Question

Find the linearization L(x,y) of the function at each point. f(x.y) = x2 + y2 + 1 a. (3,3) b. (1,3) a. L(x,y)=

Answer 1

Here, frac, y) = x+yíti fac (x, y) = a Co²+ y²+1) los e DC - 220 toto ( ək=0 for any une xe D= chotre peso luncu 20e doc cunstenti fx (x, y) 200 Also fy (x, y) = a (x² + y² + 1) he - O +24+O uhe he puro luhu (xife 2 ak 0, co oy ( for any constent k - fy(x, y) = 2y Now, we know, linearization of a function fox, y) at (alb) is given by, L(x, y) = fca,b) + (x-a) fx ca, b) + (y-b) fog Carb) =) 3²+3²+1 = - - 61= 17676 =) So, for ca,b) = (3,3) fca, b) f(3,3) f(3,3) 19 fx caib) = of (33) = 2(3)=6 fy (a, b) = fy (3, 3) = 2(3)=6 so, linearization of fcxc, y) = ?+y?+1 (aib) = (3,3) is given by, 20

[ (x,y) = f(3,3) + (x-3) fx(3,3) + (9-3) fy (3,3) 19+ (2-3) 6+ (9-3)(6) 19 + 6X-18+ 64-18 :- LOX,9) = 6x + 64-17 So, linearization of f(x, y) = x++y?+1 at (3, 3) is, Loooy)= 6x toy - 17 b) How, Here, ca,b)=(1,3) b so fca,b) f(1,3) = 2+3+1 = 1+9+1=1 also, fxcaib) = fx (1,3)= 2011=2 and fy (a,b) be fy (1,3) = 2(3) = 6 So, linearization of flor, y) = x² + y² + at ca,b) = (1,3) is given by, = L (x,y) = f(1/3) + (x-1) f2c61, 3) + (y - 3) fyC113) 11 + (0-1)2+ (9-3) (6) - 11 + 220-2+69-18 at :- Low,y). 20 + 64-9 so, linearization or f(x, y) = x² + y²+1 carb) = (1,3) is given by, LC21,9 ) = 2x + 649