Mathematical Formulas and Equations, Study notes of Mathematics

Download Mathematical Formulas and Equations and more Study notes Mathematics in PDF only on Docsity! WWW ee PLANE we ANALYTIC > Adeals wW/ the distance b/0 Zpointe knowing the coovdinotes, the GEOMETRY equa Kor ona rer fee ofthe curve or wwwweuwu#ew [speciel Case - Mid point } re Vs MF (Xm, Sn) I. pissance BAS pomia7 7 meving TH Xm Ye xr 4%) t te eee mse ($2495) x Le Mer A Toot (yh PAL Wer) “the distance formula” Example : ° > x N Gwen jg @ hne Segmert Note: for the rt P(g) from (3,2) 40 (-2,-¢)- X F distance of Pp from Find 4he coordinates of yraxis (absacca) dhe first dri-secton Pair 4 Yr! distance ©f P frew From (3,27 and the point of Kraxis (erctinote) dhe line segmend . Example : Soleben : D Find the distance from point (3,2 , pein (-2,-8) point (3,2) to a) £( 2,2) Sowhon cee we, de | Os (42-9)? Substiuje values * Thay a (6-2) d = 4.43 _! answer [+ by €¢ ~ mode complex | type chily Abs (#2) ~(-2-ci) = 4.43 —> answer I. Division or ting SEGMENT Figure: Pe xa,%e) Ya-u, +0 (Kel yo). Pidxy %2-™) by Smilax Ass Rap eres Mee RR %2- 2-W Pe - 1 Pa Fermulas : Pas (x, ¥,) Me ress Yo Yr C92-)4¥8. Pe 57969 7 r= APS _o Pie, Bal ye! 3 Pr (2,0) vse xg ¥ (h2-%)) 4% Xo = 73 (-2-3)43 = Yo = * (Y2-Y) a Yo = Ya(-6-a) ta: % herce, Pos (3, %) b) for me (Xm Ym) Xm = 4 Cx.4%) *'/a(-243)* /2 wae «a (m+n) = Va (-612)=-2 hence + Ms Ch,-2) —<> ancner 4Ja = EL wo 5 s = V- AREQ OF PoLYGoN By CoorpinatEes OF VERTICES METHOD Formula : A= lt Node. Nerfices must be traced in COW Aireekon from any vertex Vi ee. . © Find ane area of pentagon w/ vertces @ C10 %,Gi,6 ), C52) and (0.4) oe @ SP omener @ Find the area of the A wy versces a4 -4-l, 142i and geal, Sslution : C22 va Hirt) Me 04-3 note: a complex no. represent © a pent in the wey plane °o Anh =: e Azz {+ (a- 140) - cs-e-4)} A =\3 —— ~anseer [by Es method] mede matrix mMatA 3x7F type 4s -3e1F 17 ze. 7 -4err ete Ac 0.5 Shift Matix det shifts Mantis mata) =17 —— answer R.- PROPERTIES &% EQVATION OF LINE Model Functone : © Ceneral Formula Ax +By #0 +O where 4,8 and Cc ore constanis © Zpeins Form) [civen ave \ Pre x,y) Pas C 2-42) wae xy y -y, OF ecamnty Merion or Kak 92 -Y, y-4i 2 Bee ¥ Sal, (%~ *1y ® Point Slope Foewm gives wie Pr = Cx yun) y-y, em Cx -«,y i ® Stope - Indereepd Form [oe are wie: hpi bb =y-indercepd Yrmxab Site Ove @ Mndercepy rom | SX -m4. be Wing. x Ta bee b 4) What is the area bounded by the curve x?- Gy so and He \udus vechm? Solvhon: x2 - pyro er xte ey} te =4ar®? 428 V=(ee) for parabolic segment Asc? hence, a= 3 (a2) “AF 1069 ——» answer VI. THE CIRCLE >. o member of 2° degree certal come sechon w/e ic the locus foint w/eors font Aisdance Croetivs ) from a fixe point (center ) Model Equations ; ) General Form x24y2 +A + ByAc=s 2) Standard Ferm or Center - Radius Form by cts 9) c= Chk) Cx-W)2 + Cy-k ter? [sHoereur Foe 4, tc end FROM] GENERAL FoRM he “A, k= =, cte2 [Aee% 4c Preperfies : *) Area, ARte = Yar 2) Circumference, C= 27h = wd Yorgent distance From a point Po = (Xs. ¥%e) +o the circle tay 4+ Ax t+ By t+ C=O dy { M074 ¥,% 4 AX at B¥O7 Ee 4) Eauion ef Redica) Axis, ERA pardical/Britas femme were commen tangent b) c#(2,0) x2ay2er? 3) ERA is, Be. Cred Eq. of al a ~ ( ars) =o ee ortrogonul oles C,: Xt4yt*tAix + Bt =O Ge: K24S 4 Are + Bay tz = 3 5) Kf, Ag 48:B2 2 2C ter) porabela, open up As Yabh] br tR=s heqgerZ Example ) Given ig a Arcle XL 4 72-4 x toy ~ £0. Find, @) the locaten of dhe center (byitc recive (6) the targent Aistonce trom the por C86), solution : X24y%- aw 4 GyY-IT=0 (x2 47% pax teyr cro) Ar-4, BEC, cal? <= €2,-3) anewer bv) rea lataezac Ne ifs [<-497 + C7418) TE Ss answer <) Pe= (8/6) ay = J e746t aca secer-sz ars 1s answer 2) Given ore 2eirclke xXP+7%42x-Sy SFB ond W242 44x 47d TES. Find the valve ef Kk co thot they ave orthogonal . Sotwhor: Cui xteyt% 42" By-S TS teteyt ade +kyi 2G cz we? So dthed c, and C2 are orthegonal hence, 2C40*O3% ey = 20-842) B-sk =-6 ke My —_— om ser Nil; THE ELvLiIPse _> wis the member of central sonic Sectors wis ia the leocot point of Aistencec from wwe fixed peints (Fand Ff’, foct d is conctond (gmail aliameter) Figure 2 mA Awechir ee Camel) MA ~ dircvie VON eee Model Equathong a)c= Chk) | het _ - (36 ~-32 _- Te Se GO ee dy! Ax™+4 Byt+Cx +Dy dE rg 2¢a) eif AcB, MA ee a>h ez (zr) : — answer HE ap ae at aoe TT Fern a 4 “_—» anewer o f{ ADB, MAy arla na < = 2 be tp jk 28 ec) es at et, jee s 28 a az ae at%=b74 7 42-3 € = oO. 6644 2) Standard Ferm aes by crs c2 sa2-b? “3 cneeaee a) Ce (hk) d) As Nab = 0 (4003) = lat | ® (xh? 2 P= an [== —S answer — Grey | — 7 Fan age? P= 22.2) ee 3) A hallwoy . z . —P answer May arch ewiling! wns cian semt = Mig teal i @ Och? Cyn? from the floor , ye the twee gt ake cewier — +” =) eth ends of th mi Ss are Gm fail as v2 quer {> callin the ceiling ~ how high ix the oi ie, eats 3 Wm horizontally fom cither woul? : cC,o f 4 . oe Solution + cemi -ellipkeal Mat F CrCey0) ox +e 7 ceiling x? yz ae be! — ay) ar Re M4n) eo T 2m} ¥ x* x @ x* yess ars c=(e,0) 2 T eg Bey oF ae 3m 2 eS?) Bt _ cn | Bsa) Properties + | oe az eB2. i) eccentricity, e<).0 om] — oe = 3 thes Ya + ana} e+ § L— Jom ota cml ¥> 24m 2) AF sbe4yc% 8) Length of Lotus (HES ty) —>@ SvbsHIVIe © 7 A= cemin MA 2 = He G4aag lo = Semi -mA LR = ab He Sam ct Cio F a —> answer 4) Astunce of Altectrix S) Area = Talk 3) A trough har an alliptical from the center cross section which ic 1 zo Focal Radivs = distance Adee. i ake eo top eet 12 inches 8 r c) Perimeter ef she pi. Cx. ¥) an she ¢ Sur foce B inches below the 4op, Sino the Pz an | a24b62 ps ei’ oan From Width ef dhe Waser surface 2 Ate, 7 mone min = Sotvton : a . : exemple: 2? for elipse of ) Given ic on eMpse Ax? 41092 36% 432 MAY and C2 (0) -9a 70, Find, (a) the location ofits cemer, a 2 = (6) the length of Ws lotus rectum, ce) the fired eccentricity. (Jd) avea and perimeter, ant +ioyt_ 36x 1382y- 9270 Solution: PARR + By +Cx4m GEO A249, B= : CEA, BF, C 2-36 , pe 32 EE 42 .n sUbskiie valves of aae> mau} 9° fB eMC 4 sabe es ix, 8) nie oa +3 ES «= IS t-€-—=MS tt hen ee, w=2 (355) Wwe BAF inches wer BHHHHUHRUUYUEEHYNaD AY bn cMipse kas ie center mt (0,0) b) C= (2,0 te : With Hs axis horiaomal. the cietanre >) x2 yt, slags a x? ‘nav b/w the verheet is @ ane ie eccenbicity az 2 at “G2 iS O-€ — COMPUTE the length ef the lorgess een! rudive trom point (2,3) on Properties + the curve . ‘ : 1) eceenticity ,€>'-0 3) ¢7 2g? 4? Selvtien : ese use, Pmaythin 2 APEX, MOH a fer longect fecal radius , airs 26 ga-2 8) o satan! finac = ave a é ” where, v's Ba™ B A=4 PH Ps (3) 6) equaten of asymptoies i? Sid cox ny | =O aA. rsechan & aie, eso replace one in he oie ve the aarnptetes fmav = 440.82) og form e9. by zero ——> answer for y in serme of x, 8) Slope of asrrphinc, m= tb Vill. THE HyPERBoLA Dinciinea right ; OD welined Tess > Hic member of 24 degree centre! conic sectons w/ Aifference of distances from 4wo Fixed pointe (ranar’, 4) Given is a hyperbola X? -By*- 4% -6y-26 fou ) 1s © constens. \ 7 Find, @ LR. ty the <ccenticity Example: tsa) (CO) fre angle b/w the accymptoter | Figure : ugrle Anis (A) Alstance bY divechices. Selven: x*_ Byt- 4x - Gy-2 = 0 CAx?- By*~Cx -Dy-E=0) Asl, BE 73), c#-4, p=-6, -2 eee rey asfe=>Is —sat*=3 be {a = TV ay pee z aires ae" 20" was 7 IZ —> ancwer by es te 2. [s be ~ az C7 = az 4b 2 Model Equations: e= PE _ 2 a 9 ™ i) General Ferm, 4% -> TAY 3 te ici a @) Ant- By? tex4Dy aEHa} Ce Ck) o: le ~ b ©) 7 = ote & 3 be TA; he ls kee P= atewl_t Leo 2 —_—P answer by) Ay?-8x* + Cut Dy +e2 0] ty wray ao: Je ;b-lr C= (hk) 4) SBD = ads 22 WE sige tO Pens 2 =" = 26 aA TE anwer 2) Sdandora Fom. by crs a) c(h, k) rakyk ire . ) o> ~ ie (rau) L a). coer + ee 21 (TAY) — — — — _— _— — — — —, - 5 1 --) + > xample * Toon jars Feel it) * Find the area bounded by the azb ff. polar ourves. oes af(itcocse) a) r= 2 rcose — ay FF asin? Se 5) r25rlésin2ze ) cs a(ttooce) SL reacece Ady tan - y {@ golvton ; ) v= a(iscoso@lya carciod , + © A(t 4case) As ay a7 azz "OQ Y as 3m/,(2)7= 68 ~).. a) re 28NO —» reaSwmne, rose azz, n=2 ~ a-leave rose — > answer As YWa*= A c2d* = at Pike ansver 3) vea(issmoe) a) rr 2cosce > 5 Acoce , earcle ~ Y a=Z = diometer D + ” dees SRL? Oe Waste Woe x — er a. 4 * B R+cCOEe > r=a + beose , imacon —a>h z,bel Va (za*4b*) = % {2G)°+"} - a 2 Ss) rte i6sin26 + r*ca2sin2e WNeeRGcoe a? =l6, Azats \S ——> ancwer . - sino 2S ret Goname) caroli cl s rsa (i-<ine) As = 3h q = SE ay* aw ¥ 2 —> sinwer r : Perimeter > Ba D~ LEMNISCATE oF BERVOULI “infinity” Models + e:°%, ) rts atCos2& XI. Equation oF corVES ww COMPLEX VARIABLE & Cimeginc 7 ame RK ‘ > Figure 2) r?> a%sinze Y iesand 2 e=% Diagram Pt a Area, A= a® x CreaVaxsy A Node: Special Case: D2 + xayi r2= atcos® m ¥ where, As Zaz Z = complex variable x —_ <F veal para of 2 = Re fz} Y= maginary part ofs = Sm {2} . tor JF vr E- Rose wy = Unig Imaginary nO, mag nitvale | Model Equatens a) Jel= Taw Be ofc )ore sete |e 2,3,4... @ * Tar" Ye} eprren asin n . . yore +f ner} Wis weirele | Note: po? a) 1f n= oddne., there anen - | a Ae Waar if n= even no, , there are mn - leave, >) oe A= at Exomple: ) ldenkty 4he curve and find the Ores Wounded by the curve Jz-2|+3 Selhon : |z-2\-3 guest as XH LL} xe yteade s, | Ce-2) tyi]e3 he itud get the magnitucle , heren, ‘& sieole Meeayte yt 2 3 —Y Mener Areae tty? = ow square both sides wer cx-a)tagy?=9 } ©@,0) Ze 9 r>3 2) \dentfy the curve repvecented ky, Je-sl+ |z48)-10 Selvhop : Noie: if the given is \2-sl rb lzaaleic Ja-3) © Jetal=4 Ix+¥t-3) 4 lx yyisaleio [(x-s.4yil + J@wasyayil 10 A sbpaihans get the magnitude, 2) Ie-ar%4y? + Joeas yay? =o ‘ smPlifying algebraically , hence. z= lex? . +2Sy* zsaco0 fy xe yz _ ban ollipse — + > 2s \e 8 \dentify the ourve represented by Im { 27} Ke ed Selution: ( Zz x+aye a ae s . Reh 2 pax 2": Ce vide t-ytaaeys ft f ‘ y Subsets 40 given ea. He wf{e } B2xy sol Im (27) - or nor, “4 —» axy =4 *YT2 7 Ah eailederal hyper belo T. ————— ee RSW SE