Question: When the depth of liquid in a container is xcm,the volume is x(x² + 25)cm³.Liquid is added to the container at a constant rate of 2cm³s¯¹.Find the rate of change of the depth of the liquid at the instant when x=1.
V = x(x² + 25)cm³, x >= 0
dV/dt = 2 cm³s¯¹
Rate of change of depth is dx/dt, where
dx/dt = (dx/dV).(dV/dt)
Now, dV/dx = 3x² + 25, so dx/dV = 1/(3x² + 25)
And, dx/dt = (1/(3x² + 25)).(2)
dx/dt = 2/(3x² + 25)
at x = 1,
dx/dt = 2/(3 + 25) = 2/28
dx/dt = 1/14 cm³s¯¹