- #1
padraig
- 10
- 0
Sorry to bore you all with this rate of change question, I really hate this area of maths and struggle to get my hea round it. Here it is, I'd be greatful for any help you have to offer:
Fluid flows out of a cylindrical tank with constant corss section. At the time t minutes (t is greater of equal to 0), the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows, in m^3min^-1, is proportional to the square root of V. Show that the depth h metre's of fluid in the tank satisfies the differential equation:
dh/dt = -kh^(1/2) where k is a +ve constant
Cheers
Pat
Fluid flows out of a cylindrical tank with constant corss section. At the time t minutes (t is greater of equal to 0), the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows, in m^3min^-1, is proportional to the square root of V. Show that the depth h metre's of fluid in the tank satisfies the differential equation:
dh/dt = -kh^(1/2) where k is a +ve constant
Cheers
Pat