Mathematics. Author TNAU, Tamil Nadu

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2 Author TNAU, Tmil Ndu

3 Index SN Lecture Pge No Three dimensionl Anlyticl geometry 5-0 Anlyticl geometry -8 Binomil theorem 9-4 Conics -5 5 Differentition Chin rule differentition Differentil equtions Circles Chin rule differentition Integrtion 57-6 Inverse of mtrix 6-68 Mtrices Physicl nd economic optimum for single input Model Chin rule differentition 0-6 Permuttion nd combintion -6 7 Physicl nd Economic Optimum for single input 7 8 Popultion dynmics 8-9 Progressions -8 Second order differentil equtions with constnt 0 coefficients 9-0 Inverse of mtrix - Demnd function nd supply functions -7 Vector lgebr 8-4

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5 Three dimensionl Anlyticl geometry Let OX,OY & OZ be mutully perpendiculr stright lines meeting t point O. The extension of these lines OX, OY nd OZ divide the spce t O into octnts(eight). Here mutully perpendiculr lines re clled X, Y nd Z co-ordintes xes nd O is the origin. The point P (x, y, z) lies in spce where x, y nd z re clled x, y nd z coordintes respectively. Z P(x,y,z) M x O z x R Y y N X where NR x coordinte, MN y coordinte nd PN z coordinte 5

6 Distnce between two points The distnce between two points A(x, y, z ) nd B(x, y, z ) is dist AB ( ) ( ) ( ) x x y y z z In prticulr the distnce between the origin O (0,0,0) nd point P(x,y,z) is OP x y z The internl nd Externl section Suppose P(x, y, z ) nd Q(x, y, z ) re two points in three dimensions. P(x,y,z ) A(x, y, z) Q(x, y, z ) The point A(x, y, z) tht divides distnce PQ internlly in the rtio m :m is given by 6

7 A,, m m z m z m m m y m y m m m x m x m Similrly P(x, y, z ) nd Q(x, y, z ) re two points in three dimensions. P(x,y,z ) Q(x, y, z ) A(x, y, z) The point A(x, y, z) tht divides distnce PQ externlly in the rtio m :m is given by A,, m m z m m z m m y m y m m m x m m x If A(x, y, z) is the midpoint then the rtio is : A,, z z y y x x Problem Find the distnce between the points P(,-) & Q(,,) PQ ( ) ( ) ( ) 8 Direction Cosines Mthemtics 7

8 Let P(x, y, z) be ny point nd OP r. Let α,β,γ be the ngle mde by line OP with OX, OY & OZ. Then α,β,γ re clled the direction ngles of the line OP. cos α, cos β, cos γ re clled the direction cosines (or dc s) of the line OP nd re denoted by the symbols I, m,n. z P(x,y,z) β γ O α M y Result x By projecting OP on OY, PM is perpendiculr to y xis nd the POM β lso OM y Similrly, cos β cos α cos γ y r x r z r x y (i.e) l, m, r r l m n n r z x y z r ( r x y z Distnce from the origin) l m n x y z x y z l m n (or) cos α cos β cos γ. Note :- The direction cosines of the x xis re (,0,0) 8

9 The direction cosines of the y xis re (0,,0) The direction cosines of the z xis re (0,0,) Direction rtios Any quntities, which re proportionl to the direction cosines of line, re clled direction rtios of tht line. Direction rtios re denoted by, b, c. If l, m, n re direction cosines n, b, c re direction rtios then l, b m, c n (ie) kl, b km, c kn (ie) (or) l b c k (Constnt) m n l m n b c k (Constnt) To find direction cosines if direction rtios re given If, b, c re the direction rtios then direction cosines re l k l k similrly m n b k c k l m n ( b c ) k (ie) ( b c ) k k b c Tking squre root on both sides () K b c l b c, m b b c, n c b c 9

10 Problem. Find the direction cosines of the line joining the point (,,6) & the origin. Solution By the distnce formul r x y z z Direction Cosines re l cos x r 7 r (,,6) m cos β y r 7 o y z r 6 7 n cos γ x. Direction rtios of line re,4,. Find direction cosines Solution Direction rtios re,4, (ie), b 4, c Direction cosines re l b c 4 69 m b b c Note n c b c 4 69 ) The direction rtios of the line joining the two points A(x, y, z ) & B (x, y, z ) re (x x, y y, z z ) ) The direction cosines of the line joining two points A (x, y, z ) & B (x, y, z ) re r distnce between AB. x x y y z z,, r r r 0

11 Anlyticl geometry Introduction The brnch of mthemtics where lgebric methods re employed for solving problems in geometry is known s nlyticl geometry. It is sometimes clled Crtesin Geometry. Let X OX nd Y OY be two perpendiculr stright lines intersecting t the point O. The fixed point O is clled origin. The horizontl line X OX is known s X xis nd the verticl line Y OY be Y-xis. These two xes divide the entire plne into four prts known s Qudrnts. Y P (-X,Y) P(X,Y) X M O N X Q (-X,-Y) Q(X,-Y) Y All the vlues right of the origin long the X-xis re positive nd ll the vlues left of the origin long the X- xis re negtive. Similrly ll the vlues bove the origin long Y xis re positive nd below the origin re negtive. Let P be ny point in the plne. Drw PN perpendiculr to X xis. ON nd PN re clled X nd Y co-ordintes of P respectively nd is written s P (X,Y). In prticulr the origin O hs co-ordintes (0,0) nd ny point on the X-xis hs its Y co-ordinte s zero nd ny point on the Y-xis hs its X-co-ordintes s zero. Stright lines A stright line is the minimum distnce between ny two points. Slope The slope of the line is the tngent of the ngle mde by the line with positive direction of X xis mesured in the nticlockwise direction.

12 B Y A θ X O X Y let the line AB mkes n ngle θ with the positive direction of X-xis s in the figure. The ngle θ is clled the ngle of inclintion nd tn θ is slope of the line or grdient of the line. The slope of the line is denoted by m. i.e., slope m tnθ Y Y X θ (obtuse) X θ (cute) X Y Slope m tnθ Slope m tn θ Slope is negtive Slope is positive Note (i) (ii) (iii) The slope of ny line prllel to X xis is zero. Slope of ny line prllel to Y xis is infinity The slope of the line joining two points (x, y ) nd (x, y ) is Slope m tnθ y x y x Y

13 (iv) When two or more lines re prllel then their slopes re equl (v) When two lines re perpendiculr then the product of their slopes is i.e., m m - Eqution of stright line There re severl forms of stright line. They re,. Slope intercept form Let the given line meet y-xis t B (o, c). We cll OB s Y intercept. Let A be ny point on the given line. Drw AM perpendiculr to OX nd BD AM. Let this line mke n ngle θ with X xis. Then the slope, Y A y -c B (O, C) D X I o M X AD AM m tn θ BD OMDM AM OB OM y c x y c ie. m x mx y c y mx c Hence, the eqution of line with slope m nd y intercept c is given by y mx c Note: (i) Any line pssing through the origin does not cut y xis (c 0) i.e., y intercept is zero. Therefore its eqution is y mx (ii) Any line which is prllel to x xis hs slope equl to zero. Therefore its eqution is y c (Becuse m 0) (iii) Any line perpendiculr to x-xis, ie which is prllel to y-xis t distnce of K units from the origin is given by x k. Exmple : Find the eqution of stright line whose (i) Slope is four nd y intercept is (ii) Inclintion is 0 0 nd y intercept is 5

14 Solution: (i) Slope (m) 4 Y intercept (c) - Eqution of line is y mx c Y 4x Eqution of line is 4x y 0 (ii) θ 0 0, y intercept 5 Slope tn θ m tn 0 0 Eqution of line is y mx c Y x 5 y x 5 Eqution of line is x - y 5 0 Exmple : Clculte the slope nd y intercept of the line x y 0 Solution: x y 0 y x x y Compring with y mx c, we get m, c. Slope one point form Slope ; y intercept Let the line AB mke n ngle θ with x- xis s shown in the figure nd pss through the point P (x, y ). If (x, y) represents point other thn the point (x, y ), then m y y x x where m is the slope of the line or y y m (x x ). Y B P (x,y ) θ o X A Hence the eqution of line pssing through point (x, y ) nd hving slope m is 4

15 y y m (x x ). Exmple: Find the eqution of stright line pssing through (-4,5) nd hving slope - Solution: Slope - Point (-4,5) Eqution of the line is (y-y ) m(x-x ) y 5 - (x4) y 5 -x 8 Eqution of line is x y 7 0. Two points form Let P (x, y ) nd Q (x, y ) be ny two points on the given line AB. We know, the slope, m y x y x. We hve the slope-point form of line s y y m (x x ). Substituting the vlue of m in the bove eqution we get, Y B Q (x,y ) P (x,y ) θ o X y y ie y y y y y x y x (x x ). x x x x Hence, the eqution of line pssing through two points is given by y y y y A x x x x 5

16 Exmple: Find the eqution of the stright line pssing through the points (,6) nd (-,5). Solution : Eqution of the line is y y y y x x x x y x 4. Intercept form x 5y 0 0 y 6 x 5 5y 0 x Eqution of the line is x 5y 7 0 Let AB represent the given line which intersects X xis t A (, 0) nd Y- xis t B (0, b). We cll OA nd OB respectively s x nd y intercepts of the line. B (0, b) 0 A (, 0) X The two points form of the eqution is given by y y y y x x x x Substituting (, 0) for (x,y ) nd (0, b) for (x, y ), we get the eqution s y 0 b 0 x 0 ie y x b y b x - Thus, y b x 6

17 x y b Hence, the eqution of line hving x-intercept nd y-intercept b is given by x y b Exmple: Find the intercepts cut off by the line x y 5 0 on the xes. Solution x intercept: put y 0 x x This is the x intercept y intercept: Put x 0 -y y This is the y intercept Exmple : Give the mthemticl eqution of the supply function of commodity such tht the quntity supplied is zero when the price is Rs.5 or below nd it increse continuously t the constnt rte of 0 units for ech one rupee rise in price bove Rs.5. Solution A(0,5) B(0,6) C(0,7) Point B is (0,6) Point C is (0,7) Eqution of stright line joining two points is y y y y x x x x y x y 6 x (y-6) x 0 0y-60 x-0 x -0y

18 Note The four equtions we hve obtined re ll first degree equtions in x nd y. On the other hnd it cn be shown tht the generl first degree eqution in x nd y lwys represents stright line. Hence we cn tke generl eqution of stright line s x by c 0 with t lest one of or b different from c. Further, this gives by - x c x c i.e. y - b b Now, compring this with the eqution y mx c, we get slope m b coefficien - t of x coefficient of y 8

19 BINOMIAL THEOREM A Binomil is n lgebric expression of two terms which re connected by the opertion (or) - For exmple, xsiny, x x, cosxsin x etc re binomils. Binomil Theorem for positive integer: If n is positive integer then n n 0 n nr r n 0 n ( x ) nc x nc x... nc x nc x nc x ----() 0 r... n n Some Expnsions ) If we put - in the plce of in ( x ) n nc x n ( ) 0 nc x n ( )... nc x nr ( ) r nc x ( ) n nc x 0 ( ) n 0 r... n n 0 n r nr r r n r 0 n ( x ) nc x nc x... ( ) nc x... ( ) nc x... ( ) nc x 0 r n b) Put x nd x in () n r n ( x ) nc x nc x... nc x... nc x ( ) ( n ) n( n )( n ) n n n x nx x x... x!! c) Put x nd -x in () r () n r r n n ( x) nc x nc x... ( ) nc x... ( ) nc x ( ) r ( n ) n( n )( n ) n n x nx x x...!! n ( ) n x n () (d) Replcing n by n in eqution () ( ) ( n ) n( n )( n ) n n x nx x x...!! ( ) n x n n n (4) e) Replcing n by n in eqution () ( ) ( n ) n( n )( n ) n n n x nx x x... x!! Specil Cses. ( x) x x x.... ( x ) x x x.... ( x) x x 4x ( x ) x x 4x (5) n n 9

20 Note:. There re n terms in the expnsion of (x) n.. In the expnsion the generl term is T r i.e. T nr r r ncr x.. nc 0, nc, nc,... nc r,... ncn re clled binomil coefficients. nc r x nr r. Since this is the (r) th term, it is denoted by 4. From the reltion ncr nc n, we see tht the coefficients of terms equidistnt from the r beginning nd the end re equl. Note: The number of terms in the expnsion of (x) n depends upon the index n. the index is either even (or) odd. Then the middle term is Cse(i): n is even The number of terms in the expnsion is (n), which is odd. Therefore, there is only one middle term nd is given by Cse(ii) : n is odd T n The number of terms in the expnsion is (n), which is even. Therefore, there re two middle terms nd they re given by Exmples. Expnd (i). Find 7. Solution: 7 (0) 7 7C 0 6 x x T n nd ( ) ( 0) 7C( ) ( 0) 7C ( ) ( 0) 7C( ) ( 0) 7C4 ( ) ( 0) 5 6 7C ( ) ( 0) 7C ( ) ( 0) 7C ( ) 0 ( 0) T n

