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TAMU MATH 251 - Exam2B-Fall15

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Math 251-copyright Joe Kahlig, 15C Page 11. (a) False. notice fxand fyare nice polynomials in x and y. Thus fxyand fyxshould be equalbut they are not.(b) True. j = h0, 1, 0i.(c) False. Notice D < 0. thus saddle point.(d) True2. (a) ∇g(1, 2, 3) = h18, 12, 12i(b) u =13h2, −1, 2iDug(1, 2, 3) = 18 ∗23+ 12 ∗−13+ 12 ∗23= 163. path gives f (x, 5x + 1) =6x2− 5x −12x − 2limx→1f(x, 5x + 1) =72By L’Hopital method.4. the normal vector is hfx, fy, −1iat the point n = h15, 10, −1itangent plane: 15(x − 2) + 10(y − 1) − 1(z − 15) = 0 or 15x + 10y − z = 255. (a) ∇f(2, 1) = h4, 2i|h4, 2i| =√20(b) h−4, −2i6. since we know f(2, 1) = 3, thendx = ∆x = 1.7 − 2 = −0.3dy = ∆y = 2.2 − 1 = 1.2let z = f(x, y) =p4 + x2+ y2.fx=xp4 + x2+ y2and fx(2, 1) =23fy=yp4 + x2+ y2and fy(2, 1) =13dz = fx∗ dx + fy∗ dy thus dz = 0.2f(1.7, 2.2) ≈ f (2, 1) + dz = 3 + 0.2 = 3.27. Let f(x, y, z) = 2x3+ y3+ z5− 11 Thenn = ∇f (1, 2, 1) = h6, 12, 5iparametric equation of the normal line:x = 1 + 6ty = 2 + 12tz = 1 + 5t8. ga= fx∗ tat−1+ fy∗ 2e2agt= fx∗atln(a) + 4t3+ fy∗ 3t2sec2(t3)Math 251-copyright Joe Kahlig, 15C Page 29. left f = x2+ y2+ z2= 1 and g = 2x + y − 3z − 2we want ∇f = k ∗ ∇gsolving gets two possible points:2√14,1√14,−3√14or−2√14,−1√14,3√1410. zx=−FxFz= −2x sin(x3+ y2) + 3x4cos(x3+ y2)2yz + 4 sin(4z)11. x =125, y =150, z =35012. fx= x2+ 4y − 9 and fy= 4x − 2y .solving fy= 0 gives y = 2xnow sub this into fx= 0 and get x2+ 8x −9 = 0, after factoring x = −9 and x = 0.Critical points(1, 2) saddle point(−9, −18) local max13. step 1 find fxand fyand solve for any critical points in the interior. you get the point (2, 2)note this in not in the region so ignore it.Boundary 1: x=4(line)x = 4 and y = y for −2 ≤ y ≤ 2.g(y) = f (4, y) = 16 − 8y + y2.g0= −8 + 2y set equal to zero and solvehas a critical value at y = 4note this is outside the region so ignore it.Boundary 2: parabola x = y2x = y2, y = y for −2 ≤ y ≤ 2g(y) = f (y2, y) = 15y2− 2y3.g0= 10y − 6y2set equal to zero and solvehas a critical value at y = 0 and y = 5/3critical points (0, 0) and (25/9, 5/3)critical points function value(4,-2) 36(4.2) 4(0,0) 0(25/9, 5/3) 125/27abs max = 36abs min =


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TAMU MATH 251 - Exam2B-Fall15

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