Math Is Fun Forum

yonski

Member

Registered: 2005-12-14

Posts: 67

Differential equation

Hi there, i'm having a bit of trouble with this question:

Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t>0, the volume of fluid remaining in the tank is V m³. The rate at which the fluid flows in m³/min is proportional to the square root of V.

Show that the depth h metres of fluid in the tank satisfies the differential equation

dh/dt = -k√h , where k is a positive constant.

Here's what i've done so far:

Let the radius of the base of the cylinder be r metres. This gives

dV/dt = -k√V

V = pi*r²h

dh/dV = 1/(pi*r^2)

dh/dt = dh/dV * dV/dt = - k√(h/(pi*r²))

Sorry if all this looks a bit messy, i'm not very good at typing equations on here, hopefully it's legible though.

Anyway, my book says the answer is simply dh/dt = -k√h . I can't for the life of me see what i'm doing wrong though.

Any help appreciated!

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Differential equation

You started with the correct idea, but you just need to use a different constant of proportionality than k at the beginning – say, c.

Let A be the cross-sectional area (which is given to be constant). Then V = Ah.

Last edited by JaneFairfax (2007-07-04 10:39:02)

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: Differential equation

Ah yeah, I shouldn't have used the same constant of proportionality with both equations. Doh!

Thanks for your help

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."