The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Then, V = x3 and S = 6x2 where x is a function of time t.
It is given that
\begin{align} \frac{dV}{dt} = 8 cm^3 / s \end{align}
Then, by using the chain rule, we have:
∴ \begin{align} \frac{dV}{dt} = \frac{d}{dt} (x^3) . \frac{dx}{dt} = 3x^2 . \frac{dx}{dt} =8 \end{align}
⇒ \begin{align} \frac{dx}{dt} = \frac{8}{3 x^2} ……… (1) \end{align}
Now \begin{align} \frac{dS}{dt} = \frac{d}{dx}(6x^2).\frac{dx}{dt} [By Chain Rule] \end{align}
\begin{align} =12x .\frac{dx}{dt} =12x.(\frac{8}{3x^2}) = \frac{32}{x} \end{align}
Thus, when x = 12 cm, \begin{align} \frac{dS}{dt} = \frac{32}{12} cm^2 / s = 8 cm^2 / s \end{align}
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \begin{align} \frac{8}{3} cm^2 / s \end{align}.
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