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Major Maths

Bsc 2nd year

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105 views26 pages

Major Maths

Bsc 2nd year

Uploaded by

Jorabar Parasar
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2 Rank of Matrix | 27 BY x 1-12 -3 \Bxsimple 8. Reduce the matrix A = 63.9 3) tots normat form and find its VY 0 10 2 rank. Solution : Given matrix is yh -12 -3 10 2 A=\o 31 ; 0 10 2 By transformation Cy (1), Co1(—2» Ca), We have ey 45 A~lo3 1 | . 1 0 2 By transformation Ray (—»), we have 05 -8 18 5 - A~lo3 1 4 ol 0 By transformation Rx, we have 1 0 9 1 4~lo3 1 4 05 -8 14 By transformation Ca—2), we have oi 0 8 1 0 A~lo 3 1 -2 05-8 4 By transformation Riz(~s), Ra(-), We have 19 0 o4 0 A~loo 1 -2| oo -8 4 By transformation Re (a), We have onoo La Bboo A A~|o 0 By transformation Cxo(2}, we have A~ ones 28 | RP Unified Algebra and Trigonometry _ By transformation C, (a) We have 100 0) € ~ 0100) _y, oo10 0001 re™ which is normal form of the given matrix A and so rank A = 4. JT 1.14. Upper Triangular Matrix : Definition : Let 4 = [aj] be a square matrix, then A is called upper triang, — ( matrix if for alli > ja; = 0. ( ie., A square matrix A = [aj] is called an upper triangular matrix if all clem below main digonal are zero. a2 4 513 For example, 4-333 0005 is an upper triangular matrix. 1.15. Lower Triangular Matrix. Definition : Let A = [aj] be a square matrix, then A is called lower triangy matrix if for alli < j ay = 0. ie, A square matrix A = [aj] is called a lower triangular matrix if ll elem above the main digonal are zero. 1.16. Echelon Matrix Definition : A matrix is said to be Echelon form if (@ all the zero rows occur if any occur at the bottom. (i) _ the number of zero before the first non-zero element in a row is less thant number of such zeros in the next row. or Let A be a matrix, then A is called Echelon matrix, if A having the follovi T properties : (i) The zero rows, if any occur at the bottom. s (ii) If Ri, Ro, ...., Ri be the non-zero rows written in order, then leading entry' Ri+1 (ce, the first non-zero entry) lies strictly to the right of the leadi entry of Ri. For examples, 1236 7 - -1-47 - 0372 7 5 0 0 14 «0 A=/0 0 0 4 -1 -2),B=|0 0 04 -6 0000 0 6 0 0 00 0 0000 0 2 0 0 00 0 Eigen Value and Eigen Vectors | 53 Solution : If A is an indeterminate, then the characteristic matrix is A — AJ and polynomial is | A ~ Al |. 2-1 Y 100 Now, A-if={-1 2 -1;-Ajo 10 ™ | | 10 Od “pea -1 1 = -12-A -1 1° -12-a «. Characteristic polynomial is 2A 1 1 -12-A -1 1 B+ -U+4=0 = B-W+N-4=0 > @-DA-DHA-4)=0 = A=1,1,4, So, the eigen values areA = 1, land = 4. ipl ad thi ean ame plegpin eigen values and corresponding eigen vectors of the matrix Zo 34 [Ujjain 2003; Indore 2012; Jabalpur 1993, 97; Solution : The characteristic equation of matrix is | A — Ar | = 0 Rewa 97, 2007; Bhopal 2012] 8-2 -6 2 v 4 7-2 = B-AL(7-A)3-a) - 16) + d-0G- a a - + 2[24- 1442) = 9 . #180 + 