Math 251-copyright Joe Kahlig, 15A Page 11. path gives f(x, 3x2) =12x4x2+ 9x4=12x21 + 9x2limx→0f(x, 3x2) =01 + 0= 02. (a) fx= 5 cos(5x)yz+ 12x2y(b) fy= sin(5x)z ∗ yz−1+ 4x3(c) fz= sin(5x)yzln(y)3. (a) ∇g(3, 2, 1) = h6, 4, 24i(b) u =16h4, 4, 2iDug(3, 2, 1) = 6 ∗46+ 4 ∗46+ 24 ∗26=443(c) |h6, 4, 24i| =√628(d) h−6, −4, −24i4. the normal vector is hfx, fy, −1iat the point n = h12, 10, −1itanget plane: 12(x − 2) + 10(y − 1) − 1(z − 9) = 0 or 12x + 10y − z = 255. since we know f(1, 2) = 3, thendx = ∆x = 1.3 − 1 = 0.3dy = ∆y = 1.9 − 2 = −0.1let z = f (x, y) =px + y3.fx=12px + y3and fx(1, 2) =16fy=3y22px + y3and fy(1, 2) = 2dz = fx∗ dx + fy∗ dy thus dz = −0.15f(1.3, 1.9) ≈ f (1, 2) + dz = 3 − 0.15 = 2.856. Let f(x, y, z) = x2+ 2y2+ z3− 10 Thenn = ∇f(1, 1, 2) = h2, 4, 12iparametric equation of the normal line:x = 1 + 2ty = 1 + 4tz = 2 + 12t7. zy=−(3x4y2− 2)2z − 4 sec2(4z)8. wa= 2xyz3(2at) + 3xy2z2(ateat+ eat)Math 251-copyright Joe Kahlig, 15A Page 29. x =511, y =3011, z =81110. fx= 2xy − 4x and fy= x2− 16y + 16.solving fx= 0 gives x = 0 and y = 2now set fy= 0 and sub theses values in and solve for the other variable.Critical points(0, 1) local max(4, 2) and (−4, 2) saddle points11. step 1 find fxand fyand solve for any critical points in the interior. you get the point (1, 5)note this in not in the region so ignore it.Boundary 1: x-asis.x = x and y = 0 for 0 ≤ x ≤ 2.g(x) = f(x, 0) = 3x2− x.g0= 6x − 1 set equal to zero and solvehas a critical value at x = 1/6Boundary 2: vertical line x=2x=2, y = y for 0 ≤ y ≤ 8g(y) = f(2, y) = 10 − y.since g0= −1 no critical values.Boundary 3:x= x, and y = 4x for 0 ≤ x ≤ 2g(x) = f(x, 4x) = −x2+ 3xtake deriva and find critical value x = 1.5critical points function value(0,0) 0(2.0) 10(2,8) 2(1/6, 0 -1/12(1.5,6) 2.25abs max = 10abs min =
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