How Do You Find the Volume of Water in a Leaking Cylindrical Bucket Over Time?

[jlmac2001]

Problem:

A cylindrical bucket of cross-sectional area A has water in it up to an initial depth of d at t=0. The water has density p, an the gravitational acceleration is g. The water leaks out the bucket through a hole in the bottom with the rate of change of he volume of the water in the bucket proportional to the pressure in the bottom of he bucket, dV/dt=-kP, with k positive constant. Find the volume of the water in the bucket as a function of time.
I tried doing some of it but I'm not sure if I'm doing this right. Can someone help?
dV/dt=-kP = dV/dt= -k(d-g-p) = dV/(d-g-p)=-k dt = integral(dv/(d-g-p))=-kt+C

 

[Chen]

dV/dt is the derivative of V(t) function. So if:

$$V'(t) = -kP$$

You need to integrate it to find the actualy function V(t):

$$V(t) = -kPt + C$$

But what is C? You can see that the function V(t) gets the value of C at t = 0. What is the volume of the water in the bucket right before it starts to leak?

 

[himanshu121]

How it can be true pressure is not constant

 

[Chen]

Yes, it does depend on the volume.

$$P = d\rho g = \frac{\rho g}{A}V$$

$$V'(t) = -kP = -\frac{K\rho g}{A}V = -cV$$

$$V(t) = V_0e^{-ct} = V_0e^{-\frac{K\rho g}{A}t}$$

Where V₀ is d₀A. Is that more like it?

 

[himanshu121]

$$V=\pi r^2 h$$
$$P= \rho g h$$
So u can set up an equation from given condition that
$$\frac{dh}{dt}= - \frac{k\rho g }{\pi r^2} h$$

 

[Chen]

They ask for the function V(t) though, not h(t).

 

[himanshu121]

Oh Yes I believe that's right