21 . Find the coefficient of x 5 in the expnsion of Solution x x 7 In the expnsion of x x T 7 7 r r 7 Cr x 7C r x 7 4r, the generl term is x Let T be the term contining x 5 r then, 7-4r 5 r r T T r coefficient of x ( ) 7C r x 680 x 5. Find the constnt term in the expnsion of Solution x x 0 In the expnsion of T x x ( x ) 0 0 r r 0 Cr 0C r x 0r ( ) x r, the generl term is x r 05r r ( ) 0C r x r 0C r ( ) Let T be the Constnt term then, r 0 5r 0 r The constnt term 0C ( ) x r 05 x ( ) 0r r x 0 80

22 Conics Definition A conic is defined s the locus of point, which moves such tht its distnce from fixed line to its distnce from fixed point is lwys constnt. The fixed point is clled the focus of the conic. The fixed line is clled the directrix of the conic. The constnt rtio is the eccentricity of the conic. L M P F L is the fixed line Directrix of the conic. F is the fixed point Focus of the conic. FP PM constnt rtio is clled the eccentricity e Clssifiction of conics with respect to eccentricity. If e <, then the conic is n Ellipse M M L B L A S I S F O F A L I L I B I M M x y ) The stndrd eqution of n ellipse is b

23 ) The line segment AA is the mjor xis of the ellipse, AA ) The eqution of the mjor xis is Y 0 4) The line segment BB is the minor xis of the ellipse, BB b 5) The eqution of the minor xis is X 0 6) The length of the mjor xis is lwys greter thn the minor xis. 7) The point O is the intersection of mjor nd minor xis. 8) The co-ordintes of O re (0,0) 9) The foci of the ellipse re S(e,0)nd S I (-e,0) 0) The verticl lines pssing through the focus re known s Ltusrectum ) The length of the Ltusrectum is b ) The points A (,0) nd A (-,0) ) The eccentricity of the ellipse is e b 4) The verticl lines M M nd M M re known s the directrix of the ellipse nd their respective equtions re x nd x e e. If e, then the conic is Prbol. P L O S L Q ) The Stndrd eqution of the prbol is y 4x. ) The horizontl line is the xis of the prbol. ) The eqution of the xis of the prbol is Y 0 4) The prbol y 4x is symmetric bout the xis of the prbol. 5) The vertex of the prbol is O (0,0) 6) The line PQ is clled the directrix of the prbol.

24 7) The eqution of the directrix is x - 8) The Focus of the prbol is S(,0). 9) The verticl line pssing through S is the ltus rectum. LL is the Ltus rectum nd its length LL 4. If e >,then the conic is Hyperbol. B M L M L S I A I O A S L M B I M L x y ) The stndrd eqution of n hyperbol is b ) The line segment AA is the Trnsverse xis of the hyperbol,aa ) The eqution of the Trnsverse xis is Y 0 4) The line segment BB is the Conjugte xis of the hyperbol,bb b 5) The eqution of the Conjugte xis is X 0 6) The point O is the intersection of Trnsverse nd Conjugte xis. 7) The co-ordintes of O re (0,0) 8) The foci of the hyperbol re S(e,0)nd S I (-e,0) 9) The verticl lines pssing through the focus re known s Ltusrectum 0) The length of the Ltusrectum is ) The points A (,0) nd A (-,0) b ) The eccentricity of the hyperbol is e b 4

25 ) The verticl lines M M nd M M re known s the directrix of the hyperbol nd their respective equtions re x nd x e e 5

26 DIFFERENTIATION In ll prcticl situtions we come cross number of vribles. The vrible is one which tkes different vlues, wheres constnt tkes fixed vlue. Let x be the independent vrible. Tht mens x cn tke ny vlue. Let y be vrible depending on the vlue of x. Then y is clled the dependent vrible. Then y is sid to be function of x nd it is denoted by y f(x) For exmple if x denotes the time nd y denotes the plnt growth, then we know tht the plnt growth depends upon time. In tht cse, the function yf(x) represents the dy growth function. The rte of chnge of y with respect to x is denoted by nd clled s the derivtive of function y with respect to x. S.No. Form of Functions yf(x) dy. Power Formul x n n d( x ) n nx. Constnt C 0. Constnt with vrible Cy dy C 4. Exponentil e x e x 5. Constnt power x x x log 6. Logirthmic logx x 7. Differentition of sum y u v dy d du ( u v) where u nd v re functions of x. 8. Differentition of y u v dy d du ( u v) difference where u nd v re functions of x. 9. Product rule of y uv, dy d dv ( uv) u v differentition where u nd v re functions of x. 0. Quotient rule of u dy d y, u vu uv differentition v v v dv dv du 6

27 where u nd v re functions of x. where du u, dv v Exmple. Differentite ech of the following function Solution. Differentite following function Solution Here is the derivtive.. Differentite following function Solution diff. w.r.to x 4. Differentite the following functions. ) Solution 7

28 diff y w. r. to x 5. Differentite the following functions. diff f(x) w r to x Derivtives of the six trigonometric functions Exmple. Differentite ech of the following functions. Solution We ll just differentite ech term using the formuls from bove.. Differentite ech of the following functions Here s the derivtive of this function. Note tht in the simplifiction step we took dvntge of the fct tht 8

29 to simplify the second term little.. Differentite ech of the following functions In this prt we ll need to use the quotient rule. 9

30 CHAIN RULE DIFFERENTIATION If y is function of u ie y f(u) nd u is function of x ie u g(x) then y is relted to x through the intermedite function u ie y f(g(x) ) y is differentible with respect to x Furthermore, let yf(g(x)) nd ug(x), then dy dy du du There re number of relted results tht lso go under the nme of "chin rules." For exmple, if yf(u) ug(v), nd vh(x), Problem then dy dy du dv.. du dv Differentite the following with respect to x. y (x 4). y e x Mrginl Anlysis Let us ssume tht the totl cost C is represented s function totl output q. (i.e) C f(q). dc Then mrginl cost is denoted by MC dq TC The verge cost Q Similrly if U u(x) is the utility function of the commodity x then du the mrginl utility MU The totl revenue function TR is the product of quntity demnded Q nd the price P per unit of tht commodity then TR Q.P f(q) dr Then the mrginl revenue denoted by MR is given by dq The verge revenue Q TR Problem. If the totl cost function is C Q - Q 5Q. Find Mrginl cost nd verge cost. 0

31 Solution dc MC dq TC AC Q. The demnd function for commodity is P ( - bq). Find mrginl revenue. (the demnd function is generlly known s Averge revenue function). Totl revenue d( Q bq ) TR P.Q Q. ( - bq) nd mrginl revenue MR dq Growth rte nd reltive growth rte The growth of the plnt is usully mesured in terms of dry mter production nd s denoted by W. Growth is function of time t nd is denoted by Wg(t) it is clled growth function. Here t is the independent vrible nd w is the dependent vrible. dw dw dw The derivtive is the growth rte (or) the bsolute growth rte gr. GR dt dt dt The reltive growth rte i.e defined s the bsolute growth rte divided by the totl dry mtter production nd is denoted by RGR. dw i.e RGR. w dt bsolutegrowthrte totl dry mtter production Problem. If G t b sin t 5 is the growth function function the growth rte nd reltive growth rte. dg GR dt RGR. dg G dt Implicit Functions If the vribles x nd y re relted with ech other such tht f (x, y) 0 then it is clled Implicit function. A function is sid to be explicit when one vrible cn be expressed completely in terms of the other vrible. For exmple, y x x x is n Explicit function xy y x 0 is n implicit function

32 Problem gives For exmple, the implicit eqution xy cn be solved by differentiting implicitly d ( xy) d () Implicit differentition is especilly useful when y (x)is needed, but it is difficult or inconvenient to solve for y in terms of x. Exmple: Differentite the following function with respect to x Solution So, just differentite s norml nd tck on n pproprite derivtive t ech step. Note s well tht the first term will be product rule. Exmple: Find for the following function. Solution In this exmple we relly re going to need to do implicit differentition of x nd write y s y(x). Notice tht when we differentited the y term we used the chin rule. Exmple: Find for the following. Solution First differentite both sides with respect to x nd notice tht the first time on left side will be product rule. Remember tht very time we differentite y we lso multiply tht term by since we re just using the chin rule. Now solve for the derivtive.

33 The lgebr in these cn be quite messy so be creful with tht. Exmple: Find for the following Here we ve got two product rules to del with this time. Notice the derivtive tcked onto the secnt. We differentited y to get to tht point nd so we needed to tck derivtive on. Now, solve for the derivtive. Logrithmic Differentition For some problems, first by tking logrithms nd then differentiting, (i) (ii) it is esier to find dy. Such process is clled Logrithmic differentition. If the function ppers s product of mny simple functions then by tking logrithm so tht the product is converted into sum. It is now esier to differentite them. If the vrible x occurs in the exponent then by tking logrithm it is reduced to fmilir form to differentite. Exmple: Differentite the function.

34 Solution Differentiting this function could be done with product rule nd quotient rule. We cn simplify things somewht by tking logrithms of both sides. Exmple Differentite Solution First tke the logrithm of both sides s we did in the first exmple nd use the logrithm properties to simplify things little. Differentite both sides using implicit differentition. As with the first exmple multiply by y nd substitute bck in for y. PARAMETRIC FUNCTIONS Sometimes vribles x nd y re expressed in terms of third vrible clled dy prmeter. We find without eliminting the third vrible. Let x f(t) nd y g(t) then 4

35 dy dy dt dt Problem dy dt dt dy dt dt. Find for the prmetric function x cos θ, y b sinθ Solution sinθ dy b cosθ dθ dθ dy dy dθ dθ b cosθ sinθ b cotθ Inference of the differentition dy Let y f(x) be given function then the first order derivtive is. The geometricl mening of the first order derivtive is tht it represents the slope of the curve y f(x) t x. The physicl mening of the first order derivtive is tht it represents the rte of chnge of y with respect to x. PROBLEMS ON HIGHER ORDER DIFFERENTIATION dy The rte of chnge of y with respect x is denoted by nd clled s the first order derivtive of function y with respect to x. 5

36 The first order derivtive of y with respect to x is gin function of x, which gin be differentited with respect to x nd it is clled second order derivtive of y f(x) nd is denoted by d y which is equl to d dy In the similr wy higher order differentition cn be defined. Ie. The nth order derivtive of yf(x) cn be obtined by differentiting n- th derivtive of yf(x) n n d y d d y where n,,4,5. n n Problem Find the first, second nd third derivtive of. y x b e. y log(-bx). y sin (xb) Prtil Differentition So fr we considered the function of single vrible y f(x) where x is the only independent vrible. When the number of independent vrible exceeds one then we cll it s the function of severl vribles. Exmple z f(x,y) is the function of two vribles x nd y, where x nd y re independent vribles. Uf(x,y,z) is the function of three vribles x,y nd z, where x, y nd z re independent vribles. In ll these functions there will be only one dependent vrible. Consider function z f(x,y). The prtil derivtive of z with respect to x denoted by z nd is obtined by differentiting z with respect to x keeping y s constnt. x Similrly the prtil derivtive of z with respect to y denoted by z nd is obtined by y differentiting z with respect to y keeping x s constnt. Problem. Differentite U log (xbycz) prtilly with respect to x, y & z We cn lso find higher order prtil derivtives for the function z f(x,y) s follows 6

37 (i) The second order prtil derivtive of z with respect to x denoted s z obtined by prtilly differentiting x second order prtil derivtive of z with respect to x. z z x x x with respect to x. this is lso known s direct (ii)the second order prtil derivtive of z with respect to y denoted s z z y y y is is z obtined by prtilly differentiting y with respect to y this is lso known s direct second order prtil derivtive of z with respect to y (iii) The second order prtil derivtive of z with respect to x nd then y denoted s z z y x y x z is obtined by prtilly differentiting x with respect to y. this is lso known s mixed second order prtil derivtive of z with respect to x nd then y iv) The second order prtil derivtive of z with respect to y nd then x denoted s z z x y x y z is obtined by prtilly differentiting y with respect to x. this is lso known s mixed second order prtil derivtive of z with respect to y nd then x. In similr wy higher order prtil derivtives cn be found. Problem Find ll possible first nd second order prtil derivtives of ) z sin(x by) ) u xy yz zx Homogeneous Function A function in which ech term hs the sme degree is clled homogeneous function. Exmple ) x xy y 0 homogeneous function of degree. ) x 4y 0 homogeneous function of degree. ) x x y xy y 0 homogeneous function of degree. To find the degree of homogeneous function we proceed s follows: Consider the function f(x,y) replce x by tx nd y by ty if f (tx, ty) t n f(x, y) then n gives the degree of the homogeneous function. This result cn be extended to ny number of vribles. 7

