Question

The volume of a cube is increasing at a rate of $9c{m}^3/s$. how fast is the surface areas increasing when the length of an edge is $10cm$?

Answer 1

Let the side of cube is x c.m.

Then Volume of cube $=side\times {side}^2$

$V=x^3$

Given that,

Rate of change in the volume of the cube $=9c.m./s^3$

$\frac{dV}{dt}=\frac{d}{dt}{x}^3\Rightarrow 3x^2\times \frac{dx}{dt}=9$

$\Rightarrow \frac{dx}{dt}=\frac{9}{3x^2}$

$\Rightarrow \frac{dx}{dt}=\frac{3}{x^2}$ ……… (1)

Now,

Rate of change in the surface area $=6\times sid{e}^2$

$\frac{dA}{dt}=\frac{d}{dt}6x^2\Rightarrow 12x\times \frac{dx}{dt}$ ……… (2)

By equation (1) and (2) to, we get,

$\frac{dA}{dt}=12x\times \frac{3}{x^2}$

$\frac{dA}{dt}=12\times \frac{3}{x}$

When given $x=10$

Then,

$\frac{dA}{dt}=12\times \frac{3}{10}$

$\frac{dA}{dt}=\frac{36}{10}c.m{.}^2/\sec$

Or $\frac{dA}{dt}=3.6c.m{.}^2/\sec$