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finding the derivative (check my work)
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a(t) = 3cos t -2 sin t Im finding the antiderivative should i be getting v(t) = 3sin t + 2 cos t + c OR v(t) = 3 sin t - 2 cos t + C because the antideriv of sin is - cos, but we already have a neg there
first one...
The negative sign coming from the antiderivative will cancel with the negative sign initially there. So: - 2 sin t goes to - 2 (-cos t) = + 2 cos t
v(t) = 3sin t + 2 cos t + c is right
The negative in -2sin(t) gets multiplied with the negative that arises when you integrate sin(t), hence giving you a positive +2cos(t). So V(t) = 3sin(t) + 2cos(t) + C is correct.
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