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Q.
The volume of a cube is increasing at the rate of $8\, cm ^3 / s$. If the length of an edge is $12 \,cm$, then the surface area is increasing at the rate of
Application of Derivatives
Solution:
Let $x$ be the length of a side (edge), $V$ be the volume and $S$ be the surface area of the cube.
$V=x^3$ and $S=6 x^2$
( $\because$ cube has six square faces, each of side $x$ )
where, $x$ is a function of timet. It is given that $\frac{d V}{d t}=8 cm ^3 / s$.
Then, by using the chain rule, we have
$8=\frac{d V}{d t}=\frac{d}{d t}\left(x^3\right)=3 x^2 \cdot \frac{d x}{d t} \Rightarrow \frac{d x}{d t}=\frac{8}{3 x^2} .....$(i)
Now, $\frac{d S}{d t}=\frac{d}{d t}\left(6 x^2\right)=12 x \frac{d x}{d t}=12 x\left(\frac{8}{3 x^2}\right)=\frac{32}{x}$
[from Eq. (i)]
Thus, when $x=12 \,cm , \frac{d S}{d t}=\frac{32}{12} cm ^2 / s =\frac{8}{3} cm ^2 / s$.
Hence, if the length of the edge of a cube is $12 \, cm$, then the surface area is increasing at the rate of $\frac{8}{3} cm ^2 / s$.