Fórum de Matemática
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Se f(r,R)=2pi(r²-R²), onde r(t)=sen(t) e R(t)=cos(t), determine df/dt quando t= pi/3

\(\frac{df}{dt} = \frac{\partial f}{\partial r} \, \frac{d r}{ dt} + \frac{\partial f}{\partial R} \, \frac{d R}{ dt} = 4 \pi r \cdot \cos t - 4 \pi R \cdot (-\sin t) = 4 \pi \sin t\, \cos t + 4 \pi \cos t\, \sin t = 8 \pi \sin t \,\cos t\)

Assim,

\(\frac{df}{dt}(\pi/3) = 8 \pi \sin \frac{\pi}{3} \cos {\pi}{3} = 8 \pi \frac{\sqrt{3}}{2} \cdot \frac 12 = 4 \pi \sqrt{3}\)