https://class.coursera.org/calcsing-002/lecture/320
In the the above linked lecture at 5:44, we are trying to find how fast liquid leaks from a cone shaped tank. I understand the derivation but at the end it mentions that $h$, which is in $\mathcal O ( t_\mathrm{end}-t )^{ \frac{2}{5} }$, goes to zero rapidly, which I don't understand.
The derivation goes like this:
If the slope of the side of the cone is y=mx, the radius is x, the height is $y$, and the area is $$A(y)= \pi x^2 = \pi (y/m)^2$$ If this is a non-circular shape, we can just use whatever the area is as a function of height $$A(h) = \mathcal O(h^2)$$
$$\frac{dh}{dt} = -\frac{\kappa}{ h^2}\sqrt{h} = - \kappa h^{-\frac{3}{2}}$$
We can separate the variables. $$h^{\frac{2}{3}} \, dh = - \kappa \, dt $$
$$\int h^{\frac{2}{3}} \, dh = \int - \kappa \, dt $$
$$\frac{2}{5}h^{\frac{5}{2}} \, dh = - \kappa t + c $$
The 2/5ths is absorbed into the constants $$h = ( - \kappa t + c )^{\frac{2}{5}}$$
Here's my question:
From what I understand, a function $f(x)$ is in $O( g(x) )$ if $f(x)$ approaches zero as fast or faster than $g(x)$.
For example, when $x$ is close to zero, $f(x)=x^2$ is in $O(x^{\frac{1}{2}})$ because for every value of $x$, $f(x)=x^2 \le g(x)=x^{ \frac{1}{2} }$
Using a test value of $x=.01$, $x^2$ is smaller than $x^{\frac{2}{5}}$ $.01^2 - (.01)^{2/5}$
It seems like $h= ( x^{\frac{2}{5}})$ approaches zero slowly.
Initially, I thought $h= ( x_f-x)^{\frac{2}{5}}$ would be in $\mathcal O (x^{\frac{2}{5}})$
However, $ (x_f-x)^{\frac{2}{5}}$ can be written in binomial form and then expanded with a Taylor series. The leading term is ${x_f}^{\frac{2}{5}}$, which still doesn't seem to approach zero very fast. Why does the lecture say that $\mathcal O ( t_\mathrm{end}-t )^{ \frac{2}{5} }$ goes to zero quickly?
Speed is the derivative w.r.t. time. If $h=K(t_e- t)^{2/5}$, then $$ h'(t)=\frac{dh}{dt}=-(2K/5)(t_e-t)^{-3/5}. $$ So when $t\to t_e$ from below, we have $h'(t)<0$ in accordance with the observation that the surface of the water is decreasing. Note that $$ \lim_{t\to t_e-}h'(t)=-\lim_{t\to t_e-}\frac{2K}{5(t_e-t)^{3/5}}=-\infty. $$ So the rate of descent of the surface grows without an upper bound as $t\to t_e$. That's fast in my book :-)
If the depth of the remaining liquid behaved like $Kt^2$, then the depth would, indeed, reach a small value earlier (in comparison to $Kt^{2/5}$), but this means that there is very little liquid remaining towards the end, and the depth is close to constant towards the end, i.e. not changing at all. I guess it is a matter of taste, which you would call fast. Whoever wrote that book at least appears to compare the speeds of change of depth.