38 Problem Find the degree of the homogeneous function. f(x, y) x xy y. f(x,y) x y x y Euler s theorem on homogeneous function If U f(x,y,z) is homogeneous function of degree n in the vribles x, y & z then u u u x. y z n. u x y z Problem Verify Euler s theorem for the following function. u(x,y) x xy y. u(x,y) x y z xyz INCREASING AND DECREASING FUNCTION Incresing function A function y f(x) is sid to be n incresing function if f(x ) < f(x ) for ll x < x. The condition for the function to be incresing is tht its first order derivtive is lwys greter thn zero. dy i.e >0 Decresing function A function y f(x) is sid to be decresing function if f(x ) > f(x ) for ll x < x. The condition for the function to be decresing is tht its first order derivtive is lwys less thn zero. dy i.e < 0 Problems. Show tht the function y x x is incresing for ll x.. Find for wht vlues of x is the function y 8 x x is incresing or decresing? 8

39 Mxim nd Minim Function of single vrible A function y f(x) is sid to hve mximum t x if f() > f(x) in the neighborhood of the point x nd f() is the mximum vlue of f(x). The point x is lso known s locl mximum point. A function y f(x) is sid to hve minimum t x if f() < f(x) in the neighborhood of the point x nd f() is the minimum vlue of f(x). The point x is lso known s locl minimum point. The points t which the function ttins mximum or minimum re clled the turning points or sttionry points A function yf(x) cn hve more thn one mximum or minimum point. Mximum of ll the mximum points is clled Globl mximum nd minimum of ll the minimum points is clled Globl minimum. A point t which neither mximum nor minimum is clled Sddle point. [Consider function y f(x). If the function increses upto prticulr point x nd then decreses it is sid to hve mximum t x. If the function decreses upto point x b nd then increses it is sid to hve minimum t point xb.] The necessry nd the sufficient condition for the function yf(x) to hve mximum or minimum cn be tbulted s follows Mximum Minimum First order or necessry dy 0 condition Second order or sufficient d y < 0 condition dy 0 d y > 0 Working Procedure dy d y. Find nd dy. Equte 0 nd solve for x. this will give the turning points of the function.. Consider turning point x then substitute this vlue of x in d y nd find the nture of the second derivtive. If d y t x < 0, then the function hs mximum 9

40 vlue t the point x. If d y t x > 0, then the function hs minimum vlue t the point x. 4. Then substitute x in the function y f(x) tht will give the mximum or minimum vlue of the function t x. Problem Find the mximum nd minimum vlues of the following function. y x x 40

41 Differentil Equtions Differentil eqution is n eqution in which differentil coefficients occur. A differentil eqution is of two types () Ordinry differentil eqution () Prtil differentil eqution An ordinry differentil eqution is one which contins single independent vrible. Exmple: dy sin x dy 4y x e A prtil differentil eqution is one contining more thn one independent vrible. Exmples z z. x y z x y z z. 0 x y Here we del with only ordinry differentil equtions. Definitions Order ppering in it. dy The order of differentil eqution is the order of the highest order derivtive 4y e x Order d y dy 4 y 0 Order - Degree The degree of differentil eqution is defined s the degree of highest ordered derivtive occurring in it fter removing the rdicl sign. Exmple Give the degree nd order of the following differentil eqution. dy ) 5 (xy) xy x degree -, order - 4

42 ) d y dy - 6 xy 0 degree -, order - d y ) dy Squring on both sides d y 9 degree -, order 4) dy dy d y dy dy dy degree, order Note eqution. dy 6 d y If the degree of the differentil equtions is one. It is clled liner differentil Formtion of differentil equtions Given the solution of differentil eqution, we cn form the corresponding differentil eqution. Suppose the solution contins one rbitrry constnt then differentite the solution once with respect to x nd eliminting the rbitrry constnt from the two equtions. We get the required eqution. Suppose the solution contins two rbitrry constnt then differentite the solution twice with respect to x nd eliminting the rbitrry constnt between the three equtions. Solution of differentil equtions (i) Vrible seprble method, (ii) Homogenous differentil eqution iii) Liner differentil eqution Vrible seprble method dy Consider differentil eqution f(x) 4

43 Here we seprte the vribles in such wy tht we tke the terms contining vrible x on one side nd the terms contining vrible y on the other side. Integrting we get the solution. Note The following formule re useful in solving the differentil equtions (i) d(xy) xdy y x y xdy d y y (ii) y xdy y d x x (iii) Homogenous differentil eqution dy Consider differentil eqution of the form f f ( x, y) ( x, y) (i) where f nd f re homogeneous functions of sme degree in x nd y. Here put y vx dy dv v x (ii) Substitute eqution(ii) in eqution (i) it reduces to differentil eqution in the vribles v nd x. Seprting the vribles nd integrting we cn find the solution. Liner differentil eqution dy A liner differentil eqution of the first order is of the form py Q, Where p nd Q re functions of x only. To solve this eqution first we find the integrting fctor given by Integrting fctor I.F e p Then the solution is given by ye p p Qe c where c is n rbitrry constnt. 4

44 Circles A circle is defined s the locus of the point, which moves in such wy, tht its distnce from fixed point is lwys constnt. The fixed point is clled centre of the circle nd the constnt distnce is clled the rdius of the circle. The eqution of the circle when the centre nd rdius re given Let C (h,k) be the centre nd r be the rdius of the circle. Let P(x,y) be ny point on the circle. CP r CP r (x-h) (y-k) r is the required eqution of the circle. Y r P(x,y) C(h,k) X O Note : If the center of the circle is t the origin i.e., C(h,k)(0,0) then the eqution of the circle is x y r The generl eqution of the circle is x y gx fy c 0 Consider the eqution x y gx fy c 0. This cn be written s x y gx fy g f g f c (i.e) x gx g y fy f g f c (x g) (y f ) g f c [ x ( g) ] [ y ( f )] This is of the form (x-h) (y-k) r g f c The considered eqution represents circle with centre (-g,-f) nd rdius g f c The generl eqution of the circle is x y gx fy c 0 where c The Center of the circle whose coordintes re (-g,-f) r The rdius of the circle g f c 44

45 Note The generl second degree eqution x by hxy gx fy c 0 Represents circle if (i) b i.e., coefficient of x coefficient of y (ii) h 0 i.e., no xy term 45

46 CHAIN RULE DIFFERENTIATION If y is function of u ie y f(u) nd u is function of x ie u g(x) then y is relted to x through the intermedite function u ie y f(g(x) ) y is differentible with respect to x Furthermore, let yf(g(x)) nd ug(x), then dy dy du du There re number of relted results tht lso go under the nme of "chin rules." For exmple, if yf(u) ug(v), nd vh(x), Problem then dy dy du dv.. du dv Differentite the following with respect to x. y (x 4). y e x Mrginl Anlysis Let us ssume tht the totl cost C is represented s function totl output q. (i.e) C f(q). dc Then mrginl cost is denoted by MC dq TC The verge cost Q Similrly if U u(x) is the utility function of the commodity x then du the mrginl utility MU The totl revenue function TR is the product of quntity demnded Q nd the price P per unit of tht commodity then TR Q.P f(q) dr Then the mrginl revenue denoted by MR is given by dq The verge revenue Q TR Problem. If the totl cost function is C Q - Q 5Q. Find Mrginl cost nd verge cost. 46

47 Solution dc MC dq TC AC Q. The demnd function for commodity is P ( - bq). Find mrginl revenue. (the demnd function is generlly known s Averge revenue function). Totl revenue d( Q bq ) TR P.Q Q. ( - bq) nd mrginl revenue MR dq Growth rte nd reltive growth rte The growth of the plnt is usully mesured in terms of dry mter production nd s denoted by W. Growth is function of time t nd is denoted by Wg(t) it is clled growth function. Here t is the independent vrible nd w is the dependent vrible. dw dw dw The derivtive is the growth rte (or) the bsolute growth rte gr. GR dt dt dt The reltive growth rte i.e defined s the bsolute growth rte divided by the totl dry mtter production nd is denoted by RGR. dw i.e RGR. w dt bsolutegrowthrte totl dry mtter production Problem. If G t b sin t 5 is the growth function function the growth rte nd reltive growth rte. dg GR dt RGR. dg G dt Implicit Functions If the vribles x nd y re relted with ech other such tht f (x, y) 0 then it is clled Implicit function. A function is sid to be explicit when one vrible cn be expressed completely in terms of the other vrible. For exmple, y x x x is n Explicit function xy y x 0 is n implicit function 47

48 Problem gives For exmple, the implicit eqution xy cn be solved by differentiting implicitly d ( xy) d () Implicit differentition is especilly useful when y (x)is needed, but it is difficult or inconvenient to solve for y in terms of x. Exmple: Differentite the following function with respect to x Solution So, just differentite s norml nd tck on n pproprite derivtive t ech step. Note s well tht the first term will be product rule. Exmple: Find for the following function. Solution In this exmple we relly re going to need to do implicit differentition of x nd write y s y(x). Notice tht when we differentited the y term we used the chin rule. Exmple: Find for the following. Solutio: First differentite both sides with respect to x nd notice tht the first time on left side will be product rule. Remember tht very time we differentite y we lso multiply tht term by since we re just using the chin rule. Now solve for the derivtive. 48

49 The lgebr in these cn be quite messy so be creful with tht. Exmple: Find for the following Here we ve got two product rules to del with this time. Notice the derivtive tcked onto the secnt. We differentited y to get to tht point nd so we needed to tck derivtive on. Now, solve for the derivtive. Logrithmic Differentition For some problems, first by tking logrithms nd then differentiting, dy it is esier to find. Such process is clled Logrithmic differentition. (i) If the function ppers s product of mny simple functions then by tking logrithm so tht the product is converted into sum. It is now esier to differentite them. (ii) If the vrible x occurs in the exponent then by tking logrithm it is reduced to fmilir form to differentite. Exmple Differentite the function. Solution Differentiting this function could be done with product rule nd quotient rule. We cn simplify things somewht by tking logrithms of both sides. 49

50 Exmple Differentite Solution First tke the logrithm of both sides s we did in the first exmple nd use the logrithm properties to simplify things little. Differentite both sides using implicit differentition. As with the first exmple multiply by y nd substitute bck in for y. PARAMETRIC FUNCTIONS Sometimes vribles x nd y re expressed in terms of third vrible clled dy prmeter. We find without eliminting the third vrible. Let x f(t) nd y g(t) then dy dy dt dt 50

51 Problem dy dt dt dy dt dt. Find for the prmetric function x cos θ, y b sinθ Solution sinθ dy b cosθ dθ dθ dy dy dθ dθ b cosθ sinθ b cotθ Inference of the differentition dy Let y f(x) be given function then the first order derivtive is. The geometricl mening of the first order derivtive is tht it represents the slope of the curve y f(x) t x. The physicl mening of the first order derivtive is tht it represents the rte of chnge of y with respect to x. PROBLEMS ON HIGHER ORDER DIFFERENTIATION dy The rte of chnge of y with respect x is denoted by nd clled s the first order derivtive of function y with respect to x. The first order derivtive of y with respect to x is gin function of x, which gin be differentited with respect to x nd it is clled second order derivtive of y f(x) d y d dy nd is denoted by which is equl to In the similr wy higher order differentition cn be defined. Ie. The nth order derivtive of yf(x) cn be obtined by differentiting n- th derivtive of yf(x) 5

52 n d y n d Problem n d y where n,,4,5. n Find the first, second nd third derivtive of. y x b e. y log(-bx). y sin (xb) Prtil Differentition So fr we considered the function of single vrible y f(x) where x is the only independent vrible. When the number of independent vrible exceeds one then we cll it s the function of severl vribles. Exmple z f(x,y) is the function of two vribles x nd y, where x nd y re independent vribles. Uf(x,y,z) is the function of three vribles x,y nd z, where x, y nd z re independent vribles. In ll these functions there will be only one dependent vrible. Consider function z f(x,y). The prtil derivtive of z with respect to x denoted by z nd is obtined by differentiting z with respect to x keeping y s constnt. x Similrly the prtil derivtive of z with respect to y denoted by z nd is obtined by y differentiting z with respect to y keeping x s constnt. Problem. Differentite U log (xbycz) prtilly with respect to x, y & z We cn lso find higher order prtil derivtives for the function z f(x,y) s follows (i) The second order prtil derivtive of z with respect to x denoted s z obtined by prtilly differentiting x second order prtil derivtive of z with respect to x. z z x x x with respect to x. this is lso known s direct is 5

53 (ii)the second order prtil derivtive of z with respect to y denoted s z z y y y is z obtined by prtilly differentiting y with respect to y this is lso known s direct second order prtil derivtive of z with respect to y (iii) The second order prtil derivtive of z with respect to x nd then y denoted s z z y x y x z is obtined by prtilly differentiting x with respect to y. this is lso known s mixed second order prtil derivtive of z with respect to x nd then y iv) The second order prtil derivtive of z with respect to y nd then x denoted s z z x y x y z is obtined by prtilly differentiting y with respect to x. this is lso known s mixed second order prtil derivtive of z with respect to y nd then x. In similr wy higher order prtil derivtives cn be found. Problem Find ll possible first nd second order prtil derivtives of ) z sin(x by) ) u xy yz zx Homogeneous Function A function in which ech term hs the sme degree is clled homogeneous function. Exmple ) x xy y 0 homogeneous function of degree. ) x 4y 0 homogeneous function of degree. ) x x y xy y 0 homogeneous function of degree. To find the degree of homogeneous function we proceed s follows. Consider the function f(x,y) replce x by tx nd y by ty if f (tx, ty) t n f(x, y) then n gives the degree of the homogeneous function. This result cn be extended to ny number of vribles. Problem Find the degree of the homogeneous function. f(x, y) x xy y 5

