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Standard VIII

Mathematics

Volume of a Cube

The surface a...

Question

The surface area of a cube increases at the rate of $12$ $c m^{2} / s e c$ . Find the rate at which its volume increases, when its edge has length $5$ cm.

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Solution

$S$ (Surface area of cube) $= 6 a^{2}$
$V$ (Vol. of cube) $= a^{3}$
Now,
$\frac{d s}{d t} = 12 a \times \frac{d a}{d t} = 12 ⟹ a \frac{d a}{d t} = 1$
$⟹ \frac{d v}{d t} = 3 a^{2} \frac{d a}{d t} = 3 a^{2} \times \frac{1}{a} = 3 a$

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The surface area of a cube increases at the rate of $12$ $c m^{2} / s e c$ . Find the rate at which its volume increases, when its edge has length $5$ cm.

$S$ (Surface area of cube) $= 6 a^{2}$
$V$ (Vol. of cube) $= a^{3}$
Now,
$\frac{d s}{d t} = 12 a \times \frac{d a}{d t} = 12 ⟹ a \frac{d a}{d t} = 1$
$⟹ \frac{d v}{d t} = 3 a^{2} \frac{d a}{d t} = 3 a^{2} \times \frac{1}{a} = 3 a$

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The surface area of a cube increases at the rate of 12 cm2/sec. Find the rate at which its volume increases, when its edge has length 5cm.

S (Surface area of cube) =6a2V (Vol. of cube) =a3Now, dsdt=12a×dadt=12⟹adadt=1⟹dvdt=3a2dadt=3a2×1a=3a

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FREE TEXTBOOK SOLUTIONS

STATE BOARDS

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The surface area of a cube increases at the rate of $12$ $c m^{2} / s e c$ . Find the rate at which its volume increases, when its edge has length $5$ cm.

$S$ (Surface area of cube) $= 6 a^{2}$
$V$ (Vol. of cube) $= a^{3}$
Now,
$\frac{d s}{d t} = 12 a \times \frac{d a}{d t} = 12 ⟹ a \frac{d a}{d t} = 1$
$⟹ \frac{d v}{d t} = 3 a^{2} \frac{d a}{d t} = 3 a^{2} \times \frac{1}{a} = 3 a$