Math 251-copyright Joe Kahlig, 15A Page 11. path gives f(2y3, y) =10y64y6+ y3=10y34y3+ 1limy→0f(2y3, y) =00 + 1= 02. (a) fy= z tan(3x)yz−1+ 8x3(b) fx= 3 sec2(3x)y2+ 24x2y(c) fz= tan(3x)yzln(z)3. (a) ∇g(2, 4, 1) = h3, 8, 12i(b) u =13h2, 1, 2iDug(2, 4, 1) = 3 ∗23+ 8 ∗13+ 12 ∗23=383(c) |h3, 8, 12i| =√217(d) h−3, −8, −12i4. the normal vector is hfx, fy, −1iat the point n = h6, 4, −1itanget plane: 6(x − 1) + 4(y −2) − 1(z − 5) = 0 or 6x + 4y − z = 95. 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) = x3+ 5y2+ z2− 50 Thenn = ∇f(1, 3, 2) = h3, 30, 42iparametric equation of the normal line:x = 1 + 3ty = 3 + 30tz = 2 + 4t7. zy=−(3x4y2− 2)2z −4 cos(4z)8. wa= y2z3∗ 2at + 2xyz3∗ (eat+ ateat)9. x =511, y =3011, z =811Math 251-copyright Joe Kahlig, 15A Page 210. fy= 2xy −4y and fx= y2− 16x + 16.solving fy= 0 gives x = 2 and y = 0now set fx= 0 and sub theses values in and solve for the other variable.Critical points(1, 0) local max(2, 4) and (2, −4) 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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