Question

Needing help understanding part C! Thank you.

4. 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 FIGURE 2. Figure for Problem 5. 5. (5 pts) Let f(x,y) be a smooth function with Vf(0,0), V fz (DU) and Vfy(0. OẢ, OB and OČ in Figure 2. The x, and y-axes are horizontal and vertical, res- (a) Computs the directional derivative fob(0,0). (b) Is there a direction in which the rate of change of f at (0,0) is –2.5? Circle your answer: Yes No Giv a reason: ) If f(0,0) = 2, estimate f(0.01, -0.02) using quadratic approximation. 7

Answer 1

derivatives of if the first 8 second partial fcn, y) exist Point (a, b) is ne aj Quadrafic af Proximation then Ole, 9) = f(a, b) f f (a,b). (a-alt fy (a, b). (y-b) til fox (a, b) Goog! + fay (a, b). Ca -a )(y-a) thy (a, b) (4+5)* 2 Given B ce (-12) A A 3 180o C (2,-1) hiver oh = -1 +23, 0B = 37+ 23, 52= 27.3 also. If cool 5+23 - ОА af an 7 (0,0) + afco, ay 10, o) -) : 3 -> (0,0) = -1 f. 10,o) (0,0) =

and fu E Given V8 (2,y) = filthy I - f (x, y) = x + fxy 5 Ify (2,4)= fayt of (0,0) = 0B -3 7 + ail I from Co, o) =-3 ofy (0,0) = 2 = 27 - 3 - fy (0,0) = fag (0,0)=21) lug (0,0) = -1 f cool 2 and (oo) fin, y) near so Quadratic approximation fcm, y) Z (2, 4) = f(0,0) + f(0,6). (2-0) + fyko, o). (4-0) + Ico,o).(n-0² + (x-o) cy-o) fayloco) 2 t fyy (0,0). (4-0)2 Given flo, o)=2 - 0 (x, y) = f(opo) + flool at fycool .y 2 f (0,0) + f (0,0).xy y² t fyy Loo) 2 - Oln, y = 2-2 + 2y + (-3) x² +axy - y 2 2

So, f (0.ol-0.02) 2-0.0172 (-0.02 ) -360.0132 ♡ Q10.01 -0,02) - + 2(0.011(-0.02) - (0-02) 2 =) O 10.01, -0.02 ) 2 2-0.01 -0.04 - 00163 -4 -2] 2-0.05 - (0.01) (-1) a 2 -0.05 - 0.00075 a1 1.94925