finalexamsol

Z z 2 31 n f 1 3 2 1 p0 1 3 2 2 2 tangent

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Unformatted text preview: 00 = I3 = ⎝ 0 1 0 ⎠ 001 ⎞⎛ ⎞ ⎛ ⎞ ∗∗ 1 0 3 ∗ ∗∗ p 5 ⎠ ⎝ −2 1 −1 ⎠ = ⎝ −3 − 2p − 5 ∗ ∗ ⎠ ⇒ −8−2p = 0 ⇒ p = −4. ∗∗ −1 1 1 ∗ ∗∗ Problem 4. √ √ a) r� (t) = �− sin(et )et , cos(et )et , et � ⇒ |r� (t)| = et 1 + 1 = et 2 r� (t) 1 = √ �− sin(et ), cos(et ), 1�. ⇒ T(t) = � |r (t)| 2 b) 1 et T� (t) = √ �− cos(et ), − sin(et ), 0� = − √ �cos(et ), sin(et ), 0�. 2 2 Problem 5. ∂ F xz a) Fx = = 2 ⇒ Fx (1, 3, 23) = 2/2 = 1. ∂x (x + y )1/2 ∂F z 2 Fy = = + ⇒ Fy (1, 3, 2) = 3/2. 2 + y )1/2 ∂y 2(x z ∂F 2y 1 Fz = = (x2 + y )1/2 − 2 ⇒ Fz (1, 3, 2) = . ∂z z� 2 � 31 n = �F (1, 3, 2) = 1, , , P0 = (1, 3, 2) 2 2 ⇒ tangent plane equation 3 1 1(x − 1) + (y − 3) + (z − 2) 2 2 or 2x + 3y + z = 13. b) At P0 = (1, 3, 2) we have |Fy | = 3/2 > |Fx |, |Fz |. So, a change in y produces the largest change in F . 3 ΔF = Fy Δy = (0.1) = 0.15. 2 � � df � 1 1 1 � c) = u · �F (P0 ) = ± �−2, 2, 1� · �1, 3/2, 1/2� = ± (−2 + 3 − 1/2) = ± . � ds P0 ,u 3 3 6 � � dF � 1 ΔF ≈ Δs ⇒ 0.1 = Δs ⇒ Δs = 0.6 � ds 6 � P0 ,u 2 Problem 6. a) fx = 1 − 2/(x2 y ) = 0 fy = 4 − 2/(xy 2 ) = 0 � x2 y = 2 ⇒ xy 2 = 1/2 � ⇒ x = 4y 1 1 1 ⇒ y3 = ⇒ y= ⇒ x = 2. 2 8 2 There is one critical point at (x, y ) = (2, 1/2). ⇒ 4y 3 = b) fxx = 4/(x2 y ), fyy = 4/(xy 3 ), fxy = fyx = 2/(x2 y 2 ) A = fxx (2, 1/2) = 1, C = fyy (2, 1/2) = 16, B = fxy (2, 1/2) = 2 ⇒ AC − B 2 = 12 > 0, A > 0 ⇒ f has a relative minimum at (2, 1/2). Problem 7. f (x, y, z ) = dist2 = (x − x0 )2 + (y − y0 )2 + (z − z0 )2 subject to g (x, y, z ) = Ax + B y + C z...
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