Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1 tells us that y '(0) = 1 y(0) = 1 y '(1) = 0 y(1) = 0 Given that the derivative value of y(t) is 3 when t = 2 tells us that y '(3) = 2 y '(0) = 2 y '(2) = 0 y '(2) = 3 (b) Find dy dt = kcos(bt2)·b2t (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.