454 = 9 S AM - 18 445) =0 > AA—3A~15)~0 . H3As So, eigen values of matrix A are a 3,15, bn Let X = |x| be the cigen vector, correspondi fs Ben vector, corresponding to eigen value A = 0 of matrix A, then (A-0)xX=0 54 | RP Unified Algebra and Trigonometry 8-6 2) fx 07 -6 7 —4/ In! =0 = |0| 2-4 3) [x 0) Rt; — 6x; + 2x) = 0 Gr; + Ie — 4ry= 0 2ey = 4x; + Oxy = 0 (iy Ai From equations (i) and (ii), we get Mom 24-14 ~ -12 +32 ~ 56-36 HoH _ - 10~ 20 ~ 20 > X= ky x = Ue xs = Ae Above values satisfy equation (iii) for all values of k, so eigen vector: ki 1 corresponding to eigen value A = 0 is X = Ey =k E 5 The eigen vector corresponding to eigen value A =3 of the matrix 4, is the} non-zero solution of the equation (A-3)X=0 5-6 2 fr > 6 4 -4| |x| = |0 2-4 O}ls} lo Su — 6 + 2x5 = 0 > — 6x + 4%) — 4r5= 0 2x, — 4x + Ory = 0 Solving (iv) and (¥), we get mM ks W—8~ +H ~ W-3 Ho - ieee te > 1 = Ue = hays = —Uer So, eigen vector corresponding to eigen value A = 3 i X = ka[2,1, -2)' The cigen vector corresponding to eigen value A = 15 of the matrix A, is the non-zero solution of the equation (4-15) X = 0. 8-15 -6 = fr -6 7-15 Fl lea} = i 2 —4 3-15) [as 0 Eigen Value and Eigen Vectors | 55 [-7 -6 In) [oO > -6 -8 —4| |m| = |0 | 2 -4 12) [ay 0. Tx, — 6x) + 2x1= 0 _Avii) = 6x; — Br) — 44 = 0 ea (viii) 2x, — Ay — 12x = 0 ix) Solving equations (viii) and (ix), we get x 6— 16 x * 80 x - 2 > X1 = Uy x, = —Ukeayts = ks So, the eigen vector corresponding to eigen value A = 15 is X = [2k - 2s, kal’ = ka [2 -2, 1]. AC Brame 6. Find the eigen values and corresponding eigen vector of the matrix 21 A= fi 3 i 1 2 2) [Ujjain 2009; Rewa 2007; Bhopal 2004; Sagar 93, 2007; Indore 2000; Jabalpur 2007, 09] Solution : The characteristic equation of matrix A is |A-ar| =0 2-4 2. 1 > | 1 3-4 1 | 1 2 2-8 : = (2-A[GB-AQ-A-Y-AR-A-1} + 12-G-A] =0 = -2+0-1N4+5=0 = P-W+1U-5=0 > @-NA-)a-5)=0 > 4=1,1,5 fei The cigen vector X = |x2| corresponding to cigen value A = 5 is non-zero solution x3, of equation (A-5)X=0 2-5 2. 17) fH > 1 3-5 1 | |x| = |o 1 2 2-5) hn) [o > —3y, + 2 t= 0 m1 — In +25 = 0 w+ 2a — 3s = 0 ig ig ; 5 Cayley Hamilton Theorem and its use in Finding Inverse of Matrix | 91 30 0 48) 0 0 12 24 30] = |0 0 0) =9. 4 0 78] [0 0 0 = [12 24 30 48 0 78 30 0 | Now, we will find 4 ~' with the help of the above theorem. We are given with ° 5 A’ - 6A? + 7A + =O. Multiply above by A ~', we get AGA +71 +247! e, e, z ib Ww as Determine the characteristic equation of the matrix ‘ing 1 I: A= p 3 ih 1 1-1 Also verify whether this equation is satisfied by A, or not. Find A ~’. [Indore 1991; Gwalior 95, 96; Rewa 2006] Solution : We know that the characteristic equation of the matrix A is |A-a|=0 aA 1 2 | 2-3-4 0 | =0 1 4-1-2 > B+ -A-12=0. Above equations is the characteristic equation of A. Now we will verify that the above equation will be satisfied by A or not, i., whether A +t A~ I= Oornat 1 10 2 -3 » therefore, J = 2 10 1 O04 Since, fl 1 Again, a=AA “ 3 he i and 92 | RP Unified Algebra and Trigonometry 4 5) = [38 38 10 a a = A+ 44—A-127 10 z 4 5 4-1-2) [0 3 J-Bal ~ [> -35 =] +4{-4 4 te! Se 130-1 —3 00 =|0 0 0] =o. 0 0 0 we will find 4 ~ by using equation Now, A +44 —4-121=0. Multiplying above byA~' we get A’ + 44? — 124 = 0. A‘= hi +44-4 10 4~1 - 12 [10 > At=dl-6 1 4) aalo -3 ‘j-[ Rt 3 3 fb 1 - 001 3 6 oa Ay bf -1 4). 