Question

The volume of a cube is increasing at the rate of 9 cm³/sec. How fast is the surface area increasing when the length of an edge is 10 cm?

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{x}}\text{be the side and}\mathit{\text{V}}\text{be the volume of the cube at any time}\mathit{\text{t}}\text{. Then,} \\ V=x^3 \\ \text{⇒}\frac{dV}{dt}\text{=3}{\mathit{\text{x}}}^2\frac{dx}{dt} \\ \text{⇒9=3}{\left(10\right)}^2\frac{dx}{dt}\left[\because x=10\mathrm{cm}\mathrm{and}\frac{dV}{dt}={\mathrm{cm}}^3/\sec \right] \\ \text{⇒}\frac{dx}{dt}\text{=0.03 cm/sec} \\ \text{Let}\mathit{\text{S}}\text{be the surface area of the cube at any time}\mathit{\text{t}}\text{. Then,} \\ S=6x^2 \\ \text{⇒}\frac{dS}{dt}\text{=12}\mathit{\text{x}}\frac{dx}{dt} \\ \text{⇒}\frac{dS}{dt}\text{=12×10×0.03}\left[\because x=10\mathrm{cm}\mathrm{and}\frac{dx}{dt}\text{= 0.03 cm/sec}\right] \\ \text{⇒}\frac{dS}{dt}{\text{=3.6 cm}}^2\text{/sec} \end{gathered}$$