SPECTRUM MATUENMATICS-Im FOR CS 51 PARTIAL DERIVATIVES Ou u tanu sin 2u (1) Ox oy sec u (P.T.U. 2017 u- a* tan - tan', then evaluate y " tan u Sin u Cos u=sin u cos uSIn u 2 sec u COs Differentiating () partially w.r.t. x, we get Here u = a* tan - tan Cu+ = (cos 2 u) 2 Ox Cy or 2 Ox (cos 2u-1) u -W. wherer= *tan , W= r* tan Multiplying both sides by x, we get 0u 0u = (cos 2 u Now r is homogenous function of degree 2 and w is homogenous function of degree 2 in x and (2 ..) .2* 0u Similarly, (cos2 u- )y- ...3) 2) dx dy W 2W and +2x Adding (2) and (3), we get if: is homogenous function in x and y of deg n, then 2x (cos 2u ) y x = (cos 2u- 1)sin 2u 2 Iof () Subtracting (2) from (1), we get 2 (1-2sin u -1)sin 2u Tt-w)-2xy w)+y* -w) =2 (- w) 2xO.,2 -2u +2x - sin u sin 2 u. ox* : U-w= Expe 11. 1fu=xy|, find 2+2xy Cu +y2u 2 u Ox Cy ..() Sol. Given, u =xp| show that x +2xy y - sin 2u sin u. +2XxOy Example 10. If u = tan Oy (P.T.U. 2011) Here u is a homogenous function in x and y of degree We know, by Euler's theorem Sol. Given, u = tan or lan u or tan u =x 2xy 2xy ,2 0u n(n- 1) u ..(2) tan u is a homogeneous function of degree l, by Euler's theorem, x(tan u)+ y-(tan u) = I tan u dx Put n= I in (2), we get Oy Ou Tan u 2 sec x sec" u+y secs u tanu y oy OX SPECTRUNM MATHEMATICS-III FOR CSs PARTIAL DERIVATIVES 52 4. ( If: = sin - show that r y Example 12. If u-x f roprove that 21 y* 0. PTU 2012 dy X+y Sol Here tf (i) Ifz = cos show that r-+ I+y (ii) If z = tan show that x+ 2 uis a homogeneous function in r and y of degree 1 by Euler's Theorem. show that x ax Differentiating () partially w.r.t I, (2) Ifu= log +y') u_a show that 3 Cy Diferentiating (1) partially w.rty. 5. ) If u= cos ++x then prove that r ...(3) (in) If u = log| u show thatr x Multiplying (2) by x. (3) by y and adding y +2y+3z show that x +y +: -3 tan u = 0 oy 6. Ifu= sin" EXERCISE 2 (b) Ox 1. Verify Euler's theorem for the function 7. If fx, y) = 2-r2 sin show that r )u=r* log (i) u= sin tan (ii) u=xyf X 3 (iv) I/4+4 If u= y? +y 8. prove that ()u= ax +2 hxy+by2 2, au If u= prove that X 2 xy 2. If: log (x+xy+y*), then show that xy 02 - 2 (P.T.U. 2013) (i) (P.T.U. 2010) 9. Ifu= ( +y2y/3, show that 2 2y u,2 0*u 3. Uf z= sin + tan ,prove that y 0 y SP TR uMATENMATICS-111 FOR CS PARTIAL DERIVATIVES 2.10. Differentiation of Composite Functions 55 If a function /(r, y) has continuous partial derivatives w.r.t. x and y and x. y have derivatives 10. pnove that w.r... 7 then dzOz dr dy d r d y dt tan () 2 2 an Proof. Here =f(r, y) Le ox, dy be the increments in independent vari ies x. y respectively and : be the coresponding 11. If utan prove that increment in z. +d: =f(r + ox, y + dy) 2 sin 4 u sin 2 u = 2 cos u Sin Subtracting (1) from (2), we get, dz =f(x + dx. y + dy) -1(c, y) dz -ra+dx, y + dy)- fa.y +oy))-fa.y -dy)-fE. )} e that 2 +2x y y =(1-4 sin' z) sin 2 u)If= tan ay pividing both sides by ôr, a small change in 1, we get dx ax dy S+, y+ o)- fx.y + o))UT.y + ))-f(. y}] 12 Gven : = x prove that , y + ô)-f.y+)) f.y-o)-f{z 31 n': Taking limits as dt 0, dx > 0, dy » 0, we get, Lt = Lt + y+ o)-f, y + o)) à a. y+d)-f, y) A0 *0 2.8. Composite Function Let:f(1.) be a function of variable x and y and x. y are further function of another variable them is calied composite function of