51-2 . a1 1 & Example 3. Write the characteristic equation ofthe matrix A = [_} 2] Show a satisfies the characteristic equation and using this fact, expressJA° — 34° +A” — 4T in linear polynomial, Solution : We know that characteristic equation of the matrix A will be |A-ar|=0 3-a 1 > 1 2-a|=9 > G-AQ-A+1=0 > P-4+7=0, Above equation is characteristic equation of 4, (Stud c A’~ 5A +71 = O as verified in example 2.) Now, A -SA+7=0 > - A= 54—74 A‘ = SA? — 74? A’ = 54‘— 743 g Sg onticarion of Matrices to a System of Linear (Both Homogeneous...| 117 7 Exar p€3) Solve the following equations using matrix method =1 x af xtytz=6, x-ytz=2 wy [Gwalior 2007; Bhopal 2004, 08; Ujjain 2006, 08; Indore 2006, 08, 16; _Rewa 2014; Jabalpur 2003, 05, 06; Sagar 2008, 12, 13] Solution : The given cquations in matrix form can be written as, AX = By, 1 1 1 6 [x where, 4=|1 -1 1],Bi = [2] andX = ly]. 2 1-1 1 IfB is augmented matrix of A, then 1 1 1:6 B=|A:B}=|1 -1 1: 2| 2 1-1: 1 Now, we reduce the augmented matrix B in its Echelon form. : fh By Ra(—1) and Rsi(—2), We have 11 1 ap —2 0 0 -1 -3 1ii: B~\0 10 2). 013: 11 111: 6 B~|0 10: 2|. 003:9 By Rig), we have Lidl: B~ i 10 oo1: Clearly, p (A) =p(B) =3. Hence, the equations have unique solution. The matrix equation transforms in the form 11 6) 010) ly} = [2I, ook) B giving, xtytz=6 y=2 z=3. Solving these equations, we get x=Ly=2z 12 | RP Unified Vector Analysis & Geometry Solution : We know that, k 4 3) =-1- 4 + 2k 34 and aj + 7k).(- 1-29 + 2) Similarly, — Sj + 2k. Hence, be(axc)=(44+ 39+ 4k): (2i — Sj + 2K) =4-154+8=-3. : Examnte@) If vector p = xi + 3) + 2k is coplanar with vectors a= 24+ 2) +) and b = 2i + 3} + 4k, then find the value of x. Solution : Since, p, a and b are coplanar vectors, hence p:(axb)=0 : x32 > 22 3/=0 234 > 0 =x-6+4=0 Pe MAP 5 gde Becnpi a Prove shat, (a + b):[(b + 6) x (€ + a)) = label: a Solution : LHS. = (a + b):[(b + ¢) x (c+ a)] © OA = (atb):[bx (+a) tex +a)] =(a+b)-[bxc+bxatexe+exa)] (By distributive lay) an =(a+b)-(bxetbxatexa) [rexe=t) =a-(b xe) + ar(b Xa) ta-(€X a) + b-(b X ¢) + b-(b Xa) + b-(CXa) (By distributive law = [abe] + [aba] + [aca] + [bbe] + [bba] + [bea] = [abe] + [abe] [. [aba] = 0 and [bea] = [abe] ete. = 2fabe] =RHS. a Ib be Example 5. Prove that, {Imn] [abe] = | m-a_m-b m-c|. nea nb me Solution : Let, 1 = hi+ bj + hk m = mi + mj + mk n= nit tn3k and a=aitaj+ ak b = bii + bj + bak c= cii +c + oak. Then, Fea = hay + boa + Isas m+ a= mya, + may + mya ete. Scalar and Vector Product of Three and Four Vectors | 21 Forming the dot product of both sides with the vector a’, a [abe] a’-(b’ xX’) = = apa” equation (2), sec, 2-10 (3) Substituting for [a b e] in (2), we obtain ) Similarly, a'xb’ [a'b’c’]* b and ¢ = [abe] ‘showing that a, b, cis the reciprocal system toa’, b’, ¢’. Note : Equation (3) shows that [a’b’c’] is the reciprocal of [a b ¢}; consequently both have the same sign. That is, the two systems of vectors, viz., a,b,c anda’, b’,¢’, are either both right-handed or both left-handed. 