How do you find the derivative of #P(t) = 3000 + 500 sin(2πt−(π/2))#?

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Answer 1 · 2017-02-28T20:03:13

# P'(t) = 1000 pisin(2pit) #

We have:

# P(t) = 3000 + 500 sin(2pit-pi/2) #

Which we can rewrite (*) as:

# P(t) = 3000 - 500 cos(2pit) #

Differentiating wrt #t#, and applying the chain rile we have:

# P'(t) = 0 - 500 * (-sin(2pit)) * 2pi #
# " " = 1000 pisin(2pit) #

# #

Notes (*)

Using the Angle-Sum Identity:

# sin(A+-B) -= sinAcosB +- cosAsinB #

We can write;

# sin(2pit-pi/2) = sin(2pit)cos(pi/2)-cos(2pit)sin(pi/2) #
# " " = sin(2pit)(0)-cos(2pit)(1) #
# " " = -cos(2pit) #