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Sagot :
The formula for the volume of a sphere is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]Since we are asked the rate of change of the volume with repect to time, we take the derivative on both sides, taking into account the chain rule:
[tex]\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}[/tex]taking out the constants:
[tex]\frac{dV}{dt}=\frac{4}{3}\pi\frac{d(r^3)}{dx}[/tex]Now we derivate, using the chain rule, that is:
[tex]\frac{df(g(x))}{dx}=f^{\prime}(x)g^{\prime}(x)[/tex]Applying the rule:
[tex]\frac{dV}{dt}=\frac{4}{3}\pi(3r^2)\frac{dr}{dt}[/tex]Simplifying:
[tex]\frac{dV}{dt}=4\pi(r^2)\frac{dr}{dt}[/tex]We have the following known values:
[tex]\begin{gathered} \frac{dr}{dt}=\frac{2\operatorname{cm}}{s} \\ r=3\operatorname{cm} \end{gathered}[/tex]Replacing we get:
[tex]\frac{dV}{dt}=4\pi(3)^2(2)[/tex]Solving we get:
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