The figure shows level curves of a function f (I, y). (@) Draw gradient vectors at P and Q. Is V f(P) longer than, shorter than, or the same length as V f(Q)? shorter than (b) If C is the line segment from P to Q, then Vf . di = C (Click on graph to enlarge) (c) If C is any piecewise-smooth path from P to T' to Q, then (d) If C is any piecewise-smooth path from S to P, then Vf . di = C (e) If C is any piecewise-smooth closed path from P to Q to T to S to P. then Vf . di = c(a) Iin(P) i 0, then Vf(P) is perpendicular to the lever curve of f(x,y) through the point P. First draw a tangent line to the level curve of f at the point P. Draw a perpendicular S axis with its positive side in the direction in which f increase. So the vector Vf (P) points in the direction of the positive S axis. Now nd the length of the vector Vf ( P) . The S axis intersects the level curve f : 31 atR z (2.4,1.?). The change in f on the S axis from the level curve f = 26 at the point P to the point R on the level curve f231isz(P):3126: 5. And the approximate distance between the points P{2,0.4] and the point R (2.4.1.?) is As a: 1.4 The directional derivative in the direction of the unit vector in the direction of the s-axis is So, the length of the gradient vectoer(P) is 3.5 If Vf(Q) i 0, then Vf{Q) is perpendicular to the lever curve of f (x, y} through the point Q. First draw a tangent line to the level curve of f at the point Q. Draw a perpendicular S axis with its positive side in the direction in which f increase. So the vector Vf{Q)points in the direction of the positive 8 axis. s-axis u= 3.3,1.8 As = 0.6 36 ex 3,1.3 31 26 Now find the length of the vector Vf (Q). The S - axis intersects the level curve f =36 at R = (3.3,1.8). The change in f on the S - axis from the level curve f =31 at the point O to the point u on the level curve f =36 is Vf (P) =36-31 =5. And the approximate distance between the points O(3,1.3) and the point u (3.3,1.8) is As = 0.6 The directional derivative in the direction of the unit vector in the direction of the s-axis is D. f ( 2 ) = |f (0) = 4 2 5 28.33 As 0.6 So, the length of the gradient vector Vf (O) is 8(b) Fundamental theorem for Line integrals: Let C be a smooth curve given by the vector function r(r) ,a s I s I}. Let f be a differential function of two or three variables whose gradient vector Vf is continuous on C . Then IVf-dr _______ Here C is a line from P to Q. f(P)=263ndf(Q)=31 J'Vf-dr=f(Q)f(P) =3126 =3 (c) We have to find [Vf . dr Here C is a line from P to O to T. f(P) = 26 and f (Q) =31 [vf . dr = f(0)-f(P) =31-26 = 5 (d) We have to find [ Vf . dr Here C is a line from S to P. f(S) =31 and f ( P) =26 [vf . dr = f(P) -f(s) =26 -31 -5(e) We have to find [Vf . dr C Here C is a line from P to O to T to S to P. Here the initial and the terminal points are same that is the curve is closed so the line integral is zero. f (P) = 26 and f ( P) = 26 [Vf . dr = f(P) - f(P) =26 -26 0