2. The volume of a cube is increasing at the rate of 8 cm^3 /s. How fast is the surface area increasing when the length of an edge is 12 cm?

Answers (1)

The volume of the cube(V) =  x^{3}    where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of 8 cm^3 /s

we can write  \frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}       ( By chain rule)

\frac{dV}{dt} = 8 = \frac{dV}{dx}.\frac{dx}{dt}     

\frac{dx^{3}}{dx}.\frac{dx}{dt} = 8      \Rightarrow 3x^{2}.\frac{dx}{dt} = 8

\frac{dx}{dt} = \frac{8}{3x^{2}}                                                        - (i)
Now, we know that the surface area of the cube(A) is   6x^{2}

\frac{dA}{dt} = \frac{dA}{dx}.\frac{dx}{dt} = \frac{d6x^{2}}{dx}.\frac{dx}{dt} = 12x. \frac{dx}{dt}                      - (ii)

from equation (i) we know that  \frac{dx}{dt} = \frac{8}{3x^{2}}

put this value in equation (i)
We get,
              \frac{dA}{dt} = 12x. \frac{8}{3x^{2}} = \frac{32}{x}
It is given in the question that the value of  edge length(x) = 12cm 
So,
    \frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} cm^2/s
            

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