn solve for the constnt nd then simplify for the dependent vrible or vice-vers We cn use the differentil eqution to solve for x tht y will hit certin vlue; or we cn see wht the vlue of y will be t time x; nd so on. If dy/dx = g(y) then n equilibrium solution is solution to g(y) = 0. A stble equilibrium is one tht ttrcts points nerby while n unstble equilibrium is one tht repels points nerby. There re two methods to test stbility of equilibrium solution ŷ: g (ŷ) < 0 is stble; g (ŷ) > 0 is unstble Drw picture.

19 Vectors A vector x = [x 1, x 2,..., x n ] cn be used to indicte direction in spce. Given points A = ( 1, 2,..., n ) nd B = (b 1, b 2,..., b n ) the vector tht strts t A nd points to B is found by AB = [b 1 1, b 2 2,..., b n n ]. The length (or mgnitude) of vector is found by x = x1 2 + x x n 2. A vector which hs length 1 is clled unit vector (these re often used when deling with direction).

20 Dot product Given vectors x = [x 1, x 2,..., x n ] nd y = [y 1, y 2,..., y n ] the dot product is x y = x 1 y 1 + x 2 y x n y n = x y cos θ, where θ is the ngle between the two vectors (if they emnted from the sme point). We cn use this to solve for the ngle between vectors. Two vectors re perpendiculr if the ngle they form right ngle. Equivlently two vectors re perpendiculr if their dot product is 0.

21 Lines To find line we need point (ˆx 1, ˆx 2,... ˆx n ) nd direction which will be provided for us by vector [ 1, 2,..., n ]. Then points on the line stisfy Vector form: x 1. x n = ˆx 1. ˆx n + t 1. n or Prmetric form: x 1 = ˆx 1 + t 1. x n = ˆx n + t n

22 Plnes A plne cn be thought of s flt 2-dimensionl surfce embedded in three dimensionl spce. The most importnt prt bout the plne is norml vector, or vector which is perpendiculr to ll points in the plne. If plne hs the point (x 0, y 0, z 0 ) nd norml vector n = [, b, c] then the set of points (x, y, z) which re in the plne re those which stisfy Things to wtch out for (x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. Norml vectors re not unique, they cn chnge by scling by constnt. Two plnes re prllel if their norml vectors re prllel (i.e., sclr multiples of ech other). When reding coefficients to get the norml vector mke sure ll the coefficients re on the sme side.

23 Functions of severl vribles The domin of the function is the set of ll possible inputs. Generlly when looking for domin we void problems, the type of problems tht we might hve include 0, < 0, ln( 0). The rnge of the function is the set of ll possible outputs. Generlly speking this cn be very hrd to find (one of the resons Clculus ws invented ws to help nswer such questions). To understnd function we often look t slices in either x, y or z direction. In the z direction this mens f (x, y) = C which re curves in the plne clled level curves or contour lines.

24 Limits lim f (x, y) = L (x,y) (x 0,y 0 ) Limits don t tell us wht does hppen, rther they tell us wht should hppen bsed on wht is hppening nerby. For most functions we cn plug in the limit point nd evlute; if we don t get 0/0 then we re done. If we do get 0/0 then more work is hed. If limit exists then we should lwys get the sme nswer regrdless of which pth we tke to get to the limit point. So if long two different curves we get two different vlues then the limit does not exist. (Best curves re usully stright lines or curves of the form y = x k, x = y l.)

25 Continuity A function is continuous t point if wht should hppen is wht does hppen, i.e., lim f (x, y) = f (x 0, y 0 ). (x,y) (x 0,y 0 ) All polynomils re continuous, nd composition of continuous functions re continuous.

26 Prtil derivtives For multivrible functions we often use prtil derivtives (denoted insted of d ). Ide is to tret ll but one vrible s constnt nd then tke derivtives. f x (x, y) = f f (x + h, y) f (x, y) (x, y) = lim x h 0 h f y (x, y) = f f (x, y + h) f (x, y) (x, y) = lim y h 0 h We cn lso tke higher order derivtives, i.e., f xx, f xy, f yx, f yy nd so on. For nice functions (i.e., functions in our clss), 2 f (x, y) = 2 x y f (x, y). y x

27 Tngent plnes A tngent plne is higher dimensionl nlog of tngent line. For function f (x, y) the tngent plne t point (x 0, y 0 ) is given by z = f (x 0, y 0 ) + f (x 0, y 0 ) x (x x 0 ) + f (x 0, y 0 ) (y y 0 ). y Since tngent plnes mke good pproximtions loclly we cn use tngent plnes to pproximte the function. Things to wtch out for Mke sure to evlute the prtil derivtives.

28 Grdients The grdient vector is vector where the entries re prtil derivtives. f (x, y) f (x, y) = x f (x, y) y We cn use the grdient to find directionl derivtives t point in direction u ( unit vector). Nmely, D u f (x, y) = f (x, y) u. Things to wtch out for Mke sure u is unit vector!

29 Properties of grdients At point (x, y) we hve, f (x, y) D u f (x, y) f (x, y). So the grdient bounds the directionl derivtives! The grdient points in the direction of gretest increse. Opposite the grdient points in the direction of gretest decrese. The grdient vector f (x, y) is perpendiculr to the level curve tht psses through the point (x, y).

30 Locl mximum/minimum A locl mximum (or minimum) is point which is greter thn (less thn) ll other points nerby. Locl mximum/minimum occur t criticl points which re where f (x, y) = 0. To determine if it is mximum, minimum or sddle we use the second derivtive test. Let D(x, y) = f xx (x, y)f yy (x, y) ( f xy (x, y) ) 2 > 0 nd f xx < 0 then mximum, > 0 nd f xx > 0 then minimum, then D(x, y) < 0 then sddle, = 0 inconclusive.

31 Lgrnge multipliers When mximizing/minimizing f (x, y) given constrint g(x, y) = c then mximum/minimum occur when f (x, y) = λ g(x, y) nd g(x, y) = c. Equivlently, the criticl points of F(x, y, λ) = f (x, y) λ ( g(x, y) c ). Find the points, then plug them in to find the extreme vlues. Things to wtch out for The equtions re lmost lwys nonliner (i.e., tricky to solve). If nothing else, solve for λ nd get reltionships mong vribles.

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