The volume of a cube is increasing at the rate of \(3\frac{1}{2} cm ^3 s^{

Explanation

\(\frac{dV}{dt}=3\frac{1}{2}cm^3s^{-1}=3.5cm^3s^{-1}\)

L = 6cm

\(\frac{dL}{dt}=?\)

\(V = L^3\)

\(\frac{dV}{dL}=3L^2\)

\(\frac{dL}{dt}=(\frac{dV}{dL})^{-1}\times \frac{dV}{dt}\)

\(\frac{dL}{dt}=(3L^2)^{-1}\times 3.5\)

At L= 6cm

\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)

\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)

\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)

\(\therefore \frac{dL}{dt}0.032cms^{-1}\)