Calculus, water poured into a cone: Why is the derivative non-linear?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 .

Answer 6

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: $$\frac{dh'}{dt}\sim t^{-2/3}$$

Answer 7

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.

Answer 8

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.