At x = 3, the answer is 2[ 2(−2) + 5^2 ] = 42.
(A) The object is at rest when v(t) = ln(2 − t^2 ) = 0; that occurs when 2 −
t^2 = 1, so t = 1. The acceleration is .
(C) . Find x when .
implies that x = 3.
(C) represents the rate of change of the surface area; if y is inversely
proportional to x, then, .
(D) The velocity functions are
v 1 = −2t sin (t^2 + 1)
and
Graph both functions in [0, 3] × [−5, 5]. The graphs intersect four times
during the first 3 sec, as shown in the figure in the preivious question.