variable "e ie. : = fa.) and r = h ()y =g () is a composit of dh ox dt ôy dt lunction dz dr dt For example: Let u = cos (r +y): r = e.y= 21. here u is composite function of Cor. 1. If u =f(r. y, ) and x, y, z are function of t, then u is a composite function of r and Further. if x. y are function of two variables and 'e', then = is called composite function of twi variables and e". dyd d ie fx. 1) and x = h 6. e), y =g (0. #) is a composite function ofr' and '. Cor. 2. If z =f(x. y) has continuous partial derivatives w.r.tx and y and x, y both have continuous partial For example: Let : fa. ) and x=r cos 6, y =r sin 6, here z is composite function ofr and 6. derivatives w.r.t. u and v both, then, ,E G and àèà 2.9. Theorem on Total Differentials Proof. Consider v as constant so that x and y both can be regarded as functions of u only f(r. ) possesses continuous partial derivatives of the first order, then the total differential of : IS given by C d+d. Ox d Similarly (S-IM FOR CS sPECTRUM MATIIEMATICS- 6 ction of x, PARTIAL DERIVATIVES 57 Cor. 3. If u =f(r, y) where y = p () then sinceu=ven* is a composite functic 2.13. Differentiation of Implicit Function f(x, y) = c be an implicit function, then prove that duCu d, cu 2.11. Change of Variables Let u= f(x. y) ay Jay-2//,*I50, Where r =P(s. 1) and y=y{s. I), ,f, +0 provided Jyx u cu etc. to expressions involving u, s, , Ou we want to change expressions involving u, X, J. a Proof: Given function is S(, y) = c ..(1) Here, we can takef as composite function ofx Diferentiating (1) wr.t. r', we get etc Let t be regarded as a constant. So x, y, u will be functions ofs alone. .. Similarly regarding s as constant, we have ox oy dr Solving (3) and (4) as simultaneous equations in x and Cu and ,we get their values in terms dy I,0 dx of cu ou , u, S, 1. Ifinstead of the equations (2), s and t are given in terms ofr and y, say s E(x. y) and r = 7(x, J) 2 Differentiating (2) w.r.t. 'r* Then ou ou cscu o 0-124) and s Oy C y The higher derivatives ofu can be found by repeated application of the formulae (3) and (4) or of ( and (7). , 2.12. Implicit and Explicit Function A relation of the form fx. y) = c in which x and y can not be separated out, is called implicit funcuu otherwise, it is called Explicit function. using () For Example: 1. The Relation r+y° -3axy =0 is an example of implicit function. 2. The Relation y = (x* +2x) is an example of Explicit function y dr SPECTRUMMATUEMATICS-I FOR CS S8 PARTIAL DERIVATIVES Verification z-xy+xy :=ar.4 df +df.2at 4at +2 a'f Now d y 16 r+ 10 df dt Hence the verification. DHere u- sin e',y- 1? 2,1,/ u is a function of x and y and x, y are functions of u is a composite function offr iLLUSTRATIVE EXAMPLES du du dxdu dy Ox dt dy dt dt Exampfe 1. () Find when:=xy+ry.x=ar.y=2 at. Also verify by direct substitution. S (P.T.U. 2017 u (/) If u = sin x =e'. j= r5, find and verify the result by direct differentiation. du .(1) (it) If u Sin (1-1).x31. y=4r, show that (P.T.U. 2014) dt d 1- Sol. (i) Ye have Verification u sin rVXY. x=ar.y=2 a du 2 ry. 2x + X. 2al, 2a dt dt NOW = (+2 xry) 2 a 1) +(2xy +r).(2 a) di x di oy di which is same as (). (4a-2. 