# 4) An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

It is given that the rate at which edge of cube increase  $\left ( \frac{dx}{dt} \right )$  = 3 cm/s
The volume of cube = $x^{3}$
$\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$             (By chain rule)
$\frac{dV}{dt} = \frac{dx^{3}}{dx}.\frac{dx}{dt} = 3x^{2}.\frac{dx}{dt} = 3x^{2}\times 3 = 9x^{2} cm^{3}/s$
It is given that the value of x is 10 cm
So,
$\frac{dV}{dt} = 9(10)^{2} = 9\times 100 = 900 \ cm^{3}/s$
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is  $900 \ cm^{3}/s$

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