54 . f(x,y) x y x y Euler s theorem on homogeneous function If U f(x,y,z) is homogeneous function of degree n in the vribles x, y & z then u u u x. y z n. u x y z Problem Verify Euler s theorem for the following function. u(x,y) x xy y. u(x,y) x y z xyz INCREASING AND DECREASING FUNCTION Incresing function A function y f(x) is sid to be n incresing function if f(x ) < f(x ) for ll x < x. The condition for the function to be incresing is tht its first order derivtive is lwys greter thn zero. i.e dy >0 Decresing function A function y f(x) is sid to be decresing function if f(x ) > f(x ) for ll x < x. The condition for the function to be decresing is tht its first order derivtive is lwys less thn zero. dy i.e < 0 Problems. Show tht the function y x x is incresing for ll x.. Find for wht vlues of x is the function y 8 x x is incresing or decresing? Mxim nd Minim Function of single vrible A function y f(x) is sid to hve mximum t x if f() > f(x) in the neighborhood of the point x nd f() is the mximum vlue of f(x). The point x is lso known s locl mximum point. A function y f(x) is sid to hve minimum t x if f() < f(x) in the neighborhood of the point x nd f() is the minimum vlue of f(x). The point x is lso known s locl minimum point. 54

55 The points t which the function ttins mximum or minimum re clled the turning points or sttionry points A function yf(x) cn hve more thn one mximum or minimum point. Mximum of ll the mximum points is clled Globl mximum nd minimum of ll the minimum points is clled Globl minimum. A point t which neither mximum nor minimum is clled Sddle point. [Consider function y f(x). If the function increses upto prticulr point x nd then decreses it is sid to hve mximum t x. If the function decreses upto point x b nd then increses it is sid to hve minimum t point xb.] The necessry nd the sufficient condition for the function yf(x) to hve mximum or minimum cn be tbulted s follows Mximum First order or necessry dy 0 condition Second order or sufficient d y < 0 condition Minimum dy 0 d y > 0 55

56 Working Procedure dy d y. Find nd dy. Equte 0 nd solve for x. this will give the turning points of the function.. Consider turning point x then substitute this vlue of x in d y nd find the nture of the second derivtive. If d y t x < 0, then the function hs mximum vlue t the point x. If d y t x > 0, then the function hs minimum vlue t the point x. 4. Then substitute x in the function y f(x) tht will give the mximum or minimum vlue of the function t x. Problem Find the mximum nd minimum vlues of the following function. y x x 56

57 INTEGRATION d Integrtion is process, which is inverse of differentition. As the symbol represents differentition with respect to x, the symbol stnds for integrtion with respect to x. Definition d If [ f ( x) ] F( x) F x) f ( x) c then f(x) is clled the integrl of F(x) denoted by (. This cn be red it s integrl of F(x) with respect to x is f(x) c where c is n rbitrry constnt. The integrl nd the function F(x) s integrnd. F x) ( is known s Indefinite integrl Formul on integrtion ). n n x x c ( n -) n ). log x c x ). xc x x 4). log c 5). e x e x c ( u ( ) ( )) ( ) v ( x) 6). x v x u x 7). u( x) cv( x)) cu( x) ± ( c ± c v( x) 8). c c x d 9). sin x cos x c 0). cos x sin x c 57

58 ). sec x tn x c ). cosec x cot x c sec tn sec ). x x x c 4). cos ecxcot x cosec x c ). x tn x c x 4). log x x c 5). x log x x tn x 6). x c c Definite integrl If f(x) is indefinite integrl of F(x) with respect to x then the Integrl b F x) ( is clled definite integrl of F(x) with respect to x from x to x b. Here is clled the Lower limit nd b is clled the Upper limit of the integrl. b F x) ( [ f x) ] b ( f(upper limit ) - f(lower limit) f(b) - f() Note While evluting definite integrl no constnt of integrtion is to be dded. Tht is definite integrl hs definite vlue. Method of substitution Method Formule for the functions involving (x b) Consider the integrl 58

59 I ( x b ) n () Where nd b re constnts Put x b y Differentiting with respect to x 0 dy dy Substituting in () I y n. dy c y n. dy c n y c n ( x b) n n c Similrly this method cn be pplied for other formule lso. Method II Integrls of the functions of the form f put n n ( x ) x n x y, dy nx n x n dy n Substituting we get ( ) n dy I f y nd this cn be integrted. Method III Integrls of function of the type 59

60 n [ f ( x )] f ( x) when n -, put f(x) y then f ( x) dy n [ f ( x )] f ( x) dy y n y n n [ f ( x) ] n n when n -, the integrl reduces to f f ( x) ( x) putting y f(x) then dy f (x) dy log y log f(x) y Method IV Method of Prtil Frctions Integrls of the form x bx c Cse. If denomintor cn be fctorized into liner fctors then we write the integrnd s the sum or difference of two liner fctors of the form A B ( x bx c ( x b)( cx d) x b cx d Cse- In the given integrl x bx c the denomintor x bx c cn not be fctorized into liner fctors, then express x bx c s the sum or difference of two perfect squres nd then pply the formule x x tn log x x x 60

61 x log x x Integrls of the form x bx c Write denomintor s the sum or difference of two perfect squres or x bx c x nd then pply the formul log(x x ) x log(x x ) x x sin x Integrtion by prts x or x If the given integrl is of the form udv then this cn not be solved by ny of techniques studied so fr. To solve this integrl we first tke the product rule on differentition d ( uv) dv du u v Integrting both sides we get d ( uv) then we hve ( u dv du v ) u v vdu udv re rrnging the terms we get udv uv- vdu This formul is known s integrtion by prts formul Select the functions u nd dv ppropritely in such wy tht integrl vdu cn be more esily integrble thn the given integrl 6

62 APPLICATION OF INTEGRATION The re bounded by the function yf(x), xxis nd the ordintes t x xb is given by b A f ( x) 6

63 Definition INVERSE OF A MATRIX Let A be ny squre mtrix. If there exists nother squre mtrix B Such tht AB BA I (I is unit mtrix) then B is clled the inverse of the mtrix A nd is denoted by A -. The cofctor method is used to find the inverse of mtrix. Using mtrices, the solutions of simultneous equtions re found. Working Rule to find the inverse of the mtrix Step : Find the determinnt of the mtrix. Step : If the vlue of the determinnt is non zero proceed to find the inverse of the mtrix. Step : Find the cofctor of ech element nd form the cofctor mtrix. Step 4: The trnspose of the cofctor mtrix is the djoint mtrix. dj ( A) Step 5: The inverse of the mtrix A - A Exmple Find the inverse of the mtrix 4 9 Solution Let A 4 9 Step A 4 9 (8 ) (9 ) (4 ) Step The vlue of the determinnt is non zero A - exists. Step Let A ij denote the cofctor of ij in A 6

64 ( ) Cofctor of A ( ) 6 ) (9 9 Cofctor of A ( ) 4 4 Cofctor of A ( ) 5 4) (9 9 4 Cofctor of A ( ) Cofctor of A ( ) ) (4 4 Cofctor of A ( ) Cofctor of A ( ) ) ( 4 Cofctor of A ( ) 9 Cofctor of A Step 4 The mtrix formed by cofctors of element of determinnt A is dj A Step A A dj A 4 5 Mthemtics 64

65 SOLUTION OF LINEAR EQUATIONS Let us consider system of liner equtions with three unknowns x b y c z d x b y c z d x b y c z d The mtrix form of the eqution is AXB where A b c is x mtrix x d X y nd B d z d Here AX B Pre multiplying both sides by A -. (A - A)X A - B We know tht A - A A A - I I X A - B since IX X Hence the solution is X A - B. Exmple Solve the x y z, x 5y 6z 4, 9x 6y 6z 6 by mtrix method. Solution The given equtions re x y z, x 5y 6z 4, 9x 6y 6z 6 b b c c Let A , X 6 x y, B z 4 6 The given system of equtions cn be put in the form of the mtrix eqution AXB A (80 56) (08 54) (78 45)

66 The vlue of the determinnt is non zero A - exists. Let A ij (i, j,,) denote the cofctor of ij in A 5 6 A Cofctor of 6 6 ( ) A Cofctor of 9 6 ( ) (08 54) 54 5 A Cofctor of 9 6 ( ) A Cofctor of 6 6 ( ) (6 6) 0 A Cofctor of ( ) A Cofctor of ( ) (6 9) 7 A Cofctor of ( ) 6 5 A Cofctor of 6 6 ( ) (6 ) A Cofctor of 6 5 ( ) 5 The mtrix formed by cofctors of element of determinnt A is dj A A 4 dj A 54 A We Know tht XA - B

67 5 z y x )4 (. )6 ( ) (.6 0)4 ( x 0, y, z -. SOLUTION BY DETERMINANT (CRAMER'S RULE) Let the equtions be. () Consider the determinnt c b c b c b c b d c b d c b d x c d c d c d y d b d b d b z When 0, the unique solution is given by z y x z y x,, Exmple Solve the equtions x y 5z, x y 4z 6, 6x y 7z 47 by determinnt method (Crmer s Rule). Solution The equtions re x y 5z, x y 4z 6, 6x y 7z 47 Mthemtics 67

68 ; 6 4 4; y x z By Crmer s rule x x y y z z x 4, y, z

69 MATRICES An rrngement of numbers in rows nd columns. A mtrix of type (m x n) is defined s rrngement of (m x n) numbers in m rows & n columns. Usully these numbers re enclosed within squre brckets [ ] (or) simple brckets ( ) re denoted by cpitl letters A, B, C etc. Exmple 4 A B Here A is of type x 4 & B is of type 4 x Types of mtrices. Row mtrix: It is mtrix contining only one row nd severl columns.it is lso clled s row vector. Exmple: [ ] ( x 5) mtrix clled row vector.. Column mtrix: It is mtrix contining only one column. It is lso known s column vector. Exmple: 4 (x). Squre mtrix: A mtrix is clled s squre mtrix, if the number of rows is equl to number of columns Exmple The elements,, etc fll long the digonl & this is clled leding digonl (or) principl digonl of the mtrix. 69

70 4. Trce of the mtrix It is defined s the sum of the elements long the leding digonl. In this bove mtrix the trce of the mtrix is Digonl mtrix It is squre mtrix in which ll the elements other thn in the leding digonls re zero s. Eg: Sclr mtrix It is digonl mtrix in which ll the elements in the leding digonl re sme. Eg: Unit mtrix or identify mtrix It is digonl mtrix, in which the elements long the leding digonl re equl to one. It is denoted by I I Zero mtrix (or) Non-mtrix It is mtrix ll of whole elements re equl to zero denoted by O Eg: O x O x Tringulr mtrix There re two types.. Lower Tringulr Mtrix. Upper Tringulr Mtrix. 70

71 Lower Tringulr mtrix It is squre mtrix in which ll the elements bove the leding digonl re zeros. Eg: Upper Tringulr mtrix Squre mtrix in which ll the elements below the leding digonl re zeros Eg: Symmetric mtrix A squre mtrix ij ij A { ij } sid i to n ; to symmetric, if for ll i nd j. j to n sid Eg: Skew symmetric mtrix A squre mtrix A { ij } i to ij n, -j ji to n is clled skew symmetric, if for ll i & j. Here ii 0 for ll i Eg: Algebr of mtrices. Equlity of mtrices Two mtrices A & B re equl, if nd only if, (i) Both A & B re of the sme type (ii) Every element of B is the sme s the corresponding element of A. 7

72 Exmple. 4 A B Here order of mtrix A is not sme s order mtrix B, the two mtrices re not equl. A B. Find the vlue of nd b given b Solution: The given mtrices re equl, b. Addition of mtrices Two mtrices A & B cn be dded if nd only if, (i) Both re of the sme type. (ii) The resulting mtrix of A & B is lso of sme type nd is obtined by dding the ll elements of A to the corresponding elements of B. Exmple 4 5. Find 5 6 Solution Subtrction of the mtrices This cn be done, when both the mtrices re of sme type. (A-B) is obtined by subtrcting the elements of A with corresponding elements of B. Exmple 4 5. Find

73 Solution Multipliction of mtrix They re of two types :. By sclr K B.. By mtrix A x B. i) Sclr multipliction To multiply mtrix A by sclr K, then multiply every element of mtrix A by tht sclr. Exmple 4 5. Find 5 6 Solution: ii) Mtrix Multipliction Two mtrices A & B cn be multiplied to form the mtrix product AB, if nd only if the number of columns of st mtrix A is equl to the number of rows of nd mtrix B. If A is n (m x p) nd B is n (p x n) then the mtrix product AB cn be formed. AB is mtrix by (m x n). In this cse the mtrices A nd B re sid to be conformble for mtrix multipliction. Exmple. Find Solution

