MATH 230 Final: Math 230 Spring2013-FinalExam
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Let v(t) = (2t e t) i j + cos( t) k be the velocity of a particle with initial position r(0) = ( 1, 0, 2). Find the accelaration at the point (e 1, 1, 2). We have r(t) r(0) = z t. 0 v(s) ds = r(t) = (e 1, 1, 2) +z t. 0 (2s e s) i j + cos( s) k ds. = (e 1, 1, 2) + (t2 + 1 e t) i t j + sin( t) k. = (t2 + e e t, 1 t, 2 + sin( t)) Thus the point (e 1, 1, 2) corresponds to parameter t = 0. For getting acceleration, we simply di erentiate the expression of v(t) to obtain a(t) = (2 + e t) i 2 sin( t) k. Hence accelaration at point (e 1, 1, 2) is a(0) = (3, 0, 0).
