Consider the curve of the form y(t) = (a) Given that the first critical point of y(t) for positive t occurs at t = 1 tells us that Oy '(1) = 0 O y(0) = 1 Oy(1) = 0 Oy (0) = 1 Given that the derivative value of y(t) is 3 when t = 2 tells us that y (2) = 0 O y '(0) = 2 Oy (3) = 2 y (2) = 3 (b) Find dy dt :ksin(bt²). k = = (c) Find the exact values for k and b that satisfy the conditions in part (a). Note: Choose the smallest positive value of b that works. b =

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