Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

What's the derivative and critical points of s(t)=3sin(2(t-pi/6))+5?

I know the critical points should be 5pi/12 and 11pi/12, but I don't think these are the original critical points.

Sagot :

LammettHash

If s(t) = 3 sin(2 (t - π/6)) + 5, then the derivative is

s'(t) = 3 cos(2 (t - π/6)) • 2 = 6 cos(2 (t - π/6))

The critical points of s(t) occur at the values of t where s'(t) is zero or undefined. s'(t) is continuous everywhere, so we only need worry about the first case. We have

6 cos(2 (t - π/6)) = 0

cos(2t - π/3) = 0

2t - π/3 = arccos(0) + nπ

(where n is any integer)

2t - π/3 = π/2 + nπ

2t = 5π/6 + nπ

t = 5π/12 + nπ/2

If you're only looking for t in the interval [0,2π), then you have four critical points at t = 7π/12, t = 11π/12, t = 17π/12, and t = 23π/12.

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.