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calculo1:leibniz [2022/05/31 20:30] – external edit 127.0.0.1calculo1:leibniz [2022/05/31 20:30] (current) – external edit 127.0.0.1
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 e pela fórmula do volume $\frac{dV}{dh} = \frac{\pi}{4}h^2$ e portanto e pela fórmula do volume $\frac{dV}{dh} = \frac{\pi}{4}h^2$ e portanto
 $$ $$
- 2 = \frac{\pi}{4}h^2 (\frac{dh}{dt})+ 2 = \frac{\pi}{4}h^2 \times \frac{dh}{dt}
 $$ $$
 substituindo $h=6$ temos $\frac{dh}{dt} = \frac{9}{2 \pi}.$ substituindo $h=6$ temos $\frac{dh}{dt} = \frac{9}{2 \pi}.$
  
calculo1/leibniz.1654039810.txt.gz · Last modified: 2022/05/31 20:30 by 127.0.0.1