74 8 4 Note: The mtrix product AB is different from the mtrix product BA.. The mtrix AB cn be formed but not BA Eg: A is ( x ) mtrix B is ( x 5) mtrix AB lone cn be formed nd it is ( x 5) mtrix.. Even if AB & BA cn be formed, they need not be of sme type. Eg: A is ( x ) mtrix B is ( x ) mtrix AB cn be formed nd is ( x ) mtrix BA cn be formed nd is (x ) mtrix. Even if AB & BA re of the sme type, they needn t be equl. Becuse, they need not be identicl. Eg: A is ( x ) mtrix B is ( x ) mtrix AB is ( x ) mtrix BA is (x ) mtrix AB BA The multipliction of ny mtrix with null mtrix the resultnt mtrix is lso null mtrix. When ny mtrix (ie.) A is multiplied by unit mtrix; the resultnt mtrix is A itself. Trnspose of mtrix The Trnspose of ny mtrix ( A ) is obtined by interchnging the rows & columns of A nd is denoted by A T. If A is of type (m x n), then A T is of type (n x m). Eg: A A T 0 5 ( x ) ( x ) 74 6

75 Properties of trnspose of mtrix ) (A T ) T A ) (AB) T B T A T is known s the reversl Lw of Trnspose of product of two mtrices. DETERMINANTS Every squre mtrix A of order n x n with entries rel or complex there exists number clled the determinnt of the mtrix A denoted by by Aor det (A). The determinnt formed by the elements of A is sid to be the determinnt of the mtrix A.. Consider the nd order determinnt. b A b b b Eg: 4 A Consider the rd order determinnt, b c A b c b c This cn be expnded long ny row or ny column. Usully we expnd by the st row. On expnding long the st row A b c c b b c - b c c b Minors Let A ( ij )be determinnt of order n. The minor of the element ij is the determinnt formed by deleting i th row nd j th column in which the element belongs nd i j the cofctor of the element is Aij ( i) M ij where M is the minor of i th row nd j th column. Exmple Clculte the determinnt of the following mtrices. 75 7

76 () 4 Solution ( ) ( 4) (9 ) 4 4 nd it is clled non- Singulr nd Non-Singulr Mtrices: Definition A squre mtrix A is sid to be singulr if, A 0 singulr if A 0. Note Only squre mtrices hve determinnts. 4 Exmple: Find the solution for the mtrix A ( 0) 4(5 0) (5 7) Here A 0.So the given mtrix is singulr Properties of determinnts. The vlue of determinnt is unltered by interchnging its rows nd columns. Exmple Let A 4 then det ( A) ( ) ( 4) (9 )

77 Let us interchnge the rows nd columns of A. Thus we get new mtrix A. Then det ( A ) ( ) ( 9) (8 6) 4 Hence det (A) det (A ).. If ny two rows (columns) of determinnt re interchnged the determinnt chnges its sign but its numericl vlue is unltered. Exmple Let A 4 then det ( A) 4 ( ) ( 4) (9 ) Let A be the mtrix obtined from A by interchnging the first nd second row. i.e R nd R. Then det ( A ) Hence det (A) - det (A ). 4 ( 9) ( ) 4( ) If two rows (columns) of determinnt re identicl then the vlue of the terminnt is zero. Exmple Let A 4 then 77 9

78 det ( A) Hence det ( A ) 0 4 ( 4) ( 4) ( ) If every element in row ( or column) of determinnt is multiplied by constnt k then the vlue of the determinnt is multiplied by k. Exmple Let A 4 then det ( A) 4 ( ) ( 4) (9 ) Let A be the mtrix obtined by multiplying the elements of the first row by (ie. here k ) then det ( A ) ( ) ( ) ( ) ( 4 Hence det (A) det (A ). 4 [ ) ( 4) (9 ) ] [ 4 4 ] ( ) 4 5. If every element in ny row (column) cn be expressed s the sum of two quntities then given determinnt cn be expressed s the sum of two determinnts of the sme order with the elements of the remining rows (columns) of both being the sme. Exmple Let A then 78 0

79 det ( A) det ( M) det ( M ) det ( M) ( ) ( 4) (9 ) 4 4 det ( M ) det ( A) det ( M) det ( M ) Hence ( ) 4( 4) 6(9 ) det ( A) A determinnt is unltered when to ech element of ny row (column) is dded to those of severl other rows (columns) multiplied respectively by constnt fctors. Exmple Let A 4 then det ( A) 4 ( ) ( 4) (9 ) Let A be mtrix obtined when the elements C of A re dded to those of second column nd third column multiplied respectively by constnts nd. Then 79

80 4 4 (0) (0) ) (9 4) ( ) ( ) ( 4 (4) () () 4 () () 4 ) ( () 4 (4) () () () ) ( det A Mthemtics 80

81 PHYSICAL AND ECONOMIC OPTIMUM FOR SINGLE INPUT Let y f(x) be response function. Here x stnds for the input tht is kgs of fertilizer pplied per hectre nd y the corresponding output tht is kgs of yield per hectre. dy d y We know tht the mximum is only when 0 nd < 0. This optimum is clled physicl optimum. We re not considering the profit with respect to the investment, we re interested only in mximizing the profit. Economic optimum The optimum which tkes into considertion the mount invested nd returns is clled the economic optimum. dy P P x y where P x stnds for the per unit price of input tht is price of fertilizer per kgs. P y stnds for the per unit price of output tht is price of yield per kgs. Problem The response function of pddy is y x x where x represents kgs of nitrogen/hectre nd y represents yield in kgs/hectre. kg of pddy is Rs. nd kg of nitrogen is Rs. 5. Find the physicl nd economic optimum. Also find the corresponding yield. Solution y x x dy d y x 0. negtive vlue d y ie. < 0 Therefore the given function hs mximum point. Physicl Optimum dy 0 i.e x 0-0. x

82 4.4 x 4. 4 kgs/hectre 0. therefore the physicl optimum level of nitrogen is 4.4 kgs/hectre. Therefore the mximum yield is Y (4.4) -0.05(4.4) kgs/ hectre. Economic optimum dy Given P P x y Price of nitrogen per kg P x 5 Price of yield per kg P y dy Therefore x 5-0. x x 5.68 x 8. 4 kgs/hectre 0. therefore the economic optimum level of nitrogen is 8.4 kgs/hectre. Therefore the mximum yield is Y (8.4) -0.05(8.4) kgs/ hectre. Mxim nd Minim of severl vribles with constrints nd without constrints Consider the function of severl vribles y f (x, x.x n ) where x, x..x n re n independent vribles nd y is the dependent vrible. Working Rule Step : Find ll the first order prtil derivtives of y with respect to x, x, x x n.. y (ie) f x y x f 8

83 Step y x.... y x n f fn Find ll the second order prtil derivtives of y with respect to x, x, x.x n nd they re given s follows. y x x y x f f y x x y x y x x y x f f f f x y x f nd so on Step: Construct n Hessin mtrix which is formed by tking ll the second order prtil derivtives is given by 8

84 H f f..... f n f f f n f f f n... f... f... f n n nn H is symmetric mtrix. Step: 4 Consider the following minors of order,,. H f f H f f f f H..... f f f f f f f f f H n f f.... f n f f.... f n f f f n... f... f... f n n nn 84

85 Steps: 5 The necessry condition for finding the mximum or minimum Equte the first order derivtive to zero (i.e) f f..f n 0 nd find the vlue of x, x,..x n. Steps: 6 Substitute the vlues x, x..x n in the Hessin mtrix. Find the vlues of H H, H..., H n If H H H H 4 < 0 > 0 < 0 > 0... nd so on. Then the function is mximum t x, x..x n. If H >, H > 0, H 0... then the function is minimum 0 > t x, x. x n. Steps: 7 Conditions Mximum Minimum First f f f f n 0 f 0, f 0. f n 0 Second H < 0 H > 0 H > 0 H > 0 H < 0. H.. > 0. Note : Problem If the second order conditions re not stisfied then they re clled sddle point. Find the mxim (or) minim if ny of the following function. 4 y x x 4x 8x Solution Step : The first order prtil derivtives re () 85

86 f f y 4x x y x x 4 8 Step : The second order prtil derivtives re y f 8x x f x y x 0 f y x f x y x 0 Step : The Hessin mtrix is H f f f f H 4. Equte f, f 0 f 4x x x ± 8x 0 0 x, x - f x 80 x - 8 x - 4 The sttionry points re (,- 4) & (-, - 4) At the point (, - 4) the Hessin mtrix will be 8 0 H 0 86

87 H 8 > H 6 > 0 0 Since the determinnt H nd H re positive the function is minimum t (, - 4). The minimum vlue t x & x - 4 is obtined by substituting the vlues in () y 4 () (- 4) 4 () 8 (- 4) y y 4-0 y 4 60 The minimum vlue is At the point (-, - 4) H H < 0 H - 6 < 0 Both the conditions re not stisfied. Hence the point (-, - 4) gives sddle point. Economic Optimum f. f.... f n (ie) f For finding the Economic Optimum we equte the first order derivtive x to the inverse rtio of the unit prices. p y x p y f x p y x p y.. f n y x n p p x y n 87

88 where Px, Px, Px n nd Py re the unit prices of x, x.. x n nd y. These re the first order condition. The economic optimum & the physicl optimum differ only in the first order conditions. The other procedures re the sme. Mxim & Minim of severl vribles under certin condition with constrints. Consider the response function y f (x, x. x n ) subject to the constrint φ (x, x.. x n ) 0 The objective function is Z f(x, x, xn) λ[φ(x, x, xn)] where λ is clled the Lgrnge s multiplies. The prtil derivtives re z x i i fi z x x φ φ i x i f ij for i,.. n. i, j.,. n. i,.. n. Now form the Bordered Hessin Mtrix s follows. Bordered Hessin 0 φ φ H.. φ φ. f f f n φ... φ f... f n f... f n f n... f nn [Since this extr row & column is on the border of the mtrix we cll it s Bordered Hessin mtrix nd it is denoted by H ] Here minor s re [ H ] 0 φ φ f i, H 0 φ φ φ f f f f φ f f f n f f f n... f... f... f n n nn.so 88

89 H 0 f f φ f f f nd so on. φ f f f φ Problem φ φ φ f Conditions Mxim Minim First Order f f f.f n 0 f f f f n 0 Second Order H > H < 0, H 0... < 0, H < 0, H > H 4 < Consider consumer with simple utility function U f(x, y) 4xy y. If this consumer cn t most spend only Rs. 6/- on two goods x nd y nd if the current prices re Rs. /- per unit of x nd Rs./- per unit of y. Mximize the function. 89

90 Definition Model A mthemticl model is representtion of phenomen by mens of mthemticl equtions. If the phenomen is growth, the corresponding model is clled growth model. Here we re going to study the following models.. liner model. Exponentil model. Power model. Liner model The generl form of liner model is y bx. Here both the vribles x nd y re of degree. To fit liner model of the form ybx to the given dt. Here nd b re the prmeters (or) constnts of the model. Let (x, y ) (x, y ). (x n, y n ) be n pirs of observtions. By plotting these points on n ordinry grph sheet, we get collection of dots which is clled sctter digrm. y (x y ) * (x y ) * (x y ) 0 x There re two types of liner models (i) y bx (with constnt term) (ii) y bx (without constnt term) The grphs of the bove models re given below : 90

91 y y bx y ybx () x () x stnds for the constnt term which is the intercept mde by the line on the y xis. When 0 x 0, y ie is the intercept, b stnds for the slope of the line. Eg:. The tble below gives the DMP(kgs) of prticulr crop tken t different stges; fit liner growth model of the form wbt, nd find the vlue of nd b from the grph. t (in dys) ; DMP w: (kg/h) Exponentil model This model is of the form y e bx where nd b re constnts to be determined The grph of n exponentil model is given below. y i) b>0 y ii) b<0 0 x 0 y 9

92 Note: The bove model is lso known s semilog model. When the vlues of x nd y re plotted on semilog grph sheet we will get stright line. On the other hnd if we plot the points xnd y on n ordinry grph sheet we will get n exponentil curve. Eg:. Fit n exponentil model to the following dt. x in dys y in mg per plnt Power Model The most generl form of the power model is b y x y o x Exmple: Fit the power function for the following dt x 0 y Crop Response models The most commonly used crop response models re i) Qudrtic model ii) Squre root model Qudrtic model The generl form of qudrtic model is y b x c x 9

93 When c< 0 the curve ttins mximum t its pek. c< 0 o When c > 0 the curve ttins minimum t its pek. o C > 0 The prbolic curve bends very shrply t the mximum or minimum points. Exmple Drw curve of the form y b x c x using the following vlues of x nd y x y Squre root model The stndrd form of the squre root model is y b When c is negtive the curve ttins mximum x cx 9

94 C< 0 o The curve ttins minimum when c is positive. C > 0 At the extreme points the curve bends t slower rte Three dimensionl Anlyticl geometry Let OX,OY & OZ be mutully perpendiculr stright lines meeting t point O. The extension of these lines OX, OY nd OZ divide the spce t O into octnts(eight). Here mutully perpendiculr lines re clled X, Y nd Z co-ordintes xes nd O is the origin. The point P (x, y, z) lies in spce where x, y nd z re clled x, y nd z coordintes respectively. 94