2-13. Any Vector in Terms of a’, b’, c" Ifa’, b’,c’ is the reciprocal system to a, b, ¢, it is possible to express any vector r in terms of a’, b’, ¢’. Then corresponding to equation (1) in sec. 2-11, we have r= (r-a)a’ + (e-b)b! + (r-e)e". . illustrative Examples (bxe) xe, wheree=eXa = (e:b)e — (eb. Substituting back the value of e, we obtain. = (bx €) x (€X-a) = [(€Xa):b]e~ [(e x a)-efb ie, (bx) x (eX a) = [a,b,ce : (i) since the scalar triple product having a repeated factor is zero. Hence, [b x ¢,¢ X a,a x b] = {(b X ¢) x (¢ X a)} (a x b) ~ = [ab cle-(a x b), by (i) = [abe] [cab] ‘a bh be Example 2. Prove that [Imn] [abc] = | m-a m-b m-c [Rewa 2006] na mb me Solution : Let a’, b’, c’ be the reciprocal system to vectors a, b, ¢. Then by sec. 2-13, r= (r-a)a’ + (r-b)b! + (F-e)c’. Replacing r by l, m, n successively, we get 1= (L-a)a’ + (I-b)b! + (Lee; 44 | RP Unified Vector Analysis & Geometry ee, 7 Feamyle Y r= acosritasines + at tana k, then fied de atl ong | ae ae S a de | [at de dt ec [Indore 2004; Bhopal 2003; Rew, Al Solution : r=acostitasintj + attanak a —asintitacostj+atanak prot dt &. -acosti-asintj a He? a, and EE =a sinci—a costs. y i io k con ay = —asint acost atana we! | —~acost —asint 0 A 4 ? sint tanai+ (— a’ costtana)j x a°k ee el 8 ae ee eee i) = ox @* sin? tana + a" cos*t tana + a* i at at 1 Ife =a Viana +1 =a’ seca. Again, yt dr dr dr —asint acost atana se dt de dP —acost —asint 0 =a asint —acost 0 =a’ tana. 3. Fi The necessary and sufficient condition for the vector a(t) to have coy G [Bhopal 2009; Gwalior 2014; Indorez (, .d a is unit vector then aan ais unit lor 6E ( aa 7 oO [vaxa=0] The condition is necessary : Let a has constant direction and so a is a cons vector. Thus 22 = 0, Hence from (1) as It prove the necessary condition. a da e condition is sufficient : Let a x 7 = 0. Hence from (1) ap Unified Vector Analysis & Geometry 50 | =f(ey + oy2 _ fect ony tone + one) 4 feoyt One * 4 1G: 2 + 82, 2 “Taking the limits when dx, 395825.» all tend to zero and consequently 3p, \E zero, then the coefficients of 3x, 895825» iM the above expression tend to lig, values so a a a ee 2 (69,25 dr gy CGI 2 9 3e £952 se Using the notation of differentials, the above relation in limit becomes 7 at at af t= de + dy t Bet di de ta Here, the sum dfis defined as the total differential of f. Note: If the variables x,y,z, ... were expressed in terms of several new vary 5,t, .. then the vector functions f will be @ function of these new varia, Then as i . ‘en as in scalar calculus, the partial derivatives of fare given by 