The volume of a cube is increasing at the rate

Question:

The volume of a cube is increasing at the rate of $9 \mathrm{~cm}^{3} / \mathrm{sec}$. How fast is the surface area increasing when the length of an edge is $10 \mathrm{~cm}$ ?

Solution:

Let $x$ be the side and $V$ be the volume of the cube at any time $t$. Then,

$V=x^{3}$

$\Rightarrow \frac{d V}{d t}=3 x^{2} \frac{d x}{d t}$

$\Rightarrow 9=3(10)^{2} \frac{d x}{d t}$        $\left[\because x=10 \mathrm{~cm}\right.$ and $\left.\frac{d V}{d t}=\mathrm{cm}^{3} / \mathrm{sec}\right]$

$\Rightarrow \frac{d x}{d t}=0.03 \mathrm{~cm} / \mathrm{sec}$

Let $S$ be the surface area of the cube at any time $t$. Then,

$S=6 x^{2}$

$\Rightarrow \frac{d S}{d t}=12 x \frac{d x}{d t}$

$\Rightarrow \frac{d S}{d t}=12 \times 10 \times 0.03$     $\left[\because x=10 \mathrm{~cm}\right.$ and $\left.\frac{d x}{d t}=0.03 \mathrm{~cm} / \mathrm{sec}\right]$

$\Rightarrow \frac{d S}{d t}=3.6 \mathrm{~cm}^{2} / \mathrm{sec}$

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