95 Z P(x,y,z) M x O z x R Y y N where NR x coordinte, MN y coordinte nd PN z coordinte 95

96 Distnce between two points The distnce between two points A(x, y, z ) nd B(x, y, z ) is dist AB ( ) ( ) ( ) z z y y x x In prticulr the distnce between the origin O (0,0,0) nd point P(x,y,z) is OP z y x The internl nd Externl section Suppose P(x, y, z ) nd Q(x, y, z ) re two points in three dimensions. P(x,y,z ) A(x, y, z) Q(x, y, z ) The point A(x, y, z) tht divides distnce PQ internlly in the rtio m :m is given by A,, m m z m z m m m y m y m m m x m x m Similrly P(x, y, z ) nd Q(x, y, z ) re two points in three dimensions. P(x,y,z ) Q(x, y, z ) A(x, y, z) The point A(x, y, z) tht divides distnce PQ externlly in the rtio m :m is given by A,, m m z m m z m m y m y m m m x m m x If A(x, y, z) is the midpoint then the rtio is : Mthemtics 96

97 A x x y, y z, z Problem Find the distnce between the points P(,-) & Q(,,) PQ ( ) ( ) ( ) 8 Direction Cosines Let P(x, y, z) be ny point nd OP r. Let α,β,γ be the ngle mde by line OP with OX, OY & OZ. Then α,β,γ re clled the direction ngles of the line OP. cos α, cos β, cos γ re clled the direction cosines (or dc s) of the line OP nd re denoted by the symbols I, m,n. z P(x,y,z) β γ O α M y x Result By projecting OP on OY, PM is perpendiculr to y xis nd the POM β lso OM y Similrly, cos β cos α cos γ y r x r z r 97

98 x y (i.e) l, m, r r l m n n r z x y z r ( r x y z Distnce from the origin) l m n x y z x y z l m n (or) cos α cos β cos γ. Note The direction cosines of the x xis re (,0,0) The direction cosines of the y xis re (0,,0) The direction cosines of the z xis re (0,0,) Direction rtios Any quntities, which re proportionl to the direction cosines of line, re clled direction rtios of tht line. Direction rtios re denoted by, b, c. If l, m, n re direction cosines n, b, c re direction rtios then l, b m, c n (ie) kl, b km, c kn (ie) (or) l b c k (Constnt) m n l m n b c k (Constnt) To find direction cosines if direction rtios re given If, b, c re the direction rtios then direction cosines re l k l k similrly m n b k c k () 98

99 o l m n ) ( c b k (ie) ) ( c b k c b k Tking squre root on both sides K c b,, c b c n c b b m c b l Problem. Find the direction cosines of the line joining the point (,,6) & the origin. Solution By the distnce formul z y x r Direction Cosines re r l cos 7 r x m cos β 7 r y n cos γ 7 6 r z. Direction rtios of line re,4,. Find direction cosines Solution Direction rtios re,4, (ie), b 4, c Direction cosines re y z (,,6) x Mthemtics 99

100 n Note c l b c 4 69 m b c b b c ) The direction rtios of the line joining the two points A(x, y, z ) & B (x, y, z ) re (x x, y y, z z ) ) The direction cosines of the line joining two points A (x, y, z ) & B (x, y, z ) re r distnce between AB. x x y y z z,, r r r

101 CHAIN RULE DIFFERENTIATION If y is function of u ie y f(u) nd u is function of x ie u g(x) then y is relted to x through the intermedite function u ie y f(g(x) ) y is differentible with respect to x Furthermore, let yf(g(x)) nd ug(x), then dy dy du du There re number of relted results tht lso go under the nme of "chin rules." For exmple, if yf(u) ug(v), nd vh(x), Problem then dy dy du dv.. du dv Differentite the following with respect to x. y (x 4). y e x Mrginl Anlysis Let us ssume tht the totl cost C is represented s function totl output q. (i.e) C f(q). dc Then mrginl cost is denoted by MC dq TC The verge cost Q Similrly if U u(x) is the utility function of the commodity x then du the mrginl utility MU The totl revenue function TR is the product of quntity demnded Q nd the price P per unit of tht commodity then TR Q.P f(q) dr Then the mrginl revenue denoted by MR is given by dq The verge revenue Q TR 0

102 Problem. If the totl cost function is C Q - Q 5Q. Find Mrginl cost nd verge cost. Solution: dc MC dq TC AC Q. The demnd function for commodity is P ( - bq). Find mrginl revenue. (the demnd function is generlly known s Averge revenue function). Totl revenue d( Q bq ) TR P.Q Q. ( - bq) nd mrginl revenue MR Growth rte nd reltive growth rte The growth of the plnt is usully mesured in terms of dry mter production nd s denoted by W. Growth is function of time t nd is denoted by Wg(t) it is clled growth function. Here t is the independent vrible nd w is the dependent vrible. dw dw dw The derivtive is the growth rte (or) the bsolute growth rte gr. GR dt dt dt The reltive growth rte i.e defined s the bsolute growth rte divided by the totl dry mtter production nd is denoted by RGR. dw i.e RGR. w dt bsolutegrowthrte totl dry mtter production Problem. If G t b sin t 5 is the growth function function the growth rte nd reltive growth rte. dg GR dt RGR. dg G dt Implicit Functions If the vribles x nd y re relted with ech other such tht f (x, y) 0 then it is clled Implicit function. A function is sid to be explicit when one vrible cn be expressed completely in terms of the other vrible. For exmple, y x x x is n Explicit function xy y x 0 is n implicit function dq 0

103 Problem gives For exmple, the implicit eqution xy cn be solved by differentiting implicitly d ( xy) d () Implicit differentition is especilly useful when y (x)is needed, but it is difficult or inconvenient to solve for y in terms of x. Exmple: Differentite the following function with respect to x Solution So, just differentite s norml nd tck on n pproprite derivtive t ech step. Note s well tht the first term will be product rule. Exmple: Find for the following function. Solution In this exmple we relly re going to need to do implicit differentition of x nd write y s y(x). Notice tht when we differentited the y term we used the chin rule. Exmple: Find for the following. Solutio First differentite both sides with respect to x nd notice tht the first time on left side will be product rule. 0

104 Remember tht very time we differentite y we lso multiply tht term by we re just using the chin rule. Now solve for the derivtive. since The lgebr in these cn be quite messy so be creful with tht. Exmple Find for the following Here we ve got two product rules to del with this time. Notice the derivtive tcked onto the secnt. We differentited y to get to tht point nd so we needed to tck derivtive on. Now, solve for the derivtive. Logrithmic Differentition (i) For some problems, first by tking logrithms nd then differentiting, it is esier to find dy. Such process is clled Logrithmic differentition. If the function ppers s product of mny simple functions then by tking logrithm so tht the product is converted into sum. It is now esier to differentite them. (ii) If the vrible x occurs in the exponent then by tking logrithm it is reduced to fmilir form to differentite. Exmple Differentite the function. 04

105 Solution Differentiting this function could be done with product rule nd quotient rule. We cn simplify things somewht by tking logrithms of both sides. Exmple Differentite Solution First tke the logrithm of both sides s we did in the first exmple nd use the logrithm properties to simplify things little. Differentite both sides using implicit differentition. As with the first exmple multiply by y nd substitute bck in for y. PARAMETRIC FUNCTIONS Sometimes vribles x nd y re expressed in terms of third vrible clled dy prmeter. We find without eliminting the third vrible. Let x f(t) nd y g(t) then 05

106 Problem dy dy dt dt dy dt dt dy dt dt. Find for the prmetric function x cos θ, y b sinθ Solution sinθ dy b cosθ dθ dy dy dθ dθ b cosθ sinθ b cotθ dθ Inference of the differentition dy Let y f(x) be given function then the first order derivtive is. The geometricl mening of the first order derivtive is tht it represents the slope of the curve y f(x) t x. The physicl mening of the first order derivtive is tht it represents the rte of chnge of y with respect to x. PROBLEMS ON HIGHER ORDER DIFFERENTIATION dy The rte of chnge of y with respect x is denoted by nd clled s the first order derivtive of function y with respect to x. The first order derivtive of y with respect to x is gin function of x, which gin be differentited with respect to x nd it is clled second order derivtive of y f(x) d y d dy nd is denoted by which is equl to In the similr wy higher order differentition cn be defined. Ie. The nth order derivtive of yf(x) cn be obtined by differentiting n- th derivtive of yf(x) 06

107 n d y n d Problem n d y where n,,4,5. n Find the first, second nd third derivtive of. y x b e. y log(-bx). y sin (xb) Prtil Differentition So fr we considered the function of single vrible y f(x) where x is the only independent vrible. When the number of independent vrible exceeds one then we cll it s the function of severl vribles. Exmple z f(x,y) is the function of two vribles x nd y, where x nd y re independent vribles. Uf(x,y,z) is the function of three vribles x,y nd z, where x, y nd z re independent vribles. In ll these functions there will be only one dependent vrible. Consider function z f(x,y). The prtil derivtive of z with respect to x denoted by z nd is obtined by differentiting z with respect to x keeping y s constnt. x Similrly the prtil derivtive of z with respect to y denoted by z nd is obtined by y differentiting z with respect to y keeping x s constnt. Problem. Differentite U log (xbycz) prtilly with respect to x, y & z We cn lso find higher order prtil derivtives for the function z f(x,y) s follows (i) The second order prtil derivtive of z with respect to x denoted s z obtined by prtilly differentiting x second order prtil derivtive of z with respect to x. z z x x x with respect to x. this is lso known s direct is 07

108 (ii)the second order prtil derivtive of z with respect to y denoted s z z y y y is z obtined by prtilly differentiting y with respect to y this is lso known s direct second order prtil derivtive of z with respect to y (iii) The second order prtil derivtive of z with respect to x nd then y denoted s z z y x y x z is obtined by prtilly differentiting x with respect to y. this is lso known s mixed second order prtil derivtive of z with respect to x nd then y iv) The second order prtil derivtive of z with respect to y nd then x denoted s z z x y x y z is obtined by prtilly differentiting y with respect to x. this is lso known s mixed second order prtil derivtive of z with respect to y nd then x. In similr wy higher order prtil derivtives cn be found. Problem Find ll possible first nd second order prtil derivtives of ) z sin(x by) ) u xy yz zx Homogeneous Function A function in which ech term hs the sme degree is clled homogeneous function. Exmple ) x - xy y 0 homogeneous function of degree. ) x 4y 0 homogeneous function of degree. ) x x y xy y 0 homogeneous function of degree. To find the degree of homogeneous function we proceed s follows. Consider the function f(x,y) replce x by tx nd y by ty if f (tx, ty) t n f(x, y) then n gives the degree of the homogeneous function. This result cn be extended to ny number of vribles. Problem Find the degree of the homogeneous function. f(x, y) x xy y 08

109 . f(x,y) x y x y Euler s theorem on homogeneous function If U f(x,y,z) is homogeneous function of degree n in the vribles x, y & z then u u u x. y z n. u x y z Problem Verify Euler s theorem for the following function. u(x,y) x xy y. u(x,y) x y z xyz INCREASING AND DECREASING FUNCTION Incresing function A function y f(x) is sid to be n incresing function if f(x ) < f(x ) for ll x < x. The condition for the function to be incresing is tht its first order derivtive is lwys greter thn zero. dy i.e >0 Decresing function A function y f(x) is sid to be decresing function if f(x ) > f(x ) for ll x < x. The condition for the function to be decresing is tht its first order derivtive is lwys less thn zero. dy i.e < 0 Problems. Show tht the function y x x is incresing for ll x.. Find for wht vlues of x is the function y 8 x x is incresing or decresing? Mxim nd Minim Function of single vrible A function y f(x) is sid to hve mximum t x if f() > f(x) in the neighborhood of the point x nd f() is the mximum vlue of f(x). The point x is lso known s locl mximum point. 09

110 A function y f(x) is sid to hve minimum t x if f() < f(x) in the neighborhood of the point x nd f() is the minimum vlue of f(x). The point x is lso known s locl minimum point. The points t which the function ttins mximum or minimum re clled the turning points or sttionry points A function yf(x) cn hve more thn one mximum or minimum point. Mximum of ll the mximum points is clled Globl mximum nd minimum of ll the minimum points is clled Globl minimum. A point t which neither mximum nor minimum is clled Sddle point. [Consider function y f(x). If the function increses upto prticulr point x nd then decreses it is sid to hve mximum t x. If the function decreses upto point x b nd then increses it is sid to hve minimum t point xb.] The necessry nd the sufficient condition for the function yf(x) to hve mximum or minimum cn be tbulted s follows Mximum Minimum First order or necessry dy 0 condition Second order or sufficient d y < 0 condition dy 0 d y > 0 0

