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?

Answers (1)

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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