Rates of change / Differentiation (1 Viewer)

[skillz]

Their is a container in the form of an inverted cone of height 80 cm and diameter 40cm. Water is poured into the container at the rate of 20cm^3/s.

a) Find the volume, Vcm^3, of water in the container when the depth is xcm.

b) Find the rate at which the depth is increasing when x =5


Cheers.

 

[SoulSearcher]

a) The volume of the cone at any depth x, is
V = 1/3 * pi * r² * h
= 1/3 * pi * r² * x, when the height at any time is x cm
Using similar triangles, x/80 = r/40
.'. r = x/2
.'. V = 1/3 * pi * x²/4 * x
= x³pi/12
b) We are looking for dx/dt, given that dV/dt = 20
.'. dV/dx = x²pi/4
dx/dt = dx/dV * dV/dt
= 4/x²pi * 20
= 80/25pi, when x = 5
= 16/5pi cm/s