Question

(b) (i) Given the function f(x,y)= xe' + 3x, find two unit vectors who are orthogonal to the gradient of f at the point P(3,0). (7 Marks) (ii) In which direction does f(x,y)= xey + 3x increase most rapidly at the point P(3,0)? What is the maximum rate of change at that point? (3 Marks)

Answer 1

bi) f(z, ) = 10^+ 3z gradient of f is, of (0,7) i te reo of - (+3) î+(ad) A î+ 3. (es): at the point P(3,0), Vf (3,0) = (b +3) î (3 e) Let uê tu i is orthogonal to 4. + 37 lui +vſ ) :( 4 +39) = 0 ] Then 4 ut 3v = 0 → u = 3V 4 Let, 124 then US-3. then u = - 3. Then ( --3î + aj) is orthogonal to (4 î+39) If 3-4 3x(-4) 4 .. The vector (3î-aj) is orthogonal to (4î+39) & we have two vectors who are ovenogonal to gradient of fat p. To make them unit vectors, the vectors become, -3214 -3î + 4 V1-372 +42 3i - 4j - √32+42 The unit vectors (î+ 9) and (iii) are orthogonal to the gradient of fat. P(3,0) · V 25 .

(1) f(x,y) = xey+.32 + f (x,y) (eð+3) î + (redy of (3,0) = 4 î +3 lll of (3,0) || = 142 +32 +32 = 11679 125 -5. The unit direction in which f(ay) increase most rapidly at the point P(310), is Of(3,0) 4 î +3 Il vf(3,0) √ 42+32 + ر.ا Marimum rate of change of fat the point P(3,0) is, =5 ll of (3,0) ||