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


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