The volume of a cube is increasing at the rate of \(3\frac{1}{2} cm ^3 s^{ -1}\). Find the rate of change of the side of the base when its length is 6 cm .
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
\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)
\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)
\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)
\(\therefore \frac{dL}{dt}0.032cms^{-1}\)