MATH 152 Midterm: MATH 152 TAMU Exam 2a Solutions 2018a
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Here are the multiple choice answers for version a. E: so dx dt = 2e2t , dy. +(cid:16) dy dt = et + tet , and ds =r(cid:0) dx dt(cid:1)2 a 2 r ds = r 1. 2 sin and dx = 1 cos2 d . 0 2 e2tq4e4t + (et + tet )2 dt. 2 r /6 x(x 2) = a x 2 2. 3: so x+4 x + b x 2 gives x + 4 = a (x 2) + bx whence. A + b = 1 and 2a = 4. = 3 ln 2 2 ln 4+2 ln 3. dt = 1 t , dx/dt = 2t + 2t 2 = 4 for t = 1. Hence: when t = 1, (x, y) =(cid:0)4 + lnt, 2t + t 2(cid:1) = (4, 3).