2 d r)2at)- (2.2 ar+a t)(2 a) (ii) re given equations define u as a composite function oftt. (Putting values of x and y) =a' (8r8)+d(8r-2r) du u dr ôu dy dt x dt y dt =u (16r - 10f) dt -) (3) V-(r-y) r -- -12/ SPECTRUM MATHEMATICS-III FOR CS 60 PARTIAL DERIVATIVES 61 -1) (021) 3+ Sol. Given z =log (u +v), u = e*y= y I-(-) -(-) z is a composite function of x andy 3-12r) = =3(1-4) -(-) I-(31-4s)' We have, x 3(1-4) 3(1-4) -(9+16-244) -9-16° +24 u+v du 31-42) 2(*)2) 31-4) 3(1-4) +U +U a-0-8+16/) a-0-4)2 (1-4)-2 u 4ux, 2vx 2r(2u+) u+U u +v Now Exampk 2. Pnd total derivative of:= tan Sol. Here, := tan Differentiating (1) partially w.r.t.r u +U Example A. If w= x* +y*;x=r-s, y=r+s find in terms of r, s. P.T.U. 2012) Sol. Here w= x*+y; x=r-s, y=r+s w is a composite function of rand s Differentiating (1) partially w.r.t. y OwCwrw y (1) and (2) -2x,2 and .-1.1-1 Total derivative of z is given by from (1) and (2), we have dz 2x 1+2y 1=2(r+y)= 2(r-s+r+s) = 2(2r)=4r Examfe 3. If z =log (u+v), u = e**)",v= e*y, findand 2 -1)+2y I =-2(r-y)=-2(r-s-r- s)= -2 (-2 s) = 4s -4r,-4s. (P.T.U. 2017, 2018) Or OS SPECTRUM MATIEMATICS-III OR CSE. 62 PARTIAL DERIVATIVES 63 Example 5. If: - -+and =r cos , r =r sin e findand Adding (1) and (2), we get A,E- (3 -) (cos 0) +(-r+3y) (sin 6) cos _Sin 0 z Or an (cos0+sin 0) Sol. We have. 0z 3 (r cos 0)-r sin e] cos + [-r cos + 3 (r sin 8)] sin 3r cos' 0-r sin 0 cos -r sin cos +3 sin' 0 - 3 r (cos' 6+ sin' 8)-2 r sin cos e Cos 0 Sin6 2 x or r Example 8. Ifz = ear +by S(a-by), show that b a -2 abz. CE aC = (3 x'-y)(-rsin 0) +(-x + 3 y) r cos 6 Sol. Here z =e S(ax-by) Put ar+by= u, ax-by =v [3r* cos 0-r sin e] (-r sin 6) + (Tcos 4+ 3r sin 0) r cos 3r cos e sin +r sin' e-P cos0 +3 r sin? 0 cos e -3r sin 0 cos 0 (cos 0- sin 8) - r (cos0- sin* 6) z is a function of u, v and u, v are functions of x, y z is a composite function of x, y. Example 6. Show that a bif a and b are constants and w =f (a x+by) is a differentiable dx =e'fv).a + ef(v) a function of u = ar + by. Sol. Here, w=f(ax+by) and u = ax + by z a e'f0)+a.e'f(v) x ) .(1) w =f(u) where u isa function ofr and y ain ) -Z I, Now [of (1)] E êu y ej(o). b+ef')-b) ou of (1)] be'f(v)-be'f'(v) ...(2) du Multiplying (1) by b, (2) by a and adding, ab nd a b ,a=2abe" fw) ou OX ou +a2ab O Example9. z is a function of x and y. Prove that if x = e"+e",y=e'-ë then, Example 7. If :' is a function of two variables r and y where x =r cos 6, y =r sin 6, prove that (P.T.U. 2016) cos e 2 sin 6 Cr Sol. Herez is a function of x, y and x, y are functions otf u, v. OX z can be regarded as a composite function of u, v. Sol. We know that .cos0+sin [:r=rcos 0,y=r sin 6] Now x=e" +e",y=e"-e" cos =cos 0+cos 0 sin 0 - and -- (1) Or E - -rsin )(rcos ) Now îy Now udy cu Also .(1) ON sin 6 oz Sin6 -cos 0sin 0 oy ...2) FOR CS SPECTRUM MATHEMATICS-III Eo 64 PARTIAL DERIVATIVES 65 Now u is a function of X, Y, Z and further X, Y, Z are functions of x, y, z Again E À, ¢à (2) Subtracting (2) from (1), we get, x ' 0 Exampe 10. Ifu=f0-:-N,X -J). prove that Sol. Here u =fr-.:-X, I -V) Put y- X. : -x = Y, x-y=Z u=f(X. Y, Z) where X., Y, Z are functions of x, y, z. 00.0-0-RHS. u is a composite function of x, y, z Gu_Cu X u OY Cu CZ Exae 12. Ifu=x log (ry), where x + +3ry=1, find (P.T.U. 2016) OY (1) Sol. Here u =x log (xy) ..(1) and +y' +3xy=1 ...