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assume that tge volume,v, of a sphere is expanding at a rate of 100 inches cubex per minute, the volume of a sphere is given by v= 4/3 pir*3.Determine the rate at which tge radius is changing with respect to time when r=8

Respuesta :

[tex]\bf \textit{volume of a sphere}\\\\ V=\cfrac{4}{3}\pi r^3\\\\ -----------------------------\\\\ \cfrac{dv}{dt}=\cfrac{4\pi }{3}\cdot 3r^2\cfrac{dr}{dt}\implies \cfrac{dv}{dt}=4\pi r^2\cfrac{dr}{dt} \\\\\\ \cfrac{\frac{dv}{dt}}{4\pi r^2}=\cfrac{dr}{dt}\qquad \begin{cases} \frac{dv}{dt}=100\\\\ r=8 \end{cases}\implies \cfrac{100}{4\pi 8^2}=\cfrac{dr}{dt}[/tex]

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