2.8 o.oo Fo om and simil, 5 sax as ay as 2 These, peat for other new variables. formulae will clearly change from one set of independent variable another. * at Qxe® "Se Illustrative Examples ample If f= (ety — xi + (e9 — ysinx)j +7008 )K- verify that ve at + axdy ~ ayax" Solution : We have f= ey — xi + @ —ysinyj + cos yk hig ay = E+ [e+ x sinx]j — x sinyk ) = Axi + [2-1 + €? » yx — cosx]j — 2 sin yk of _ 4 axay = St + {e% +e - xy — cosx}j — 2x sinyk. Again, f= (2xy —2x4)i + ( —ysinxj + Zcosyk at HL ay — 42) ag BY — 42 Si + (C= y —y co8x)} + 2 cosyk ) = t+ eye cos) Being x er (oye ea eee Gradient, Divergence and Curl | 51 From (i) and (ii) itis evident that (he LALO, Solution : We ce a Be [a le] Now, gp =x f=xi— yj t’k of = yi — xyz] + yk 2 60 = 29% - 294) + 30K 2 {55 oo} = 4y4i - 294) +3)°7k afta alee }] eean2 ae 2 aan = Hak 4 At(2,-1,0), xyz and f = xzi — xyj + y2’k, then show that “ (2, = 1, 1) is 4i + 2p. a aatag OD = 4 (IVE 2(-1)5 = 4i+2j 3. a = xy2i — 2x2} + xz%k and b= 2i + yj — 17k, find (axb) 0, -2). Jabalpur 2003; 11998] Solution : We have Uabalpur 2003; Bhopal 1998] a=xyzi — 2a} +k and b= di t+yj—k ij axb=|xy2 203 as = (2823 —ye?)i + Qe) +242)j + 0772 + Ack = -2i+ 40 + dock. ACL 0, 2, mae Bray Xb) = = (= + AD-2 + ADO 2)k = -4i-§j. e Gradient, Divergence and Curl | 73 [Tawa 6° 7 44 — 8k, WO. 10. (a) 1, (b) -1, (c) V3 11.1023, 12, 2V14; 4i — 6j + 2k. 13,37 b 4-26. Gradient in Polar Let r be the position vecto; Uthe sense of r increasing), Let along perpendicular toy increasing). Then the distance 28 is dr and directional derivative along Gis @- Vy ¢ where 4 (*,y, 2) = 0, isthe level surface ordinates, ¥ of a point PG, 8). Let be the unit vect al 9). tor alon, ii ¢9 be the unit vector 4 ae (in the sense of 9 *R 4s in the direction of Hence, & V9, (1) Again, the distance ds in ‘perpendicular to r is ra. Also, directi along eis e9- V9. the direction Ta ional derivative Pole or Mal Line X Origin, Hence, 79 = Vo .Q) ~ Clearly, the components of vg along ¢, and along @& are respectively @ Vand 0: V9. Hence, VO=(6-V9) +(e Vo) @ a, Lag a > Vou Series .Q) {using (1) and (2) Equation (3) expresses gradient in polar coordinates, ~ B- b TAs Examples Epphrole 1. Prove that V () =- z =~ i [Jabalpur 2016; Gwalior 1993] Solution: Vv. () =2 g F 2 B Second method : r=xi+yj+zk os Paveye? en x pelo cael rind or or _y were set 74 | RP Unified Vector Anatysis & Geometry a oz Similarly, aH ewe 2 Hence. v @) = (2 Wier) () (* ai a 4) $s 13 () ed (F 1 x |r to er, 1 Oey ~ Pal Pal Fa = = -ak BiB B 1 =—3 Gy £2) oe aa [Indore imple 2. If = log | r |, show that V $ = 75: ¥ Solution : > ¢ 2 2 Second method : r i - we Tp ay Sra Because, a= r if =5. mple 3. Show that grad f(r) x r = 0. Solution : grad f(r) X r= £ fr) t xe re, —= (r Xr) [erxt Second method: r=xi+yj +2k Gradient, Divergence and Curl Similarly arate eradfi)=Vf) = ( ge thy th 53) /O AARC = FFA ae tH aeIO ay * aI ae =rori4soritsork = £0 (ai + yj + 2k) r Hence, ‘grad f(r) x r= fo (Fxr) asrxr=0. 