Water drips out of the bottom of a cylindrical bucket that is initially full. The rate of dripping is proportional to the height of water column in the bucket. If the rate of dripping at half height is R, then the average rate of dripping until the bucket becomes almost empty is:
1.greater than R
2.R
3. Between R/2 and R
4. Less than R/2
How to find the average rate of dripping, please help. Thanks in advance.
My attempt (Modified as per my trivial understanding of the comments)
Let $$V=\pi r^2 h$$
The rate of water dripping from the bucket is:
$$-\frac{dV}{dt}=\pi r^2 \frac{dh}{dt}$$
Now we have $-\frac{dV}{dt}=kh$ for some constant of proportional $k$.
Also for half height we have $R=kH/2$ which gives $k=2R/H$. Substituting back we get:
$-\frac{dV}{dt}= 2Rh/H$
When the bucket becomes almost empty, the instantaneous rate is: $-\frac{dV}{dt}= 2R$
But I need to find the average rate of dripping.
Is it the right way to do it?
What should I do next, please suggest.
It's actually quite simple after you get the differential equation in terms of R. Let us first define the equations properly. You have: $$\frac{dV}{dt} = -kh$$ Where h is the height of water inside the bucket at any time. (Note that when your bucket becomes almost empty h is very close to 0. So your rate is negligible. Your conclusion of instantaneous rate = 2R is a result of another confusion, I'm afraid)
H is the maximum height, also the height at t=0 (convince yourself it cannot become greater than height at t=0 formally)
The differential equation has a solution of the form $$h = C\exp\{-kt\}$$ for some constant C.
From the condition of rate at half-full, you have $k = \frac{2R}{H}$ and since at t=0 height must be H, we have $H = C.e^0 \implies C=H$. So you finally have $$h = H\exp\{-\frac{2Rt}{H}\}$$
Intuitively, this bucket will be always remain full as the function asymptotically approaches 0(i.e., it will be 0 only when time $t\rightarrow \infty$). So your average rate(i.e., $\frac{H}{\text{total time}}$) should be very very low with respect to R.
If you get this part, you just have to prove this formally. There are many ways to do this. I'll take an intuitive approach:
Observe that we need the same time, $\Delta t$, to reach from h1 and h1/2, for any $h1 \in (0,H]$:
By modifying(take log and multiply stuff) the general equation of h we get the time taken to reach any height is
$$t = \frac{H}{2R}ln(\frac{H}{h})$$
Hence we know the difference of times to reach h1 and h1/2:
$$\Delta t = \frac{H}{2R}ln(\frac{2H}{2h1}) - \frac{H}{2R}ln(\frac{H}{h1}) =\frac{H}{2R}ln(2)$$
The above property is sometimes called memorylessness and is quite important in many places.
Let us assume almost empty to be one-sixteenth(roughly 5%) of the original volume(and since it is proportional to height, one-sixteenth of its original height).
The average rate is the ratio of total height covered to total time taken to cover it. So I can write it like this:
$$R_{avg} = \frac{\frac{H}{2} + \frac{H}{4} + ...\frac{H}{2^n} }{n(\Delta t)}$$
$$ = \frac{H(1-\frac{1}{2^n})}{n. \frac{H}{2R}ln(2)}$$ $$ = \frac{(1-\frac{1}{2^n}). R}{nln(2)}$$
This obviously goes to 0 as n goes to infinity, but we assumed 1/16th of the original height to be almost empty. Plug in n=4 here to get 0.33R which is less than R/2.