111 Working Procedure dy d y. Find nd dy. Equte 0 nd solve for x. this will give the turning points of the function.. Consider turning point x then substitute this vlue of x in d y nd find the nture of the second derivtive. If d y t x < 0, then the function hs mximum vlue t the point x. If d y t x > 0, then the function hs minimum vlue t the point x. 4. Then substitute x in the function y f(x) tht will give the mximum or minimum vlue of the function t x. Problem Find the mximum nd minimum vlues of the following function. y x x

112 PERMUTATION AND COMBINATION Fundmentl Counting Principle If first job cn be done in m wys nd second job cn be done in n wys then the totl number of wys in which both the jobs cn be done in succession is m x n. For exmple, consider cities Coimbtore, Chenni nd Hyderbd. Assume tht there re routes (by rod) from Coimbtore to Chenni nd 4 routes from Chenni to Hyderbd. Then the totl number of routes from Coimbtore to Hyderbd vi Chenni is x 4. This cn be explined s follows. For every route from Coimbtore to Chenni there re 4 routes from Chenni to Hyderbd. Since there re rod routes from Coimbtore to Chenni, the totl number of routes is x 4. The bove principle cn be extended s follows. If there re n jobs nd if there re m i wys in which the i th job cn be done, then the totl number of wys in which ll the n jobs cn be done in succession ( st job, nd job, rd job n th job) is given by m x m x m x m n. Permuttion Permuttion mens rrngement of things. The word rrngement is used, if the order of things is considered. Let us ssume tht there re plnts P, P, P. These plnts cn be plnted in the following 6 wys nmely P P P P P P P P P P P P P P P P P P Ech rrngement is clled permuttion. Thus there re 6 rrngements (permuttions) of plnts tking ll the plnts t time. This we write s P. Therefore P 6. Suppose out of the objects we choose only objects nd rrnge them. How mny rrngements re possible? For this consider boxes s shown in figure. I Box II Box Since we wnt to rrnge only two objects nd we hve totlly objects, the first box cn be filled by ny one of the objects, (i.e.) the first box cn be filled in wys. After

113 filling the first box we re left with only objects nd the second box cn be filled by ny one of these two objects. Therefore from Fundmentl Counting Principle the totl number of wys in which both the boxes cn be filled is x 6. This we write s P 6. In generl the number of permuttions of n objects tking r objects t time is denoted by npr. Its vlue is given by ( n )( n )...( n ) n Pr n r n ( n )( n )...( n r ) ( n r)( n r ) ( n r)( n r ) n! i. e n Pr ( n r)! Note: ) npn n! (b ) np n. (c) np 0. Exmples:. Evlute 8P Solution: 8! 8! ! 8P 6! 5! 5! ( 8 ). Evlute P Solution:!! 0 9! P 0! 9! 9! ( ). There re 6 vrieties on brinjl, in how mny wys these cn be rrnged in 6 plots which re in line? Solution Six vrieties of brinjl cn be rrnged in 6 plots in 6P 6 wys. 6P 6 6! (6 6)! 6! 0! 6! [0! ] 6 x 5 x 4 x x x 70. Therefore 6 vrieties of brinjl cn be rrnged in 70 wys.

114 4. There re 5 vrieties of roses nd vrieties of jsmine to be rrnged in row, for photogrph. In how mny wys cn they be rrnged, if (i) ll vrieties of jsmine together (ii)all vrieties of jsmine re not together. Solution i) Since the vrieties of jsmine re inseprble, consider them s one single unit. This together with 5 vrieties of roses mke 6 units which cn be rrnged themselves in 6! wys. In every one of these permuttions, vrieties of jsmine cn be rerrnged mong themselves in! wys. Hence the totl number of rrngements required 6! x! 70 x 440. ii)the number of rrngements of ll 7 vrieties without ny restrictions 7! 5040 Number of rrngements in which ll vrieties of jsmine re together 440. Therefore number of rrngements required Combintintion Combintion mens selection of things. The word selection is used, when the order of thing is immteril. Let us consider plnt vrieties V, V & V. In how mny wys vrieties cn be selected? The possible selections re ) V & V ) V & V ) V & V Ech such selection is known s combintion. There re selections possible from totl of objects tking objects t time nd we write C. In generl the number of selections (Combintions) from totl of n objects tking r objects t time is denoted by n Cr. Reltion between npr nd ncr We know tht n Pr ncr x r! npr (or) ncr r! () But we know n! npr () ( n r)! 4

115 Sub () in () we get ncr n! ( n r)!r! Another formul for ncr We know tht npr n. (n-). (n-) (n-r) ( )( ) ( ) n. n. n... n r ncr..... r Exmple. Find the vlue of 0C. Solution: ( 9 -)( 9 - ) C Note - ) nc 0 b) nc n c) ncn d) ncr ncn-r Exmples. Find the vlue of 0C 8 Solution We hve 0C 8 0C 0-8 0C How mny wys cn 4 prizes be given wy to boys, if ech boy is eligible for ll the prizes? Solution Any one prize cn be given to ny one of the boys nd hence there re wys of distributing ech prize. Hence, the 4 prizes cn be distributed in 4 8 wys.. A tem of 8 students goes on n excursion, in two crs, of which one cn ccommodte 5 nd the other only 4. In how mny wys cn they trvel? Solution There re 8 students nd the mximum number of students cn ccommodte in two 5 4

116 crs together is 9. We my divide the 8 students s follows Cse I: 5 students in the first cr nd in the second Cse II: 4 students in the first cr nd 4 in the second In Cse I: 8 students re divided into groups of 5 nd in 8 C wys. Similrly, in Cse II: 8 students re divided into two groups of 4 nd 4 in 8 C 4 wys. Therefore, the totl number of wys in which 8 students cn trvel is 8 C 8 C How mny words of 4 consonnts nd vowels cn be mde from consonnts nd 4 vowels, if ll the letters re different? Solution 4 consonnts out of cn be selected in C 4 wys. vowels cn be selected in 4 C wys. Therefore, totl number of groups ech contining 4 consonnts nd vowels C 4 * 4 C Ech group contins 7 letters, which cn be rrnging in 7! wys. Therefore required number of words 4 * 4 C * 7! 6 5

117 Physicl nd Economic Optimum for single input Let y f(x) be response function. Here x stnds for the input tht is kgs of fertilizer pplied per hectre nd y the corresponding output tht is kgs of yield per hectre. dy d y We know tht the mximum is only when 0 nd 0 This optimum is clled physicl optimum. We re not considering the profit with respect to the investment, we re interested only in mximizing the profit. Economic optimum The optimum which tkes into considertion the mount invested nd returns is clled the economic optimum. dy P P x y <. where P x stnds for the per unit price of input tht is price of fertilizer per kgs. P y stnds for the per unit price of output tht is price of yield per kgs. 7

118 Popultion Dynmics Definition Model A model is defined s physicl representtion of ny nturl phenomen Exmple:. A miniture building model.. A children cycle prk depicting the trffic signls. Disply of clothes on models in show room nd so on. Mthemticl Models A mthemticl model is representtion of phenomen by mens of mthemticl equtions. If the phenomen is growth, the corresponding model is clled growth model. Here we re going to study the following models.. Liner model. Exponentil model. Logistic model Liner model The generl form of liner model is y bx. Here both the vribles x nd y re of degree. In liner growth model, the dependent vrible is lwys the totl dry weight which is noted by w nd the independent vrible is the time denoted by t. Hence the liner growth model is given by w bt. To fit liner model of the form ybx to the given dt. Here nd b re the prmeters (or) constnts to be estimted. Let us consider (x,y ),(x, y ) (x n, y n ) be n pirs of observtions. By plotting these points on n ordinry grph sheet, we get collection of dots which is clled sctter digrm. In liner model, these points lie close to stright line. Suppose y bx is liner model to be fitted to the given dt, the expected vlues of y corresponding to x, x x n re given by (bx ), (bx ), (bx n ). The corresponding observed vlues of y re y, y y n. The difference between the observed vlue nd the expected vlue is clled residul. The Principles of lest squres sttes tht the constnts occurring in the curve of best fit should be chosen such tht the sum of the squres of the residuls must be minimum. Using this for liner model we get the following simultneous equtions in nd b, given by Σy nbσx () Σxy ΣxbΣx () 8

119 where n is the no. of observtions. Equtions nd re clled norml equtions. Given the vlues of x nd y, we cn find Σx, Σy, Σxy, Σx. Substituting in equtions () nd () we get two simultneous equtions in the constnts nd b solving which we get the vlues of nd b. Note: If the liner eqution is wbt then the corresponding norml equtions become Σw n bσt () w Σtw Σt bσt () x y x y 0 y * * x y bx t x There re two types of growth models (liner) (i) w bt (with constnt term) (ii) w bt (without constnt term) The grphs of the bove models re given below : w wbt w wbt () () t 0 9

120 stnds for the constnt term which is the intercept mde by the line on the w xis. When t0, w ie gives the initil DMP( ie. Seed weight) ; b stnds for the slope of the line which gives the growth rte. Problem The tble below gives the DMP(kgs) of prticulr crop tken t different stges; fit liner growth model of the form wbt, nd lso clculte the estimted vlue of w. t (in dys) ; DMP w: (kg/h) Exponentil model This model is of the form we bt where nd b re constnts to be determined Growth rte dw bt be dt Reltive growth rte dw. w dt be e bt bt b Here RGR b which is lso known s intensive growth rte or Mlthusin prmeter. To find the prmeters nd b in the exponentil model first we convert it into liner form by suitble trnsformtion. Now w e bt () Tking logrithm on both the sides we get log e w log e (e bt ) log e w log e log e e bt logw log e bt log e e log w log e bt Y A bt () Where Y log w nd A log e Here eqution() is liner in the vribles Y nd t nd hence we cn find the constnts A nd b using the norml equtions. ΣY na bσt ΣtY AΣt bσt After finding A by tking ntilogrithms we cn find the vlue of 0

121 Note The bove model is lso known s semilog model. When the vlues of t nd w re plotted on semilog grph sheet we will get stright line. On the other hnd if we plot the points t nd w on n ordinry grph sheet we will get n exponentil curve. Problem: Fit n exponentil model to the following dt. t in dys W in mg per plnt Logistic model (or) Logistic curve follows: The eqution of this model is given by w kt ce () Where, c nd k re constnts. The bove model cn be reduced to liner form s ce kt w ce kt w ce kt w Tking logrithm to the bse e, log w log e w log c kt log c log e Y A Bt () Where Y log e w A log e c kt B - k Now the eqution () is reduced to the liner form given by eqution () using this we cn determine the constnts A nd B from which we cn get the vlue of the constnts c nd k.

122 Problem The mximum dry weight of groundnut is 48 gms. The following tble gives the dry mtter production w of groundnut during vrious dys estimte the logistic growth model for the following dt. t in dys DMP w in gm/plnt

123 PROGRESSIONS In this section we discuss three importnt series nmely ) Arithmetic Progression (A.P), ) Geometric Progression (G.P), nd ) Hrmonic Progression (H.P) Which re very widely used in biologicl sciences nd humnities. Arithmetic Progressions Consider the sequence of numbers of the form, 4, 7,0. In this sequence the next term is formed by dding constnt with the current term. An rithmetic progression is sequence in which ech term (except the first term) is obtined from the previous term by dding constnt known s the common difference. An rithmetic series is formed by the ddition of the terms in n rithmetic progression. Let the first term on n A. P. be nd common difference d. Then, generl form of n A. P is, d, d, d,... n th term of n A. P is t n (n - ) d Sum of first n terms of n A. P is S n n/ [ (n - ) d] or n/ [ first term lst term] Exmple : Find (i) The n th term nd (ii) Sum to n terms of the A.P whose first term is nd common difference is. Answer: ) ( n ) n t n n n ) S n ( ( n ) ) ( n ) Exmple : Find the sum of the first n nturl numbers. Solution The sum of the nturl numbers is given by S n n This is A.P whose first term is nd common difference is lso one nd the lst term is n. n n S n ( First term lst term) ( n ) Exmple Find the 5 th term of the A.P 7, 7, 7,

124 Solution In the A.P 7, 7, 7, 7, d nd n 5 t n ( n ) d t 5 (5 ) d 4d 7 4(0) 47. Geometric Progression Consider the sequence of numbers ),, 4, 8, 6 b),,, In the bove sequences ech term is formed by multiplying constnt with the preceding term. For exmple, in the first sequence ech term is formed by multiplying constnt with the preceding term. Similrly the second sequence is formed by multiplying ech term by 4 to obtin the next term. Such sequence of numbers is clled Geometric progression (G.P). A geometric progression is sequence in which ech term (except the first term) is derived from the preceding term by the multipliction of non-zero constnt, which is the common rtio. The generl form of G.P is, r, r, r, Here is clled the first term nd r is clled common rtio. The n th term of the G.P is denoted by t n is given by t n n r The sum of the first n terms of G.P is given by the formul n ( r ) S n if r> r n ( r ) S n if r< r Exmples. Find the common rtio of the G.P 6, 4, 6, 54. Solution t 4 The common rtio is t 6 4

