r(t) = cos^3 t + sin^3 t

How do you find r'(t)? Would it be -3sin^2 t + 3 cos^2 t or something else?

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- Sep 22nd 2009, 05:31 AMtdat1979derivitive
r(t) = cos^3 t + sin^3 t

How do you find r'(t)? Would it be -3sin^2 t + 3 cos^2 t or something else? - Sep 22nd 2009, 06:17 AMProve It
You need to use the chain rule.

$\displaystyle \frac{d}{dt}(\cos^3{t}) = -3\sin{t}\cos^2{t}$

$\displaystyle \frac{d}{dt}(\sin^3{t}) = 3\cos{t}\sin^2{t}$.

Therefore

$\displaystyle r'(t) = 3\cos{t}\sin^2{t} - 3\sin{t}\cos^2{t}$

$\displaystyle = 3\cos{t}\sin{t}(\sin{t} - \cos{t})$.