0-4/4) log Vx2 + y2, prove that _ r—k(k-r) grad = (F— ker) - (F— Ker) (Jabalpur 1990, 93, 97, 98; Indore 2003] Solution : V= log V2 + y? > V=logr [er=xityj] a a > grad V = © (log)? Haye 1A inet Tr mol > xin r-k(k-r) Ne tS en oe Gut RHS = Co) Fok) aot [ek-r=k-+ @i+yj) =0) Second method : V = log Vi? + y? = 5 log (x2 + y) grad V = ; V log (x2 + y2) 1f,2 4,0 ype P =} ie thi the ee 49 1 . . i (2 +255 + 04} a tityy @ r+y raaityj tk 76 | RP Unified Vector Analysis & Geometry Hence, i+yjtzk-2k =xit yj Also, {r — k (k= 1)} aoa Gi +98) wu So, grad V= rer? zen 2d" V[rabj=axb [Rewa 2006; Jabal,” so or grad [rab See Solution : Let ae solic and 9.D! then rheor [rab]=r-(axb) x oy @ e Fi =|a a 4% dinate by bp bs = (aabs—aphy)x + (abs — bs) Y + (abo grad [ra b] = V [rab] ing 52452442) {(aobs — asb2)x_— Proo = (rye eg) + (asbi — aibs) y + (aib2~« = (@sb3—asby) i + (ash, — a3) 5 + (abo ~ ij k =la a a; by by by =axb Exercise 4 (D) 1.Find Vr. [Jabalpur 1996) 2. Find V/" [Ujiait 3. Find we. 4. Find V |r |3, 30) . é : 5. Find V a . 6. Find V(7e~"), Pre 7. Find V(sin@ + rcos6). 8. Find V G? ~ ave 4 62) Answers Lr Qn 2p, 3. -3-5,, J@r-1 4.3rr, 5 OM gan Nery. e a 1 in) o ll (c 7 —rsin®) @%. 9 7. (cos) @ + 7 (Cos @ ~ rsin@) e Ar avy 7 ‘vers ea ee A Gradient, Divergence and Curl \ 15 Pexrty+2 ar ar _x =k 2.2 ax ax ar_yo_z ardor grad f() = V(r) = (i hy ty) =i py 2D HK TIO ar =rori re snot =f (ity +) Hence, »grad fir) x r=" (r x r) Wor QB.d=0, asrxr=0. o-h (4) le 4. 1f V-E5,2) = log Vx? + y?, prove that _ r= ker) grad V= o ker): (¢— ker)’ [Jabalpur 1990, 93, 97, 98; Indore 2003] Solution : log V2 +2 oa logr [er =xi+yj] > aes ia grad V = © (logr) t £ ar Pr ure, Now, eke (F—kk- rn): (F—kk nr) [ok-r=k> Gi +yj =o) Second method : Vs log V+ 92 = Flow (2+y) zt log (22 + y2) (231 + 24 + 0k} (i) ‘yo | RP Unified Vector Analysis & Geometry keres —zk yj + zk — 2) Tee —k(k nr) =a ty Hence, r xi ty Also, {r — (k= 8)} (= a a - i + yi) axty. ave ait BL So, grac gry kde Z - {r— K(k yy pork a show that rem ok Example 5. If'a and b are constant vectors; then si Ret V[rab]=axb [Rewa 2006; Jabal. T° or , 8. solt Solution : Let aves potical! and 9. piv then re co? frabj=rs@Xxb) | = y ala a 4% eFur by Bs ind! ie rp) + (ashi — 103) 9 + (abo = (abs grad [rab] = V(rab] me 0k ee ke = (thy by — a1bs) Y ie — aybs)J + (aiba~ ing 3) {(a2b3 — a32)* “Ergot a. bs) y + (aib2 ~« = (azb3—asb2) ij k =|a a 4 by bz bs =axb. Exercise 4 (D) 1.Find Vr. [Jabalpur 1996] 2. Find V7" 3. Find Vr-3, 4, Find V | r |°. 30, 5. Find V a . 6. Find V(?e™’). 7.Find V(sin@ +rcos6). 