125 8. Find the 0 th term of the G.P, 5,, 5 5 Solution: 8 4 Here nd r Since t 0 n t n r we get ( ) 0 5 Sum to infinity of G.P 9 0 Consider the following G.P s ).,,, ).,,,, In the first sequence, which is G.P the common rtio is r.in the second G.P the common rtio is r. In both these cses the numericl vlue of r r <.(For the first sequence r nd the second sequence r nd both re less thn. In these equtions, ie. r < we cn find the Sum to infinity nd it is given by the form S provided -<r< r Exmples. Find the sum of the infinite geometric series with first term nd common rtio. Solution Here nd r S 4 5

126 . Find the sum of the infinite geometric series / /4 /8 /6 Solution: It is geometric series whose first term is / nd whose common rtio is /, so its sum is Hrmonic Progression ( ) Consider the sequence,,,,....this sequence is formed by d d d tking the reciprocls of the A.P, d, d, For exmple, consider the sequence,,,, Now this sequence is formed by tking the reciprocls of the terms of the A.P, 5, 8,. Such sequence formed by tking the reciprocls of the terms of the A.P is clled Hrmonic Progression (H.P). The generl form of the hrmonic progression is,,,,... d d d The n th term of the H.P is given by t n ( n ) d Note There is no formul to find the sum to n terms of H.P. Exmples. The first nd second terms of H.P re nd 5 respectively, find the 9 th term. Solution t n ( n ) d Given nd d t 9 (9 ) 6

127 (8) 9 Arithmetic men, Geometric men nd Hrmonic men The rithmetic men (A.M) of two numbers & b is defined s A.M b (. ) Note: Arithmetic men. Given x, y nd z re consecutive terms of n A. P., then y - x z - y y x z y is known s the rithmetic men of the three consecutive terms of n A. P. The Geometric men (G.M) is defined by G.M b (. ) The Hrmonic men (H.M) is defined s the reciprocl of the A.M of the reciprocls ie. H.M b b H.M b (. ) 7

128 Exmples.Find the A.M, G.M nd H.M of the numbers 9 & 4 Solution: 9 4 A.M 6. 5 G.M H.M Find the A.M,G.M nd H.M between 7 nd Solution: 7 0 A.M 0 G.M H.M If the A.M between two numbers is, prove tht their H.M is the squre of their G.M. Solution Arithmetic men between two numbers is. b ie. b Now H.M b b b G.M b ( G.M ) b H.M ( G.M) 8

129 Second order differentil equtions with constnt coefficients The generl form of liner Second order differentil equtions with constnt coefficients is (D bd c ) y X (i) d Where,b,c re constnts nd X is function of x.nd D When X is equl to zero, then the eqution is sid to be homogeneous. Let D m Then eqution (i) becomes m bm c 0 This is known s uxiliry eqution. This qudrtic eqution hs two roots sy m nd m. The solution consists of one prt nmely complementry function (ie) y complementry function Complementry Function Cse (i) If the roots (m & m ) re rel nd distinct,then the solution is given by m x m x where A nd B re the two rbitrry constnts. y Ae Be Cse (ii) If the roots re equl sy m m m, then the solution is given by ( A Bx) e mx y where A nd B re the two rbitrry constnts. Cse (iii) If the roots re imginry sy m α iβ nd m α iβ αx Where α nd β re rel. The solution is given by y e [ Acos βx Bsin βx] where A nd B re rbitrry constnts. Prticulr integrl The eqution (D bd c )y X is clled non homogeneous second order liner eqution with constnt coefficients. Its solution consists of two terms complementry function nd prticulr Integrl. (ie) y complementry function prticulr Integrl Let the given eqution is f(d) y(x) X 9

130 X y(x) f(d) Cse (i) Let X x e α nd f( α ) 0 x Then P.I e α f(d) Cse (ii) f( α) e αx Let X P(x) where P(x) is polynomil Then P.I P(x) [f(d)] - P(x) f(d) Write [f(d)] - in the form ( ± x ( ± x nd proceed to find higher order derivtives ) ) depending on the degree of the polynomil. Newton's Lw of Cooling Rte of chnge in the temperture of n object is proportionl to the difference between the temperture of the object nd the temperture of n environment. This is known s Newton's lw of cooling. Thus, if θ is the temperture of the object t time t, then we hve dθ α θ dt dθ -k(θ ) dt This is first order liner differentil eqution. Popultion Growth The differentil eqution describing exponentil growth is dg KG dt This eqution is clled the lw of growth, nd the quntity K in this eqution is sometimes known s the Mlthusin prmeter. 0

131 INVERSE OF A MATRIX Definition Let A be ny squre mtrix. If there exists nother squre mtrix B Such tht AB BA I (I is unit mtrix) then B is clled the inverse of the mtrix A nd is denoted by A -. The cofctor method is used to find the inverse of mtrix. Using mtrices, the solutions of simultneous equtions re found. Working Rule to find the inverse of the mtrix Step : Find the determinnt of the mtrix. Step : If the vlue of the determinnt is non zero proceed to find the inverse of the mtrix. Step : Find the cofctor of ech element nd form the cofctor mtrix. Step 4: The trnspose of the cofctor mtrix is the djoint mtrix. dj ( A) Step 5: The inverse of the mtrix A - A Exmple Find the inverse of the mtrix 4 9 Solution Let A 4 9 Step A 4 9 (8 ) (9 ) (4 ) Step The vlue of the determinnt is non zero A - exists. Step Let A ij denote the cofctor of ij in A

132 ( ) Cofctor of A ( ) 6 ) (9 9 Cofctor of A ( ) 4 4 Cofctor of A ( ) 5 4) (9 9 4 Cofctor of A ( ) Cofctor of A ( ) ) (4 4 Cofctor of A ( ) Cofctor of A ( ) ) ( 4 Cofctor of A ( ) 9 Cofctor of A Step 4 The mtrix formed by cofctors of element of determinnt A is dj A Step A A dj A 4 5 Mthemtics

133 DEMAND FUNCTION AND SUPPLY FUNCTIONS Demnd Function In the economics the reltionship between price per unit nd quntity demnded is known s demnd function. Generlly when the price per unit increses, quntity demnded decreses. Therefore if we tke quntity demnded long x xis nd the price per unit long the y xis then the grph will be curve sloping downwrds from left to right s shown in figure. The demnd function is generlly denoted s q f (p). The following observtions cn be mde form the grph.. The slope of the demnd curve is negtive.. Only the first qudrnt prt of the demnd function is shown, since the price p nd the quntity demnded q re positive. Price Demnd Curve Quntity Figure : Demnd Supply Function In economics the reltionship between price per unit nd the quntity supplied by the mnufcturer is clled supply function. Generlly when the price per unit increses, the quntity supplied lso increses. Therefore if we tke the quntity supplied long the x xis nd price per unit long the y xis then the grph will be curve sloping upwrds from left to right s shown in following figure.

134 Price Supply Curve Quntity Figure: Supply Curve Following observtions cn be mde form supply curve.. The slope of the supply curve is positive.. Only the first qudrnt prt of the supply function is shown, since the price p nd the quntity supplied re non negtive. Equilibrium price The price t which quntity demnded is equl to the quntity supplied is clled equilibrium price. Equilibrium quntity The quntity obtined by substituting the equilibrium price in ny one of the given demnd nd supply function is clled equilibrium quntity. In the figure the point E is the equilibrium point in which, the x coordinte of the point E is Equilibrium quntity nd the y coordinte of the point E is Equilibrium price. 4

135 Price Demnd Curve E Supply Curve Equilibrium price Equilibrium point Equilibrium quntity Quntity Exmple : As simple exmple let us ssume tht both the demnd nd supply functions re liner. Let us ssume tht the demnd function is given by q bp Since the demnd function slopes downwrds, b is negtive. Also let us ssume tht the supply function is given by q c dp where d is positive. The grphs of these functions re shown in the following figure 5

136 Price Demnd Curve E Supply Curve Equilibrium price Equilibrium point Equilibrium quntity Quntity At the equilibrium point both the demnd nd supply re equl. bp c dp i.e. p(d-b) -c c P d b This is the equilibrium price. Exmple : Let the demnd function be q p nd supply function be q p then the equilibrium price is given by p p i.e p 5 which is the equilibrium price nd the equilibrium quntity is obtined by substitution this vlue of p either in the demnd or in the supply function equilibrium quntity x 5 4 Exmples : The supply nd demnd curves for commodity re known to be q s p- nd q d (qs quntity supplied; q d quntity demnded).find the equilibrium price. p 6 4

137 Solution Equilibrium price is q s q d p- p or, p -p- 0 or, (p)(p-4) 0 p 4 or - Hence, equilibrium price is 4 units. 7 5

138 Vector Algebr A quntity hving both mgnitude nd direction is clled vector. Exmple: velocity, ccelertion, momentum, force, weight etc. Vectors re represented by directed line segments such tht the length of the line segment is the mgnitude of the vector nd the direction of rrow mrked t one end denotes the direction of the vector. B A A vector denoted by AB is determined by two points A, B such tht the mgnitude of the vector is the length of the line segment AB nd its direction is tht from A to B. The point A is clled initil point of the vector AB nd B is clled the terminl point. Vectors re generlly denoted by, b, c...(red s vector, vector b, vector c, ) Sclr A quntity hving only mgnitude is clled sclr. Exmple: mss, volume, distnce etc. Addition of vectors If nd b re two vectors, then the ddition of from b is denoted by b This is known s the tringle lw of ddition of vectors which sttes tht, if two vectors re represented in mgnitude nd direction by the two sides of tringle tken in the sme order, then their sum is represented by the third side tken in the reverse order. b B b O A 8

139 Subtrction of Vectors If nd b re two vectors, then the subtrction of b from is defined s the vector sum of nd - b nd is denoted by - b Types of Vectors - b (- b ) Zero or Null or Void Vector A vector whose initil nd terminl points re coincident is clled zero or null or void vector. The zero vector is denoted by O. Proper vectors Vectors other thn the null vector re clled proper vectors. Unit Vector A vector whose modulus is unity, is clled unit vector. The unit vector in the direction of is denoted by â. Thus ˆ. There re three importnt unit vectors, which re commonly used, nd these re the vectors in the direction of the x, y nd z-xes. The unit vector in the direction of the x-xis is i, the unit vector in the direction of the y-xis is j nd the unit vector in the direction of the z-xis is k. Colliner or Prllel vectors Vectors re sid to be colliner or prllel if they hve the sme line of ction or hve the lines of ction prllel to one nother. Coplnr vectors Vectors re sid to be coplnr if they re prllel to the sme plne or they lie in the sme plne. Product of Two Vectors There re two types of products defined between two vectors. They re (i) Sclr product or dot product (ii) Vector product or cross product. 9

140 Sclr Product (Dot Product) The sclr product of two vectors nd b is defined s the number θ cos b, where θ is the ngle between nd b. It is denoted by. b. Properties. Two non-zero vectors nd b re perpendiculr if π θ. b 0. Let k j i,, be three unit vectors long three mutully perpendiculr directions. Then by definition of dot product,... k k j j i i nd 0... i k k j j i. If m is ny sclr, b m. m b. b m.. 4. Sclr product of two vectors in terms of components Let k j i : k b j b i b b. Then k b j b i b k j i b.. k i b j i b i i b... k j b j j b i j b... k k b j k b i k b... b b b 0... nd... i k k j j i k k j j i i 5. Angle between the two vectors b nd. b θ cos b Mthemtics 40

141 cos θ. b b b b b b b b Work done by force: Work is mesured s the product of the force nd the displcement of its point of ppliction in the direction of the force. Let F represent force nd d the displcement of its point of ppliction nd θ is ngle between F nd d. F. d F d cos θ Vector Product (Cross Product) The vector product of two vectors nd b is defined s vector θ is the ngle from tob nd, b, nˆ form right hnded system. It is denoted by 0 θ π, nˆ is the unit vector perpendiculr to nd b b. (Red: cross b ) b sin θ nˆ, where such tht A nˆ b θ Properties. Vector product is not commuttive b b sin (π θ ) nˆ b sinθ nˆ [ sin (π θ ) sinθ ] b b B 4

142 b b. Unit vector perpendiculr to nd b b b sinθ nˆ (i) b b sinθ ( nˆ ) (ii) b (i) (ii) gives nˆ b. If two non-zero vectors nd b re colliner then θ 0 or80. b b sinθ nˆ b (0) nˆ 0 Note If b 0 then (i) 0, b is ny non-zero vector or (ii) b 0, is ny non-zero or (iii) nd b re colliner or prllel. 4. Let i, j, k be three unit vectors, long three mutully perpendiculr directions. Then by definition of vector product i i j j k k 0 nd i j k j i k j k i k j i k i j i k j 5. (m ) x b x (m b ) m( x b )where m is ny sclr. 6. Geometricl Mening of the vector product of the two vectors is the re of the prllelogrm whose djcent sides re nd b 4

143 Note Are of tringle with djcent sides nd b ( x b ) 7. Vector product b in the form of determinnt Let i j k, b bi b j bk Then b ( i j k ) x ( b i b j bk ) i j k 8. The ngle between the vectors ndb b b sinθ b b b sinθ b Moment of Force bout point b θ sin b b The moment of force is the vector product of the displcement r nd the force F (i.e) Moment M r F 4

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