(2) From (1), u is a function of r and y. Cu From (2).y is a function of x. ou Ou OX Also x itself is a function ofx u is a composite function ofx. du ou d cu dy du cu, cu dy ..3) Cu ou (4) uCu =0 From (1), +log (xy)-1 =1 + log (r y) Adding (1), (2) and (3), we get, (5) Exampl1. If u= f then prove that y and (P.T.U. 2010, 2014) Differentiating (2) w.r.t. x, we get Sol. Here u= 33 Lat XY2 y=0 ufX, Y, Z) SPiCTRUM MATIEMATICS-II| FORCS 66 PARTIAL DERIVATIVES 67 (r ) Example 14. () Find if r"+yT - dr ) (ii) Ifz = +y and r' + y +3 axy= 5d, find the value ofwhen u. y= From (3). (4). (5) and (6), we get Sol. () Here /(. y) = r+y-c=0 (1) Differentiating partially w.r.t. r -1 log (r1)+| yy log y Differentiating partially w.r.t. y 1+ log (ry)- and =r log x+ xy (P.T.U. 2011 Example 13. If o . y. :)=0, show that ylogy y logr+xy- cin Here, y andx+y+3 axy = 5d From (1), z is a function of x and y and from (2), y is a function of r. Also r itself is a function ofr (1) Sol. Here o a.y.:)=0 0 (2) =0 when y is constant OA z is a composite function of x. Oz dr z dy x dr y dr dz .(3) r 2x (4) From( x 22 +y? x+y 2y 5 and 2 P. ( Similarly Differentiating (2) w.r.tx, 333l - oy x+ +ax+ay=0 d (y+ax) -(r +ay) Multiplying. (2). (1) and (3), we get x* +ay (6) dx y+ax sPECTRUM MATHEMATICS-II1 FO. FOR CS 68 PARTIAL DERIVATIVES (ii) Here f 2 ax+ 2hy+2g,. =2hx+ 2by+2f, S -2a, ,2h, 1y2b From (3). (4). (5) and (6), we get, 69 fa, y)= ar + 2hry + by +2gx +2/y+ c-0 +ax| Now ay -ay-2 f,+1,0, At a,V= dr 2a4(hx+by + f-2.2h2(hx +by +2(ax+ hy+ g)+2b4(ax+ hy+ gf 8(hx+by +f Exapple 15. Find if a (hx+by+f-2h(ax+hy+ g)hx+by+ fD+blax+hy+ g (in ax+2 hxy +by4 +2 gr+2 fy+c=0 8(hx+by+ f ( +=3 axy () r' +y' =6y abc+2fgh- aft - bg? - ch2 (hx +by+f Sol. ) Here fla. ) = xr+-3 axy=0 (on simplifying) f-3-3 ay.f = 3 y* -3 ax: f6x, sy-3 a. f» 6y ExampleI 6.= fa, y),x=rcos ,y=r sin show that Now Sol. Here, w =f(x, y), x = r cos 6, y=r sin 0 6xy-3ax-2r-3ay3y-3ax)-3) +6y3r-3ay 3-3ax w is a function of x, y and x, y are function of r,0. w is a composite function of r and 0. s4x 2axy)+ 54a(ry-ax-ay'+dxy)+54y( +dy-2ayr) OW cOs sin y ox or oy r ox 27-axP () cos+sin oy sX+a*x-2axy2 +ax?y2 - a?r3 -a'y3 +a'xy+x'y+a?y3 -2ax?y2 27(-ax OL Cw oxCw oy ax 00 y 00 (-r sin 8)+ cose) Also - -2 3axiy' + a'ay+r'-2xy(r+y-3axy+ ad) ..2) sin 8cos oy -ax)P o-ax)° or r -2xy(0+a°) y - ax) Squaring and adding (1) and (2), we get r+y-3 axy=0) 2a'xy & (cos +sin ) +(sin +cos 8) (ax- y) 16xy (i)Similar to () Ans. by puttinga =2. (2x-y)3 SPECTRUM MATHEMATICS-III FORC 70 PARTIAL DERIVATIVESs 0 into polar co-ordinates. PTU.20 Eample 17. Transform the equation - cos 0 cos sin Cu sin 9 &u Sol. The relation between cartesian co-ord1nates (x. I) and polar co-ordinates (r, 0) are r cos 6. =r sin Sin -Sin cos A_C"u Or Squaring and adding. r* = **+ 1 Cos Cu Sin Dividing. tan e = cos? U,2cos b sin d u sin Cu 2cos sin u sine u 2 r rx+.0 tan in ASin + cOS 6 êu Also, X 2 2y r Ou Cos& ouCcos6 o. uCOs6 cu 2y -sin in r sin 6 Sin & 8u cos ou cos6 ducose - sin sin 6 00 ôr 0 Sin 6 cu cos 6 u rce r c cos sin 4 0u rsin 6 2 sin x r sin? au 2cos0sin & ou, cos ou, 2 cos sin 6 8'u cos0 d*u 2) or2 Or r cos6 cos6 Adding (1) and (2), ., 1à,.1 r u =0 transforms into r2ce cu1 ou Cu sin u cos - Or Now = 0. Ox or Cxoe x Example 18. If u =fx, y, :) and r = r sin & cos p, y =r sin & sin o, :=r cos 8 then show that a_Sin 6 u O sin 6 