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The side of a cube is increasing at a rate of 6cm per seconds.Find the rate of increase in volume when the length of the side is 9cm

User Dherik
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1 Answer

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17 votes

..

Given


\begin{gathered} (dl)/(dt)(rate,of,increase,in,side)=6cms^(-1) \\ l=9cm \end{gathered}

To Determine: The rate of increase in volume

The volume of a cube is given as


V=l^3

The rate of change of volume with respect to the length is the derivative of the volume function. So the derivative is as calculated below


\begin{gathered} (dV)/(dl)=3l^2 \\ =3*(9cm)^2=3*81cm^2=243cm^2 \end{gathered}

The rate of increase in volume would be


(dV)/(dt)(rate,of,increase,in,volume)=(dV)/(dl)*(dl)/(dt)
\begin{gathered} (dV)/(dt)=243cm^2*6cms^(-1) \\ (dV)/(dt)=1458cm^3s^(-1) \end{gathered}

Hence, the rate of increase in volume is 1458cm³ per seconds

User Campey
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