MATH 151 Midterm: MATH 151 TAMU Y2011 2011a Exam 2a Solutions
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Spring 2011 math 151: d f (x) = 3(x2 + 1)2(2x), so f (1) = Exam ii version a solutions: a di erentiate to nd the velocity: s (t) = 6t2 42t + 72, so s (1) = 36: c di erentiate to nd the velocity: r (t) = h1 + et, 1 + 2ti. At the point (1, 0), t + et = 1 and t + t2 = 0, so t = 0. The velocity is r (0) = h2, 1i, so the speed is |r (0)| = 5. you limit: c if key jugate: don"t just multiply remember by the con- 2x(1 + cos x) sin x x sin x. = 1 0 = 0: c switch x and y and solve for y: x = y 5, so x2 = y 5 and y = x2 + 5.
