Find a formula for the following function

A function of the form y = csin(bt^2) whose first critical point for positive t occurs at t = 1 and whose derivative is 11 when t = 2.

c= ?

b= ?

Since sine reaches its first critical point at pi/2, then b* 1^2= pi/2, so b= pi/2.

Since y'(2)= 11, and y' = c* cos(bt^2) * (2bt)

11= c* cos (pi/2 * 4) * (2* pi/2 * 2)

11= c*cos(2pi) * (2pi)

11= c*2pi

11/(2pi)= c

Y= [11/(2pi)] sin(Pi*t^2/2)

Hoping this helps!