Mathematics of Engineering II SS 2016 Prof. Dr. R. Reemtsen M. Sc. M. Bähr Exercise Sheet 11 - Solution Problem 11.1 Problem 11.2 Consider each sequence separately 1+k 2 k→∞ k a) lim xk1 = lim k→∞ = 0 and lim xk2 = lim k + k→∞ k→∞ 1 k k b) lim xk1 = lim 1+ k1 = e1 and lim xk2 = lim k→∞ k→∞ k→∞ k→∞ e1 with lim ~xk = 0 k→∞ = ∞, therefore ~xk is divergent 1 k 2 = 0, therefore ~xk is convergent Problem 11.3 y3 =0 y→0 y 2 lim f (0, y) = lim y→0 lim f (x, 0) = lim x→0 x→0 0 =0 x2 t2 + t3 1+t 1 = lim = 6= f (0, 0) 2 2 t→0 t + t t→0 2 2 lim f (t, t) = lim t→0 therefore f is not continuous at (0, 0) Problem 11.4 a) 2 (2x + 3) ln(y + 3) ∇f = 2y(x2 +3x) y 2 +3 H= 3y 2 x4 − 2y x3 H= ∇f = (2x+3)2y y 2 +3 2 2 ln(y + 3) − x63 + b) 18 x4 (x2 +3x)(6−2y 2 ) (y 2 +3)2 (2x+3)2y y 2 +3 − 6y x4 12y 2 x5 6y x4 − x23 c) z3 y2 x 3 ∇f = 2yz ln(x) 3z 2 y 2 ln(x) d) ∇f = Q~x + ~b = 3 2 − zxy2 z 3 2y H= x x1 + 2x2 − 1 2x1 + 4x2 + 4 3z 2 y 2 x z 3 2y x 2z 3 ln(x) 3z 2 y 2 x 2 6yz ln(x) 6yz 2 ln(x) 6zy 2 ln(x) 1 2 H=Q= 2 4 ! Homework (due April 13/18, 2016) Problem H 11.5 Problem H 11.6 Consider 2 =0 k→∞ k lim xk1 = lim k→∞ 4k − 3 =2 k→∞ 2k + 10 lim xk2 = lim k→∞ k2 =0 k→∞ k 3 lim xk3 = lim k→∞ (−1)k k = divergent k→∞ k + 1 lim xk4 = lim k→∞ therefore ~xk is divergent Problem H 11.7 y2 0 =0 +2 x→0 x2 0 =0 +2 lim f (0, y) = lim y→0 y→0 lim f (x, 0) = lim x→0 t2 = 0 = f (0, 0) t→0 t2 + t2 + 2 lim f (t, t) = lim t→0 therefore f is continuous at (0, 0) Problem H 11.8 eyx+3x (y + 3) a) ∇f = b) ∇f = c) eyx+3x x H= 2 sin(2x) − 2xe5y −5e5y x2 + 6y 3x2 y 5 z 2 3 4 2 4 ∇f = 5x y z + z 2 2x3 y 5 z − 8y z3 eyx+3x (y + 3)2 eyx+3x (x(y + 3) + 1) eyx+3x (x(y + 3) + 1) H= eyx+3x x2 −4 cos(2x) − 2e5y 6xy 5 z 2 H = 15x2 y 4 z 2 6x2 y 5 z −10xe5y −10xe5y −25e5y x2 + 6 15x2 y 4 z 2 6x2 y 5 z 20x3 y 3 z 2 10x3 y 4 z − 8 z3 8 3 4 10x y z − z 3 2x3 y 5 + 24y z4
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