d )co = cos 0 Ox Or 0u cos6 ou Cu O sin 00 Also Sol. Here, u =f(x, y, :) and x = r sin & cos o, y=r sin & sin p. : = r cos 6 O 4)sin 6 cos Now or sin 9+ COS6 (sin & cos o)+(sin sin o)(cos ) or r 06 Ou NoW x sin or r 00 o cos in @ ou ,I, orr00 cos cos Sin6 cu or - (r cos 0 cos ) +rcos sin p) r sin ) OT or cos @ sin u FORC SPECTRUM MATHEMATICS-| PARTIAL DERIVATIVES Now z is a function of x, y and x, y are functions of u, v. 72 73 (cos cos p) (cos sin )+-sin ) or r z is a composite function of u, v. Now -rsin sin o)+rsin cos p) +0 u 3) - - sin o)+ (cos p) r sin o or Ou Squaring and adding (1). (2) and (3), we get 2 sin? 6(Sin " d cos * d+ cos O cos p+ sin 2 Also +(sin sin o+ cos 0 sin d+cos +(cos+sin2e 5) ) [(sin cos 0) cos - sin 61 I(sin+ cos 0) sin* d+ cos* o 6) (cos + sin o (sin+ cos of ) and (4)] 12 Example 19. If: is a function x and y and u and v be two other variables, such that 21m0 .(7) u = lx +my, v =ly - mx, show that Sol. Here, u =lr + my and U - mx +ly Multiplying (1) by m, (2) by l and adding, we get of (5) and (6)] mu+lv = (m +12)y or = mu+lv Multiplying (1) by l. (2) by m and subtracting, we get, 2m ...(8) lu - m v lu mv = (1 +m*)x or x= (+m2)2 a? SPECTRU NM MATNEMATICS-IL. PARTIAL DERIVATIVES 75 Adding (7) and (8). we get ( If z =*-y. x=e cos .y= é sin t, find dt +m +m) +m o +m? (i)Ifz= tan = log/. y= e, find +m +m 8: 0*E,_8 +m> (+m) a +m 2+m? d Ifz log (uf + v), u = e" * =r+y. findand dx y 4.If u=r-y,x=2r-3s+4,y= -r+ 8s-5, find 5. (i)If u=s{r. S), r =x+y, s =x-y, show that (i) Ifx =u+u,y=uv and z is a function of r, y ; show that u - Exupple 20. If y-3 ar-r-0., prove that -0. Sol. Let fr. )=y-3 ar+r'=0 (iin) If u =fr. s), r =x+at, s =y +b1, show that f-6ax+3 x.=3y* I-6a+ 6x. fh =0, 1y =6y 6. Ifz =r+y and y =r +x, find differential coefficients of first order when Now we know that () xis independent variable (i) y is independent variable fU,-2/, 1,1,, (i) is independent variable 7. Findand if: =ir + v+ where u=ye,v=re", w = , Putting values of ff.f.fy n 1n in (1). we get cx 54 x-a3ar-x)+-2ax']y 27y dy 6t-a).9-0-6y.(3x -6ax) () If u=z sin y where x = 3r +3s,y=4r-2s,: =2 r-3s, findand 27 Bax--3ar+ax+x -Aax' +4a?r*]- --ar*) 8. If V is a function of two variables x and y and x =u cos a -v sin a, y =u sin a +v cos a. where a y is a constant. Show that dr EXERCISE 2(c) 9. If x=u+e " sin u, y =v +e cos u, prove that Cu 1. () Findwhen u= x+ y, x = ar-, y =2 at Also verify by direct substitution. 10. Given that F is a function of x and y and that r = e"+e".ye +e -21 (i) If u=x +y', where x = a cos I. y =b sin 1, find and verify the result by dired di differentiation. du (in) lfu= e sin y, x = log 1, y= I, findby partial differentiation. Also verify the result D) I. If z =f(u, v) where u =e' cos y, v = e sin y, show that direct calculations. SPECTRUM MATHEMATICS-IL to 6 Cs PARTIAL DERIVATIv 2. (u) Findif sin (a+ ). where a* +b*y =c 21. Prove that ,a2 where x = u cos (a - Sin a, y = u sin d 1 Cos (6) Find the partial derivatives of ry with respect to X and y, and its total d: +xy+y=1 cocticient w ith respect to rwhere x and y are connected by the relation x+ r en (PTU. 2011) C Or 3 13 If - l- =c.prove that By changing the independent variables u andv tox and y by means of the relations (1-x)2 Y = u COS d-U Sin a, y = u sina +v cos a, show that transforms into )lf -y +yl-x* =1, show that (1-)2 1f by substitution u=x* -y*,v=2xy, fx, y) = ¢ (u, v). Show that 14. If r =y, then yU-r log y) dr x(r-y log x) using partial derivative. fx+yy 4(r +y*)(+) 3. Ifx = u+V+ w, y =U W*wut u w, z = u v w and F is a function of x. y, z show that 15. (If: = tan-.prove that dz = -y dr 23 (i) If:= "y. prove that m ANSWERS 1. ) 4a't (+2) (i) -3a cos r sin t +3b° sin t cost (i sin r 2r cos r 16. If flx. y) = 0. ©0. :) =0. show that P E = 0 2. () 2e (cos 21-sin 2 t) (in log-1) [(log) +e xytx 17. If= n 1and: is a constant, show that 2x(2u +1) z4 yu +1 3. y +v 4. 4x+2y dy u+ (in 2: l+2x dy 1+4x: 18. ( 1fu =f(r. s, ) and r=,s 1-2 2 prove that (n If u- f , then show that x y2 Cu2 cu (P.T.U. 201I6) 0Z 2y+2xe?y_2y =2ye-2re + 19f u=f(2x-3 y. 3y-4 2,4 -2 x) prove thatouOu ou 0. 2 Ox 3 y 4 (n cos cos+6r* sin, cos cos-9 sin x (P.T.U. 201) 20. If u f +2 y,y +22). prove that (y -2x)+- y)+(-xy)0. 12. () [2rcos (+y a) (6)y+4xy-2r x+2y Ou - 0 SPECTRUM MATHEMATIO FoR PARTIAL DERIVATIVES 79 2.14. Tangent Plane and Normal to a Surface ILLUSTRATIVE EXAMPLES Let Pr. :) and Q dr. + dv.: + 8:) be two neighbouring points on the surface Fr. .)0 campletFind the equation of the tangent plane and the normal to r y+z =af at (1. 1. 1). Let the ar PQ be ds and the chord PQ be de, so that Lli QP (PTU. 2017 Sol. The equation of surtace is F (x, y. 2) = xy+z -a re Let P so that ds - 0 and PQ tends to tangent line PT. So the direction cosines of PT aro At (1, 1, 1), Differentiating (1) with respect to s, we get F dF dct = 0 equation of the tangent plane at (1, 1, 1) is (x-1) (1) +(y-1)(0)+( - 1)(1) =0 the tangent line whose direction cOsines are orx-1 +y-1+z-1 =0 ds dsdperpendicular to the line having direction ratios or x+y +z-3 =0 The equations of normal to the surface at (1., 1, I) are 3) O x-1 As there can be different curves joining Q to P, we get a number of tangent lines at P and the line having direction ratios will Example 2. Find the equation of the tangent plane and the normal line to the surface r y : =6 at the pont 1,2,3). be perpendicular to all these tangent lines at P. the equation of the tangent plane to (1) at the point P is Sol. The equation of surface is F(r, y, 2) = xy z -6=0 X-)Y-y) - -o oy where (X, Y, Z) are the current co-ordinates of any point on this tangent plane. pein (1.2.3- The equation of the normal to the surface at P is the equation of the tangent plane at (1,2, 3) is Z- OF n)-)- ox Cy Oz SO SPECTRUM MATHEMATICS-I| FoR C PARTIALPERIVATIVES ample 4. Show that the plane : 4. Show that the plane 3 x+ 12y-6z- 17-0 touching the conicoid 3 -6y2 92-17 0 Find also the point of contact. or-1) 3(-2) 2( - 3) = 0 81 6r 6 3y - 6+2 -6 0 6r+3y+ 2 18 =0 Equations of the normal line at (1. 2, 3) are Sol. The equation of conicoid is F,y.) 3r*-6)2 +9:2 +17-0 The equation of plane is 3x+12y-62- 17 =0 Let plane (2) touch conicoid (1) at (x,. y,. z) Exanple 3. Find the equation of the tangent plane and the normal line of the surface Now (y71) lies on conicoid (1) point(2, 3, -1). (P.TU. 2013 3x-6 +9 z +17 =0 Sol. The equation of surface is F(x, y. 2)= - z = 0 Now 6x, 12 y 18 Oy z S OF At ) -12 18 At the point (2,3, -1), The equation of tangent plane to conicoid () at (rY) is the equation of the