8, Find V (3 - 4vr + 6”). Ler. 4.3rr, 106 | RP Unified Vector Analysis & Geometry dy yk _g * a” de dy & = ime . logy = logx + loge a . y = ox. aor _ ity tk Hence, PTE Waar ad + cyxj + c2tk > Vea cpr + cf x? itej+ok “Viwed +e which is independent of x, y,z so it is a constant vector QyQy”, OT tustrative Examples Si + tj — Bk, prove that =~ 14i + 75j- 15k. ea ° [Bhopal 2003, 05, Os, 14; Gwalior 2010, 14; Jabalpur 2004, 03, 05,08, 11, 0% Indore 2003, 05, 15; Rewa 2005, 14; Sagar 2004, 14; Ujjain 2000, 04) Solution : r= 5+ —Pk dr 2 \ & = 101 +j - 37k \ a7 itd & Fira 10i — 6tk \ 1X Fa = (Si + i — Pk) x (101 — 6rk) , ijk * =|se ¢ -6 10 0 -6¢ = (— 67) i + (— 108 + 300) j + (—104) k = — 61 + 2083) — 10k , ——— £ (eS) a= foe m5 — ya =f -62dt-i+ f 20Pdt-j-f 10rde-k -_# ff =-Fi+20fj- wok = = 28 + Sef — 5k. 2 ex PE) ae = [ — 2 + Sey — 5K)?” Si dt ll ik ae *], = (~16i + 80j - 20k) - (—21 + 5j — 5k) =~ Mi+75j- 19k : 1 (ex @f] amex tt Aliter ; We know that Vector Integration | 107 > 2. dr) lex be £ (rx os) a= [rs at |, alr. ar)? ~ dt Ir een Now, ij k set -8 10 1-3? = — i+ 545-5 kK 2 =[- Wit SA j— Sek — 161 + 80j - 20k) ~ (— 21 + 5j- 5k) \ = 14i+ 75j-15k. Example 2. Integrate the vector equation a ih, de Og! Solution : Since, we do not know the value of r in terms of t, hence, itis not "K possible to integrate r with respect to t. Hence, to solve the purpose, taking the scalar product of each side with 2 # re > = -nr, get Now, integrating both sides w.r.t. t, we get 2 dr = -Wr+ dt ae whergc is the scalar arbitrary constant of integration. (Ben 3. Given that {a jt 2katt=2 LO) a3 4+ 3k att =3 rr | a= 0. [Bhopal 1993, 95; Jiwaji 88; Rewa 94; Vikram 97; Jabalpur 2004] Solution : When, Let | r | =VOFTFE A300 } Whent = 3,Let |r| = =VI6F449 = vO. eZ! me Sedat ea a1 3d eh Sgt t= YS 8th nt ) . Theorem of Gauss’s | 123 ¢ I / pif A) Evaluhte OEE 205 + ay k)-dS, where S is the surface of the sphere + ye PAI, in the Jirst octant. [Sagar 2012; Jabalpur 2000; Bhopal 2009] Solution : Let {=f +9" +2 — then XxX Let tbe the outward drawn unit normal vector, Then, a= Brady | grad f | ; pf, ;%4,4 ; tig ted vig YY (h? fa) *(3) + @) _ 2 (ci + yj + zk) © Va Gt ye 2d | =xityj+zk © . bee +y+22= 1] [zi + 2j tayk)-dS__ OO =f, (ri + zxj +ayk)-tas = J Ozi+ aj + k)-Gi tyi + 2k) ds Now, polar coordinates of any point P on the unit sphere are x= sind cos ~ y= sind sing 2 = cos8, +» Required intégral Ct 124 | RP Unified Vector Analysis & Geomet ry of o Also, clementary surface afea “ ds = sin 9 dO dp. a : yh Required integral 5 f oA =f orto a" =3f o2dS 7 i «ng sing) cos @ sin 8 do en 57777 (and exe g) (HOO)? 4) 4s “to 70 3 7 sin cos din #005 00 P - 070 a sint 8)? [_ ot] = lea, 2 Jo r 11_3 oe and C i +2k is a cyl Evaluate f Fide where F =a +925 | =f PRY Pk where t varias 1001. —~tphopal 2006; Indore ny ~ " lol. ai aris For alpar 2013; Rewa 1996: Sagar 96; Ujjain 95,5, ‘olution : As ir where raitej+ek dr = (i+ 21j + 32K) dt. Gane eee wt 1=J Fede ae (yi tyzj + ak) (+ 25 + 37K) at =F), C+ 8) + ehy- G4 245 430K) at =f) @ #26 +36 at Example 8. x = cost,y = sint,z = ¢ which joins the points (1, 0, 0) and (— the length of arc of the curve. \8 Solution : Let He iF (Qy +22) ds. Parametric equations of the curve is given as, x=cost,y =sint,z=¢. "ond Degree and Tracing of Conies|225 as 2 if B and tan. Yoo Pe ee . (iv) ee Solvi . ng Cquations (vy Wye -£ Nd (v), we Av) Bet