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If water is poured into a cone at a constant rate and if $\frac {dh}{dt}$ is the rate of change of the depth of the water, I understand that $\frac {dh}{dt}$ is decreasing. However, I don't understand why $\frac {dh}{dt}$ is non-linear. Why can't it be linear?

I am NOT asking whether or not the height function is linear. Many are telling me that the derivative of height is not a constant so thus the height function is not linear, but this is not what I am asking.

This is my mistake, because I had used $h(t)$ originally to denote the derivative of height which is what my book used. Rather I am asking if $\frac {dh}{dt}$ is linear or not and why. It would be nice if someone could better explain what my book is telling me:

At every instant the portion of the cone containing water is similar to the entire cone; the volume is proportional to the cube of the depth of the water. The rate of change of depth (the derivative) is therefore not linear.

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    $\begingroup$ The higher the depth of water in the cone, the more “room” it has to spread out. Layers of water near the top will be much thinner than layers of water of equal volume near the bottom. $\endgroup$ – Matthew Leingang Jun 27 at 0:54
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    $\begingroup$ One way to understand it visually is with martini glasses. They're only half full (in volume) when they look almost full (79% of the height). youtube.com/watch?v=Ut_0LvLQmI4 $\endgroup$ – Eric Duminil Jun 27 at 12:23
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    $\begingroup$ What an unexpectedly great question. $\endgroup$ – Randall Jun 29 at 23:15
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I understand that $\frac {dh}{dt}$ is decreasing. However, I don't understand on an intuitive level why $\frac {dh}{dt}$ is non-linear.

Because any strictly decreasing function that is linear eventually becomes negative, but you already know that $\frac {dh}{dt}$ is always positive.

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    $\begingroup$ Oh there must be an asymptote right? $\endgroup$ – user532874 Jun 27 at 9:52
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    $\begingroup$ yes i did say intuitive before my bad. This answer is just one sentence and my question has been answered :) $\endgroup$ – user532874 Jun 27 at 9:55
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    $\begingroup$ @user532874: That's in fact correct. A strictly decreasing positive real function must have an asymptote. This is a consequence of the fact that any non-empty subset of reals that has a lower bound has a limit, which you will encounter and prove in a real analysis course. $\endgroup$ – user21820 Jun 27 at 9:56
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    $\begingroup$ @user532874: Sorry I accidentally wrote wrongly in my above comment; any non-empty subset of reals that has a lower bound has an infimum (greatest lower bound), and hence any decreasing positive function must have a limit. $\endgroup$ – user21820 Jun 27 at 10:16
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    $\begingroup$ @Acccumulation: No, that statement is true when you look only on the positive half line. The full meaning of the statement "the function eventually becomes negative" is "there exists a constant $K \in \mathbb R$ such that the function is negative on the half-line $(K,+\infty)$", and if this is true then the following stronger statement is also true: "there exists a POSITIVE constant $K > 0$ such that the function is negative on the half line $(K,+\infty)$". $\endgroup$ – Lee Mosher Jun 29 at 16:48
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The volume of water is changing linearly, but the height and volume are related nonlinearly. That is why $h(t)$ is non-linear.

\begin{eqnarray*} V &=& \frac{1}{3} \pi r^2 h\\ r &=& h \tan \theta\\ V &=& \frac{1}{3} \pi \tan^2 \theta h^3\\ \frac{dV}{dt} &=& \frac{1}{3} \pi \tan^2 \theta 3 h^2 \frac{dh}{dt}\\ \end{eqnarray*}

$\frac{dV}{dt}$ is a constant because the constant filling rate. So you can see from the above equation that $\frac{dh}{dt}$ is not constant.

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    $\begingroup$ +1, but if you could finish the calculation of $\frac{dh}{dt}$ as an explicit function of $t$ then this answer would be even better :) $\endgroup$ – Pedro A Jun 27 at 1:59
  • $\begingroup$ No, he means "not constant" in that dh/dt is a function that changes with time. Treat dV/dt and dh/dt as variables and solve for dh/dt... $\endgroup$ – Ron Jensen Jun 27 at 3:40
  • $\begingroup$ Sorry this doesn't answer my question I should have used clearer notation $\endgroup$ – user532874 Jun 27 at 8:44
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    $\begingroup$ It may be helpful to note that the proportion dV/dh is exactly equal to the area of the top surface of the water, and when we're talking about a cone, that area is proportional to h^2 (so not linear, or inverse linear). $\endgroup$ – Brilliand Jun 27 at 23:01
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The top radius $r$ of the cone is proportional to $h$: We have $r(t)=c\>h(t)$ for some constant $c$. Therefore $V(t)=c\> h^3(t)$ with some other $c$, or $h(t)=c\> V^{1/3}(t)$. This implies $${dh\over dt}=c\> V^{-2/3}(t) V'(t)=c\> V^{-2/3}(t)\ ,$$ since $V'(t)$ is constant. Now $t\mapsto V(t)$ is linear; hence $t\mapsto V^{-2/3}(t)$ is a "root function", and not linear.