tangent plane at (2, 3, 1) is (- -- dF OF -)0-) =) =0 or-x,)61,)+0-y)-12,)+(-) (18-) =0 2r-2)+(-2)(0-3)+ -)E--1)) =0 or or -))-20-y)o)+3(-:)G) =0 or 2x-4-2y+6-z- 1=0 or XXX-2 y y +2y+3zz -3 z =0 or 2x 2y-2+l =0 The equations of the normal line at (2, 3, 1) are or X-2yy, +3 : z - (-2y+3 ) =0 or x2yy, +3 z z :of (3)) Or or 3x -6y, y+9 z, z +17 =0 (4) 82 SPECTRUM MATHEMATICS-JL. TICS-IIl FON C Now (4) and () represent the same plane PARTIAL DERIVATIVES ANSWERS - 1. =3,x (r -r) =yiy- yi)= z,(t- :) 1.2 (i) x- 3y- 2z +3 0, Putting these values of x. in (3), we get ( 4r +y +z -6=0; -I=z+3 (in 4x +4y-3: -2 =-0: (-D-6 (2) +9 | +17 =0 or 3-24 + 4 + 17=0 or 0 =0, which is true. 3 7r-3y+8z = 26 y+1:-2 -1, 2, which is point of contact (i) 15x +10 y-6z = 30;2 given plane touches the given conicoid at EXERCISE 2 () 1. Find the equation of the tangent plane and normal line to the surface xy z =d at (ri, y, z) (P.T.U. 2016 (i) Find the equation of the tangent plane for the surface x +y'+3 xyz =3 at (1, 2, -1). 2 ) Find the equation of the tangent plane and the normal line to the surface 2x+=3 -2za point (2, 1, -3). (i) Find the equation of tue tangent plane and the normal line to the surface zs4 (1 +x+y)a the point (2, 2, 6). 3. Find the equation of the tangent plane and normal line to the surface () 2x:4-3xy-4x =7 at (1, -1, 2) (i) =I at (2, 3, 5) i +y + :* = at at ( ) (P.T.U. 2010) Show that the plane a x + by + cz + d=0 touches the surface p x t qy + 2 : =0, +2cd 0. 5. Show that the surface x - 2yz +y' = 4 is perpendicular to any member of the family of surtaces x+1 (2- 4a)y + az at the point of intersection (1, - 1,2). MAXIMA AND MINIMA 3 35 1LLUSTRATIVE EXAMPLES MAXIMA AND MINIMA pple 1. Find stationary points and the maxima and minima of the function f.y)=* +y- 63 (x + y)+ 12xy. 3.1. Maximum Value and Minimum Value Sol. Here fT. y) =x *y -63 (x+y)+ 12xy (P.T.U. 2016) A function f(x, v) is said to have a maximum value at x = a, y = b. Step I.3x* -63 +12y.3-6312x Iffla. b)>f(a + h, b+ k) for small values of h and k, positive or negative. A function flx. y) is said to have a minimum value at x = a, y=b. If fla. b)0 and A < 0, then fx, y) is a maxima for this p and maximum value is f1,y). + 7)x-3) =0 or y-8y+16-4y-21 =0 3,-7 or -4y-5 =0 (i) If for a point (x1, y), we have AC - B >0 and A >0, then f (x, y) is a minimum for this pa and minimum value is f(1,yi). from (3), y = 3, - 7 -5) (+ 1) =0 in If AC- B <0 for a point (x1, Yi), then there is neither max. nor minimum of flx, y). In this ca fx, y) is said to have a sddle point at (ri,y). y=5,-1 From (5), x- 1,5 four critical points are (3, 3), (-7,- 7).(-1,5) and (5, - 1) (v) If AC-B'=0 for some point (a, b), the case is doubtful. In this case, 6. B-2, c- if fla, b) -f(a+ h, b +k) >0 for small values of h and k, positive or negative, then fis max. at (a, if fla, b) -f(a+ h, b +k) < 0 for small values of h and k, positive or negative, then fis min. at (a, b AC-B = 36 xy- 144 if f(a, b) -f(a + h, b + k) does not keep the same sign for small values of h and k, then there neither max. nor minimum value. Step IV. At (3, 3) Stationary Values : The points (x1, yi), (r2, y), ... are called stationary or critical points and values fa, y) at these points are called stationary values. AC -B = 36 (3) (3)- 144 324 144 = 180>0 A 6 (3)= 18>0 fa,y) is minimum at (3, 3) It may be nojd that every extreme value is stationary value but the converse may not be true. and minimum value =f(3, 3) = 27+27 63 (3+3)+ 12 (3) (3) = 27+27-378 +108= -216 84 IS- OR SPECTRIM MATIEMATIOS-I 86 87 At -7) ACB 6(7)-7)- 144 1764 144 1620 > 0 MAXIMA AND MINIMA At (a, a AC B-36 a -9a - 27 d>0 so 67) 42 0 ( )Is maximum at (-7, - 7) Also A =6a and maimum value i f7. 