Answered

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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.

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