MAT-2130 Final: MATH 2130 App State Fall2014 Final Exam
Document Summary
Z yz + 2xy x2 + xz xy + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = hx x, (y y), (2x + z) (z + 2x)i = h0, 0, 0i. Now that we know one exists, we will construct a potential function. Z yz + 2xy dx = xyz + x2y + c1(y, z) Z x2 + xz dy = x2y + xyz + c2(x, z) Z xy + 1 dz = xyz + z + c3(x, y) Since r(t) = h2 cos(t), t, 2 sin(t)i, we have x(t) = 2 cos(t), y(t) = t, z(t) = 2 sin(t), and r (t) = h 2 sin(t), 1, 2 cos(t)i. Plugging this data into the line integral, we get. 0 ht 2 sin(t) + 2 2 cos(t) t, (2 cos(t))2 + 2 cos(t) 2 sin(t), 2 cos(t) t + 1i h 2 sin(t), 1, 2 cos(t)i dt.
