Question

Find a formula for the following function. A function of the form y = csin(bt2) whose first critical point for positive t occurs at t = 1 and whose derivative is 5 when t = 2.

Answer 1

y' = 2bc*t*cos(bt^2)
y' = 0 when t=1
--> 2bc*cos(b) = 0
assuming b and c are not zero, then cos(b) =0 --> b = pi/2

y' = 5 when t=2
--> 4bc*cos(4b) = 5
plug in b= pi/2 and solve for c
2pi*c*1 = 5
c = 5/2pi

y = (5/2pi) sin(pi/2 t^2)

Answer 2

did that make sense?

Answer 3

thank you so much

Answer 4

welcome