A function of the form y = csin(bt2) whose first critical point for positive t occurs at t = 1 and whose derivative is 9 when t = 2. Find c and b.
I took the derivative, plugged in 9 for y prime and 2 for t and solved for c. I ended up with c= 9/(4bcos(4b)) however, this is incorrect. Where am I going wrong?
Thank you for your help :)
I took the derivative, plugged in 9 for y prime and 2 for t and solved for c. I ended up with c= 9/(4bcos(4b)) however, this is incorrect. Where am I going wrong?
Thank you for your help :)
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y' = 2bct cos(bt^2)
Critical point at t =1 -> 2bc cos(b) = 0 Either b or c = 0, in which case you don't have the function or cos(b) = 0
cos (b) =0 -> b = (2n+1)π/2, n = any integer (this gives you odd values of π/2, where the cosine is 0)
You can finish the calculation and find c =9/2π(2n+1)
Check my algebra :-)
Critical point at t =1 -> 2bc cos(b) = 0 Either b or c = 0, in which case you don't have the function or cos(b) = 0
cos (b) =0 -> b = (2n+1)π/2, n = any integer (this gives you odd values of π/2, where the cosine is 0)
You can finish the calculation and find c =9/2π(2n+1)
Check my algebra :-)