Water is poured at a constant rate of \(27\pi\) cm\(^3\)/s into a hemispherical container. When the depth of water directly below the centre is \(x\) cm, the volume \(V\) cm\(^3\) of water in the container is given by \(V=\dfrac16\pi x^2(54-x)\).
Find the rate of change of the depth of the water directly below the centre when \(x=9\) cm.[3]
Sign in to view the step-by-step solution
Need help? Join our JC Math tuition classes.
Learn more