7) Two cases arise 588- 7)(-7)-63(-7-7)+ 12-7) -7)= - 343 343 882 + Case I. If a > 0, then A 6 a>0 At-1.5) fx. y) has a minimum value at (a, a) AC B- 36 (-1) (5) - 144 = 180 144 324 <0 and minimum value =J(a, a) =d +a-3 a = - at(-1. 5).fx, ) has neither maximum nor minimum. -1.5) is a saddle point. Case I. If aso, then A = 6 a<0 fx y) has a maximum value at (a, a) At (5, - 1) AC B 36 (5) (- 1)- 144- 180 144 324 <0 and maximum value =f(a, a) =- a at (-1 5).f(x. v) has neither maximum nor minimum. Tsample3. Discuss the extreme values of x-y -7 x +4 y+15 x-13 (PTU.2016) G5. ) is a saddle point. Sol. Here fa, y) = *' -ys -7 x* +4 y+ 15-13 Exampk 2. Find the maximum and minimum values of x +y°-3 axy. (P.TU. 201 so Here fa. 1)=x+-3 axy Step-L. = 3x* - 14x-15, = -2y+4 y Step I.-3x-3 ay. 3-3 oax S Step-Il. Let us solve 0 and =0 Step II. Let us solve and0 Now 0 3.:-14 x+15 = 0 i.e, r ay=0 14t196-180 1416 14t4 8 -3. and y- ax = 0 6 From (1). ay =r or y= and = 0 -2y+4 =0 2y=4 y-2 oy r from (2), -ar = 0. or r-a'x =0 xt - a') = 0 x=0. a critical points are (3. 2) 0.a critical points are (0. 0) (a, a) Step-Il. A -6x- 14, B 0.c- -2 Step I11. A= =6x,. B- C Ox Oy 3a. C-6y CX Step-IV. At (3, 2) AC- B = (6 x) (6y) -(-3 a = 36 xy- 9a A -6 (3)- 14 18- 14 4, B = 0, C= -2 Step IV. At (0, 0) AC B=0-9 a = -9 d <0 AC B =(4)(-2) - (0)=-8 <0 r. y) has neither maximum nor minimum at (0, 0) at (3, 2), f E) has neither maximum nor minimum SPECTRUM MATHEMATICS-Il OR C S8 MAXIMA AND MINIMA A - 14 10- 14 - 4, B 0, C =-2 AC B* =(-4)-2) (0) = 8 >0. Also A=-4 < 0 AC f(. v) is maximum at and ma. value 4 2)-113- 254 +8+25-13 9 3 16 125-525+43232 27 27 72 144 144 Exapple 4. Find the extreme vatues (if any) of the following function: fr,)=x'(1-x-) Sol Here f. )=x'y (1-x-v) =x'-*'y-x'y. Also Step I.-3x-4 x'-3ry. 2'y-2ry-3xy fxy) has a max. value at of Step I1. Let us solve =0 and =0 OX and max. value = - ie.. 3x-4 x'-3xy=0 and 2ry-2xy-3 xr =0 From (1), xr y(3-4 x -3 y) =0 From (2). r'y (2-2 x-3y) = 0 Example. Find the extreme value (if any) of the following functions fa.y)-+y-2* *4xy-2 Sol. Here f(x. )=*+y-2*+4 xy-2y Solving these, we get. two points as (0. 0).| which are critical points. Step. 4r-4x+4y.4+4x-4y Ox Step Il1. A- 6xy- 12r-6xy', B = =6xy-8xy-9y Ox y Step I1. Let us solve 0 and =0 y -2-2*-6?y x-x+y =0 ie. .(1) and +x-y=0 (2) Step IV. At (0, 0) AC B (0) (0)-(0)= 0 Adding (1) and (2), x* +y' =0 or y'=-x further investigation is needed at (0, 0). -X Now f(0, 0)-fh. k) = - hk1-h-k)= - HR +Hk +*È - hk from (1), r -x -x=0 or x-2x =0 (neglecting h'k+hk which is small as compared to R x(*-2) 0 *0, 2-2 J0 ,if h>0 >0 , if h<0 y0,-2.2 S0, 0)-fh, k) does not keep the same sign for all small values of h and k. critical points are (0, 0), (v2.-/2 ).(- 2.2) there is no extreme value at (0, 0).

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