MATH 151 Midterm: MATH 151 TAMU Y2009 2009c Exam 2b Solutions
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Solutions-form b: b: x = t2 + 5, y = t. The parameter t = 4 cor- dy/dt dx/dt responds to the point (21, 2). Thus m = evaluated at t = 4. sin(x 2) (x + 4)(x 2) m = 32: b: lim x 2 sin(x 2) x2 + 2x 8. 1 sin(x 2) (x 2: d: f (x) = e x2. By the chain rule, f (x) = 2xe x2 and chain rule, f (x) = 4x2e x2. 2 f (1) = 4e 1 2e 1 = 2e 1 = e. 2e x2: c: first we will nd the tangent vector at t = 3 and then make it a unit vector by dividing by the magnitude: r(t) = h4 cos t, 2 sin ti thus r (t) = h 4 sin t, 2 cos ti. 1: b: h(x) = xf (x3), thus by the product and chain rule, h (x) = f (x3) + 3x3f (x3).
