Solutions for PSet 6

Solutions for PSet 6 1. (8.22:14) (a) f (x, y) = f1 (x, y)i + f2 (x, y)j, where f1 (x, y) = ex+2y , and f2 (x, y) = sin(y + 2x). Computing all the partial derivatives ∂f1 ∂x ∂f2 ∂x ∂f1 ∂y ∂f2 ∂y = ex+2y = 2 cos(y + 2x) = 2ex+2y = cos(y + 2x) So the matrix for the total derivative is:   ex+2y 2ex+2y Df (x, y) = 2 cos(y + 2x) cos(y + 2x) Similarly for g(u, v, w) = g1 (u, v, w)i + g2 (u, v, w)j, where g1 (u, v, w) = u + 2v 2 + 3w3 and g2 (u, v, w) = 2v − u2 we have: ∂g1 ∂u ∂g2 ∂u ∂g1 ∂v ∂g2 ∂v =1 = −2u = 4v =2 ∂g1 ∂w ∂g2 ∂w = 9w2 =0 And the total derivative is:  Dg(u, v, w) = 1 4v 9w2 −2u 2 0  (b) The composition h(u, v, w) = f (g(u, v, w)) = exp(u+2v 2 +3w3 +4v−2u2 )i+sin(2v−u2 +2u+4v 2 +6w3 )j (c) The total derivative at a point (u, v, w) can be computed using the chain rule: Dh(u, v, w) = Df (g(u, v, w))Dg(u, v, w)    eg1 +2g2 2eg1 +2g2 1 4v 9w2 = −2u 2 0 2 cos(g2 + 2g1 ) cos(g2 + 2g1 ) Now we evaluate at (u, v, w) = (1, −1, 1) and thus g1 = 6, g2 = −3. As a result g1 + 2g2 = 0 and g2 + 2g1 = 9 and    2e0 1 −4 9 e0 Dh(1, −1, 1) = = 2 cos 9 cos 9 −2 2 0   −3 0 9 0 −6 cos 9 18 cos 9 1 2. (8.24:12) 1 r : R3 → R (a) We can compute ( √ r ) using the chain rule for the functions 1 defined by r(r) = r · r and g : R → R defined by g(t) = t . With these functions 1r = g ◦ r thus √ 1 2r A −1 A  ( ) = A · g  (r)  ( r · r) = A · 2 · ( √ )=− 3 ·r r r (r ) 2 r · r (b) To evaluate the left hand side in question, we need to first evaluate  A  ( 1r ) . Using part (a), this is equivalent to     −A · r f (r)  = r3 h(r) where f (r) = −A · r and h(r) = r3 . Both are real-valued functions, thus we can apply the rule for their fractions:     f (r) A·r (f (r))h(r) − f (r)  (h(r)) = =  − 3 r h(r) (h(r))2 But Therefore Now 3 1 3 h(r) = r3 = (r · r) 2 = (r · r) 2 2r = 3rr 2   A·r Ar3 − A · r3rr  − 3 = r6 r   3B · rA · r A · B A·r B· − 3 = − r r5 r3 3. To compute the gradient of multivariate function f (x, y), compute the partial derivatives:  xy  (x+h)y g(u)du − g (u)du ∂f (x, y ) 0 = lim 0 h→0 ∂x h  (x+h)y g(u)du xy = lim h→0 h 2  xy+hy If we define a function m(h) = xy g(u)du then the limit above is m (h). Using the fundamental theorem of calculus, we determine m (0) = yg(xy). The partial deriv ative with respect to y follows similarly. Thus the gradient  xy of f (x, y) = 0 g(u)du is ∇f (x, y) = (yg(xy), xg(xy)). A level set (x, y) can be described as f −1 (c). If both (x0 , y0 ) and (x, y) lie in the same level set, then:  xy  x0 y0 g(u)du = g(u)du = c 0 0  xy g(u)du = 0 Or, x0 y 0 As g is a positive function, its integral can only be 0 if the integration interval is empty or: or ∃ a value b s.t. xy = x0 y0 = b  t As g is positive, the function