if the surface area of a cube is increasing at a rate of 3⋅6cm2/sec, retaining ifs shape, then the rate of change of its volume (in cm3 / sec), when the length of a side of the cube is 10 cm, is

A
20
B
10
C
18
D
9

Solution:

Let the length an edge of the cube at any moment $t$

be $x c m .$ Then, its surface area $S$ and volume $V$ are given by

$S = 6 x^{2}$ and $V = x^{3}$

$\Rightarrow \frac{d S}{d t} = 12 x \frac{d x}{d t}$ and $\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t}$

$\Rightarrow 3 \cdot 6 = 12 \times 10 \frac{d x}{d t}$ and $\frac{d V}{d t} = 3 \times 10^{2} \times \frac{d x}{d t}$

$\begin{array}{l} \Rightarrow \frac{d x}{d t} = \frac{3}{100} and \frac{d V}{d t} = 300 \frac{d x}{d t} \\ \Rightarrow \frac{d V}{d t} = 300 \times \frac{3}{100} = 9 . \end{array}$

if the surface area of a cube is increasing at a rate of $3 \cdot 6 {cm}^{2} / \sec$ , retaining ifs shape, then the rate of change of its volume (in $c m^{3}$ / sec), when the length of a side of the cube is $10 c m$ , is

Solution:

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