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.


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.