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  • $\begingroup$ It might be a bit confusing that the $c$ in $cV^{-2/3}(t)$ is $cV'(t)$ in the expression $cV^{-2/3}V'(t)$. $\endgroup$ – N. F. Taussig Jun 27 at 8:53
  • $\begingroup$ best response yet. I understand everything except for how do I know the radius is proportional to the height? In order for radius to be proportional to height, wouldn't volume have to stay constant? $\endgroup$ – user532874 Jun 27 at 9:00
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    $\begingroup$ @user532874 $r = h \tan \theta$, and $\theta$ is constant, so the radius is proportional to height. $\endgroup$ – muru Jun 27 at 9:26
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Imagine slicing the cone into circular cross-sections. These cross-sections get bigger as you go up the cone. Let's think about two cross-sections in particular: one slice spanning between 1 cm and 1.01 cm from the bottom, and another slice spanning between 5 cm and 5.01 cm from the bottom. Since the higher "slice" has a larger volume, and the water is entering the cone at a constant rate of volume per time, it must have taken longer for the water to rise from 5 cm to 5.01 cm than it took for the water to rise from 1 cm to 1.01 cm. Thus, $dh/dt$ (the rate at which the water is rising) is decreasing as $h$ increases.

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    $\begingroup$ Almost there - you've shown that dh/dt is decreasing as h increases, but not that it's a nonlinear relationship. The accepted answer nails this in an intuitive way - the water level in the cone is always rising, so dh/dt > 0 for all h. If dh/dt was a linearly decreasing function, it would become negative at some point, so that cannot be the case. Therefore, dh/dt is a nonlinear decreasing function. $\endgroup$ – Nuclear Wang Jun 27 at 14:47
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    $\begingroup$ Ah, yes, upon re-reading the question I realize that I was answering the question "why is $h$ not a linear function of $t$?" rather than "why is $dh/dt$ not a linear function of $t$?" $\endgroup$ – Michael Seifert Jun 27 at 14:54
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You can write the volume of the cone as $V = \frac{1}{3}\cdot\pi\cdot \left(h\tan\theta\right)^2\cdot h $. Here, we are writing $r =h\cdot\tan(\theta)$.

Coming to your question, the radius increases with an exponent of 2 and on the whole volume increases with the height cubed. So by intuition you can say that height must be varying with $t^{1/3}$ so as to balance things out since the rate of water being filled is constant.

You can also work it out as $h(t) = At^{1/3} + C$ ; $A = (3k/\pi)^{1/3}$, $k = $ rate of change of vol. and $C$ is arbitrary constant .

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  • $\begingroup$ You should read the tutorial on how to format mathematics on this site. $\endgroup$ – NickD Jun 27 at 3:52
  • $\begingroup$ You have interpreted my question correctly but I can't understand why height varies with t^1/3 and how this proves the derivative is nonlinear. $\endgroup$ – user532874 Jun 27 at 4:41
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The notion by Mike is nearly correct. Christian points out the correct result without being overly specific.

If you consider the volume of a cone of maximum height $h$ and maximum radius $R$ but only calculate it to the height $h'<h$, you can shuffle the equation to give that height depending on the volume of that fraction of the cone: $$h'=\left(\frac{3V h²}{\pi R²}\right)^{1/3}$$ We assume the tip of the cone pointing downwards and the opening in positive $z$-direction with the water pouring in from above that, so no sideway filling.

If you than put in $\frac{dV}{dt}\cdot t$ as the volume depending on the time, where $\frac{dV}{dt}$ is the constant that describes how much water is added to the cone per time element, you get: $$\frac{dh'}{dt}\sim t^{-2/3}$$

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Mathematically speaking, for a cone you have $$V=\frac {\pi}{ 3} r^2 h$$

$$\frac {dV}{dt} = \frac {\pi}{3} [ 2rh \frac{dr}{dt} +r^2 \frac{dh}{dt}]$$

Since $$\frac {dr}{dt}=c\frac {dh}{dt}$$ we get

$$\frac {dV}{dt} = \frac {\pi}{3} [( c +r^2) \frac{dh}{dt}]$$

While $\frac {dV}{dt}$ is constant, $\frac {dh}{dt}$ depends on $r$ and does not seem to be linear.

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  • $\begingroup$ Sorry this doesn't answer my question I should have used clearer notation $\endgroup$ – user532874 Jun 27 at 8:43
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Phrasing it $\frac {dh}{dt}$ is a bit wrong, as the $h$ is not a direct function of $t$. You may be considering a situation where water is being poured into the cone at constant rate, in which case volume is proportional to time. But then $t$ is just a proxy for $V$, and it would be more instructive to deal with $\frac {dh}{dV}$. That in turn is the reciprocal of $\frac {dV}{dh}$. And if we take an infinitesimal slice of the cone, that slice will be a frustrum. However, the surface area of the top of that frustrum will be almost the same as the surface area of the bottom (since it's an infinitesimal slice), and so we can approximate it with a cylinder. The volume of the cylinder is the surface area of the base time the height: $V=Ah$, so $\frac V h = A$. It therefore follows that $\frac {dV}{dh}$ at a particular height is the cross sectional area at that height.

As an intuitive explanation of this, suppose $m$ is the height of one molecule of water (strictly speaking, how much height a molecule takes up isn't well defined, but since this is just an intuitive explanation, I'll put that aside). To make the height of water go up $m$, you have to put in one layer of water molecules, and the amount of water that you need is going to be proportional to the surface area.

Thus, $\frac {dh}{dV}$ is the reciprocal of the area, so for it to be linear, the area would have to be the reciprocal of a linear function. Besides the fact that there is no reason to suppose that the area is such a function, we can directly see that it isn't by calculating it; it is proportional to height squared. We can also see that the area is zero at the base of the cone, so $\frac {dh}{dV}$ is infinity there, so it's not linear.

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