Question

Q1. Let z = f(x,y) -√4x² – 2y² Find (i). domain of f(x,y) (ii). range of f(x,y) (iii). f(1,1) (iv). The level curves of f(x,y) for k = 0,1,2

Answer 1

BirsGiven, z=f(24) - 14-342-242 Square root funchon is defined when the argument is pasitworo forgiven fun chion argument is 4-22-2y2 4-02-2y20 50<? +2y? <4 (s'oum of squares 20) + y2 = 1 → Ellipse inside area 2 a 2 2 2 ( 2کارہ) Sketch domain Domain Region (-2,0) - r4-o = 2 K=2 0,-) (i) Domain of f(x,y) is a E E2, 2], Y E[r2, rz] , x²+2y² = 4 {(x,y) | XCE [-2, 2 and yE(-12, 5] and os x² + 2y²54} (i) for range ſu-(x² +242) values Now mas value of x²+2y = 4 : min value of 14-cz?4292)=14-4 zo min value of +2y = 0 1, max value of ne 14-(x² +24²) = i Range [0, 2] or ZE[02] (iii) F(1,1) V4-42)-2012) = 14-1-2 - V = 1 (iv) The level carne at KEO 3) 0 = 14-x+2y2 = 4-3?-24' = 0 Kal -) 1 = 14-x+2y =) 3-32-2y = 0 2= V4-x²2y² = 2 - 2²-24 2=0 (All three are allipses) Ani 3 f(x,y) = 42'y if (36,4 +0,0) lif fa,y) = (0,0} Ic²ty (i) lim f(xjy) > Transform to polar coordinates xercoso 24,4)=(0,0) Yersino Чx r r² cos²orsino - lim 4X² costosino - lim Yrososino r? (vino) tr?Cojo Now as ro lim 4rcos²o sino =10 (levee that limit is dependent on r and not on o so if we if we ) approach (0,0) from any direction o we get result 20 f(x,y) o (24,41710,0) 2 - lim rto n10 no noo (i) lim

2 1 k=0 k=1 k=2 -1 0 1 -2

(i) lim flory) =D E F(0,0) = 1 iilot1 (31,41719,0) i flasy) is not conh'nuous at 10,0) " lim limon 0,442,4) + f(2,4)at 24)=1930) ☺ largest set s on which flary) is conhnuous is determined bý 600,0) or (0,00) because only is at (0,0) strange discontinuity xrange .: Set S = {(1,y) / (94) € 1RXR-(0,01%