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The volume of water in a bowl is given by V = 1/3(pi)(h)^2(60 - h) cm^3, where h is the depth of the water in centimeters. If the depth is increasing at the rate of 4 cm/sec when the water is 10 cm deep, how fast is the volume increasing at the instant?

Respuesta :

[tex]\bf V=\cfrac{\pi }{3}h^2(60-h)\implies V=\cfrac{\pi }{3}(60h^2-h^3)\\\\ -------------------------------\\\\ \cfrac{dV}{dt}=\cfrac{\pi }{3}\cdot 120h-3h^2\cdot \cfrac{dh}{dt}\quad \begin{cases} \frac{dh}{dt}=4\\ h=10 \end{cases} \\\\\\ \cfrac{dV}{dt}=\cfrac{\pi \cdot 120\cdot 10\cdot 3\cdot 10^2\cdot 4}{3}\implies \cfrac{dV}{dt}=480000\pi [/tex]

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