The task is about filling containers of different shape (cone, sphere, etc) – to find a function of depth $h(t)$ when filling a container with a constant rate of liquid supply.
The problem is that all the tutorials I've seen split into two groups: the ones that show calculations but no charts, and the ones that show just charts, with no math background.
So, let's start with an inverted cone. All the pictures that show up in google are showing non-linear functions, which is natural to me – the time to fill the cone increases as the height of the cone increases: image. That chart looks like some kind of $f(x)=\sqrt{x}$ function.
But in tutorials with math only, folks come up with some constant rate (for instance, $\frac{dh}{dt} = \frac{1}{\pi}$ $ft/sec$) at some instance of time. But how would I find a function that represents that rate?
Currently, I'm stuck in here:
We are given several initial constants, such as radius and height – $r$ and $h$. Also we are given the constant liquid supply rate $c$ ($ft^3 s^{-1}$). Cone volume function is $V = \frac{1}{3}\pi r^2 h$
To find $h'(t)$, one has to solve an equation: $\frac{dV}{dt} = \frac{dh}{dt} \frac{d}{dh} [\frac{1}{3} \pi r^2 h]$.
Let's say $r=h$, then the equation would look like:
$\frac{dV}{dt} = \frac{dh}{dt} \frac{d}{dh} [\frac{1}{3} \pi h^3]$
If $\frac{dV}{dt} = c$, then
$c = \frac{dh}{dt} \pi h^2$
Therefore $\frac{dh}{dt} = \frac{c}{\pi h^2}$ where h is also a constant.
So we come up with some constant value, which is a derivative of $h(t)$. We need to find a function that represents that rate. Thus, by integrating that value, we get $h(t) = \int{h'(t)dt}$. But by doing so, we receive a linear function $h(t) = \frac{c}{\pi h^2} t$, which is obviously some kind of miscalculation – it can't be linear, because the time to fill the volume increases with height of cone.
Perhaps, I mixed up all the calculus concepts and did it all wrong, but the question remains: how to find a function that represents the rate of depth with respect to time in cone, and how the other folks came up with that kind of non-linear charts (image)?
In addition, what about the other shapes, such as sphere? Is the concept alike?
$\endgroup$ 21 Answer
$\begingroup$Remember that $h$ is a function of $t,$ so we can't assume that $$\int\frac{c}{\pi h^2}\,dt=\frac{c}{\pi h^2}t,$$ as would be the case if $h$ were not a function of $t.$ Instead, we need to rearrange a bit, to get $$h^2\frac{dh}{dt}=\frac{c}{\pi}.$$ Observing that $h^2=\frac{d}{dh}\left[\frac13h^3\right],$ then by Chain Rule, we can rewrite the equation above as $$\frac{d}{dt}\left[\frac13h^3\right]=\frac{c}{\pi},$$ so that $$\frac13h^3=\frac{c}\pi t+K$$ for some constant $K.$ Assuming that $h(0)=0,$ the $K$ must be $0,$ so that $$h(t)=\sqrt[3]{\frac{3ct}\pi}.$$
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