Question

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10. Let S (x,y) = x2 + xy (a) Find the equation of the tangent plane at the point (1,0). (b) Use linear approximation at the point (1,0) to estimate f(1.1, -.1) (c) Find the derivative of fat (1,0) in the direction of the vector < 3,4 > (d) At the point (1,0), what direction is the function increasing most rapidly does not need to unit vector)? (e) How fast is it increasing in the direction of part (d)?

Answer 1

derivative of f at (1,0), in the direction of v is TO het z=f() = atay, (110) 70 = Z = 07160) =lto = 1 The Guction of the tangent line or plane ait (26. Yo)=(160) is 7-20 = fx (x=80) + ty(9-40) :'(X0,40,70) = (1,0,1)] f = dt = I x (x²+ ay) sex+y= 2aty fy = the H = fr(1,0) = 2Wto se dy o't nong y = 0+ x[i] = x fy(1,0) = 1 Z-1 = 2(2-1)+1(4-0) 2x-2 tyt 22ty-1 f(hy) =z=(2xty-1 Equation of the tingent plane Z = OR) (217y-1-750] (OR) 2074-7 = 1 6 H1,-0.1) Z= 2(1.1)+(-0.1) -) = 2.2-0.1-1 flll,-0.1) æfil gradient f = Vf=<frfy>=(2,1) veel v= < 3,4), (VI = {(3)*+(432 =9+16 = 5 © Let = 263/5)+1(415) <2,1). <3147 = VfV IVI 6th =

Direction of the function increasing most rapidly in VI(110) < 2,1) stets > | VF (10)] Tee??? .. Direction =< tsito @ IVF(1,0) = 15 Hf denvetine at < $its> is = tf(10) · < tsita! < 2,1%. Tato fts = 7 : At is the finution is increasing in the direction of Sitsits>