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;

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

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