the coordinates Of the vertex of the Parabola Ns Focus is given by Directrix : jp is given by = 2zty43=0 Tracing : Take the venex 4 (4 Through A, draw a line AC inclined at an angle ta (3), given by (4), to X-axis This tne is the axis of the parabola. Throug line at right angle to AC. It is vertex. ‘We can very easily show that the parabola meets the axis of X at (31.2, 0) and (8, 0), approximately and it meets the axis of Yn imaginary point. th a draw another the tangent at the ¥ Hence, the shape of the conic will be as given in the fi iple 3. Trace the parabola, ()-S-@~ 16x? - 2dry + 9y? + 71x - 64y +95 = 0. igure. eC ‘oo find the coordinates of its vertex and focus. [Bhopal 2004] Solution : The equation to the given conic is 16x? - 2dxy + 9y? + 77x - 64y + 95 = 0. (i) Comparing with the general equation of second degree, we have a=16,h=-12,b=9,g=4, = - 32,¢= 95, 1 Analysis & Geometry 226 | RP Unified Vecto - ; i 12 ab= 144 - 144 = 0 ff Nature + dani , and A = abe + 2fgh - of \/ my >) -95« 144 = = 13680 + 29568 - 16384 - 9 ( 2 6, bola. Therefore, the equation (i) represents @ Pars i tion as Write the given equal Perea . Introducing an arbitrary constant A, the equation (fi) is written in the fag! form , = (81-77) x + (64 - BA) y +R ‘ (ax — By +? = (81-77) * + Cl 95 “ Choose A such that the lines , y- wt and (BL - 77) x + (4 - A) y+ are at right angles, ‘The condition for this is (ibaa or 4&4 ~ 77) ~ 3 (64 - @)=0 or 504 = 308 + 192 = 500 a ies 1=10. Substituting this value of 2 in (iii), we have 4 (x ~ 3y + 10)? = 3x + 4y +5. a” This equation can be written as 2 fa (ez) os. V4? +32) 1 (tees) 5 5 ° ‘Writing this equation in the form of standard equation of a parabola, ¥? = day, we see that 4x-3yt10 J+ Bives the axis Bxt4y+5 and x = 0, gives the tangent at the vertex, ie,, the equation to the axis of the parabola is 4x — 3y + 10=0, General Equation of Second Degree and Tracing of Conics |227 ion to the tangent at the vertex is - equa wn 3x4 dy + 5=0, oA) sre lntus eet of the parabola, da = 1 yertex ¢ Solving (iv) and (v), the coordinates of vertex are ( : The focus is obtained by solving the equations Y=Oand X=a 4x —3y + 10=0 Focus : Sxtdy+5 1 5 2 4x — 3y + 10 = 0, and 12x + 16y + 19 =0. i ence coordinates of focus are e pirectrix + The equation to directrix is x+a=0 3x+Sy+5 1 ie 3 *m7° 12x + ly + 21 =0. Construction : The trace the parabola, find the points of intersection of the equation fowite coordinate axes, Now, putting y = 0 in (i), we get 16x? + 77x + 95 =0. Itgives two imaginary roots, therefore, the parabola intersects the X-axis in imaginary © S “gnn, putting x = 0, we get oy? - 6dy + 95 =0 (y - 19) @-5)=0 ity a its yap andy =5, ie, the curve cuts the Y-axis in two points |") and (0, 5). 11 ‘Now plot the vertex A (-2 ‘Trace the axis (iv) and tangent at vertex (v). . 19 Plot the points (02) and (0, 5) and then trace the parabola as shown in the igue.

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