m(t) = g(u)du is strictly increasing in t. 0  b Therefore, there exists a unique b such that g(u)du = c = 0. In other xy = x0 y0 0 b where b is unique. x b A level set (x, y) can be parametrized as r(x) = f −1 (c) = (x, ). This level x b  set has slope r (x) given by (1, − 2 ). The gradient of f (x, y), ∇f at any point x b (x, y) is g(b)( , x). The dot product ∇f · r (x) is: x words, the level set is parametrized by y = h(x) = b b b bx g(b)( , x) · (1, − 2 ) = g (b) − g(b) 2 = 0 x x x x Hence, ∇f is orthogonal to the level set at each point on the curve.  4. f (x, y) = x2 −y 2 xy x2 +y 2 0 3 if (x, y) =  (0, 0), if (x, y) = (0, 0). The partial derivatives: ∂f f (h, y) − f (0, y) h2 − y 2 (0, y) = lim = lim 2 y = −y h→0 h→0 h + y 2 ∂x h ∂f f (x, h) − f (x, 0) x2 − h2 (x, 0) = lim = lim 2 x=x h→0 h→0 x + h2 ∂y h Using the above derivation, the second partial derivatives can be evaluated at point (0,0): ∂2f ∂ ∂f (0, 0) = (x, 0) ∂x∂y ∂x ∂y and ∂ 2f ∂ ∂f (0, 0) = (0, y) ∂y∂x ∂y ∂x Thus = (0,0) = (0,0) ∂ (x) ∂x ∂ (−y) ∂y =1 (0,0) = −1 (0,0) ∂ 2f ∂2f (0, 0) = (0, 0)  ∂x∂y ∂y∂x This means that in general, ∂ 2f ∂ 2f = ∂x∂y ∂x∂y 5. We can write F (t) = f (r(t)) where r(t) = (3t2 , 2t + 1, 3 − t3 ). Then F (t) = f ◦ r, thus F  (t) = f (3t2 , 2t + 1, 3 − t3 ) · r (t) = f (3t2 , 2t + 1, 3 − t3 ) · (6t, 2, −3t2 ) At t = 1 this evaluates to: F  (1) = f (3, 3, 2) · (6, 2, −3) The gradient of f : R3 → R is f (x, y, z) = ( ∂f , ∂f , ∂f ). Thus ∂x ∂y ∂z F  (1) = 6 ∂f ∂f ∂f (3, 3, 2) + 2 (3, 3, 2) − 3 (3, 3, 2) ∂x ∂y ∂z 4 Let Hessf denote the second derivative matrix of f : ⎛ ∂2f ∂2f (x, y, z ) ∂x∂y (x, y, z) ∂x2 ⎜ ∂2f ∂2f Hessf (x, y, z) = ⎝ ∂y∂x (x, y, z) ∂y2 (x, y, z) ∂2f ∂2f (x, y, z) ∂z∂y (x, y, z) ∂z∂x ∂2f (x, y, z) ∂x∂z ∂2f (x, y, z) ∂y∂z ∂2f (x, y, z) ∂z 2 ⎞ ⎟ ⎠ Then F  (t) = r (t)Hessf (3t2 , 2t + 1, 3 − t3 )r (t)T + f (3t2 , 2t + 1, 3 − t3 ) · r (t) = (6t, 2, −3t2 )f  (3t2 , 2t + 1, 3 − t3 )(6t, 2, −3t2 )T + f (3t2 , 2t + 1, 3 − t3 ) · (6, 0, −6t) ∂ 2f ∂ 2f ∂2f ∂2f ∂2f ∂2f ∂ 2f = 36t2 2 + 12t + 12t + 4 2 − 6t2 − 18t3 − 18t3 ∂x ∂x∂y ∂x∂z ∂y∂x ∂y ∂y∂z ∂z∂x 2 2 ∂ f ∂ f ∂f ∂f − 6t −6t2 + 9t4 2 + 6 ∂x ∂z ∂z∂y ∂z where all partial derivatives are taken at (3t2 , 2t + 1, 3 − t3 ). Substituting t = 1 we get ∂ 2f ∂2f ∂ 2f ∂2f ∂2f ∂ 2f ∂2f − 6 − 18 + 12 − 18 + 12 + 4 ∂y∂z ∂z∂x ∂x∂y ∂x∂z ∂y∂x ∂y 2 ∂x2 2 2 ∂ f ∂ f ∂f ∂f −6 −6 +9 2 +6 ∂x ∂z ∂z∂y ∂z F  (t) = 36 where the partial derivatives are taken at (3, 3, 2). 6. (a) h(x) = g (f (x)). Thus Dh((0, 0)) = ⎛ ⎞ ⎞ ⎛  1 1  5 5 −1 2 Dg(f ((0, 0)))Df (0, 0) = Dg(1, 2)Df (0, 0) = ⎝ 0 1 ⎠ =⎝ 6 3 ⎠ 6 3 2 1 4 7 (b) Let k = f −1 , then Dk(0, 0) =  −1 Df (f −1 (0, 0)) = (Df (1, 2))−1 Thus,  Dk(0, 0) = −1 3 −2 4 5 −1 1 = 2  4 −3 2 −1