Cone with height $9$ cm and radius $3$ cm is filled at a rate of $1.2~\text{cm}^3$. Find the rate of change when $h=3$. A cone with radius $3$ cm and height $9$ cm is filled with water at a rate of $1.2~\text{cm}^3$. Find the rate of change of the height of the water when the height of the water is $3$ cm.
I differentiated both sides to get $$\frac{dV}{dt}= \frac{1}{3} \pi \cdot 2(3)~\frac{dh}{dt}$$ Solving for $dh/dt$ I got $5.23599$. My textbook says to use similar triangles but I didn't, I am wondering if there is another way to solve of the rate of change of the height of the water at $h=3$?
 A: $$ V =\dfrac{\pi}{3}r^2 h $$
Note that there are two terms when you differentiate a product.
$$ \dfrac{dV}{dt} =\dfrac{\pi}{3} (2r\dfrac{dr}{dt} h + r^2 \dfrac{dh}{dt})\tag1$$
If $\alpha $ is semi-vertical angle (constant) of cone then by similar triangles of differential growths $(dr,dh)$ and quotient rule differentiation
$$\dfrac{dr/dt}{dh/dt} =\dfrac{r}{h}= \tan \alpha \tag2$$
I assume you have been asked to use similar triangles in order to appreciate that there is proportion between $(r,h)$ and their differentials $ (dr,dh)$ as shown.
Plug into (1) for $\dfrac{dr}{dt}$ from (2) as other quantities are known, in order to find out $\dfrac{dh}{dt}$. 
To find $dV/dt$ $ (r=1,h=3) $ .. should be finally plugged into (1) 
alternatively from the above we can write $r$ items into $h$ items to directly obtain $ ( t= \tan \alpha) $
$$ \dfrac{dV}{dt}=\frac{\pi}{3}(2ht\dfrac{dh}{dt} th+h^2t^2\dfrac{dh}{dt}) =\pi h^2 t^2\dfrac{d h}{dt} =\pi r_{max}^2 \dfrac{dh}{dt}= \pi \dfrac{dh}{dt};\,\, dh/dt= \frac{1.2}{\pi} \tag3 $$
EDIT1:
Using the disc method of finding differential volumes for integration we have at start as commented by Ross Millikan:
$$ dV = A dh =\pi r^2 dh,\, (dh/dt) = (dV/dt)/(\pi r^2) = \frac{1.2}{\pi} $$ 
obtained directly regardless of profile of the rotated solid. If grey shaded vessel has same top area (comparing with a conical vessel with its apex pointing down inverted) then also the same level/drain rates relation holds good. It is independent of second order volume quantities, differential triangle similarity is conceptually all built-in, what was calculated needlessly roundabout. 

A: The volume of a right-circular cone is given by the formula
$$V = \frac{1}{3}\pi r^2h$$
The volume is a function of two variables.  If we differentiate with respect to time, we obtain
$$\frac{dV}{dt} = \frac{2}{3}\pi rh~\frac{dr}{dt} + \frac{1}{3} \pi r^2~\frac{dh}{dt}$$
We wish to solve for $dh/dt$ given $dV/dt$ and $h$.  However, we do not know $r$ or $dr/dt$.
However, we can relate $r$ and $h$ by using similar triangles to eliminate $r$ from the equation and write the volume as a function of $h$.  Consider the diagram below.

By similar triangles,
$$\frac{r}{h} = \frac{3}{9} = \frac{1}{3} \implies r = \frac{h}{3}$$
Hence, we can express $V$ as a function of $h$.
\begin{align*}
V(h) & = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2h\\
     & = \frac{1}{27}\pi h^3
\end{align*}
Differentiating implicitly with respect to time yields
$$\frac{dV}{dt} = \frac{1}{9}\pi h^2~\frac{dh}{dt}$$
Since we are given $dV/dt$ and $h$, you can now solve for $dh/dt$.
If we had instead used the equation
$$\frac{dV}{dt} = \frac{2}{3}\pi rh~\frac{dr}{dt} + \frac{1}{3}\pi r^2~\frac{dh}{dt}$$
we would need to find $r$ and $dr/dt$.  To do so, we use similar triangles to obtain the relationship
$$r = \frac{h}{3}$$
which allows us to calculate $r$.  Differentiating with respect to time gives
$$\frac{dr}{dt} = \frac{1}{3}~\frac{dh}{dt}$$
We use similar triangles since we need to know the relationship between $r$ and $h$ in order to substitute for the variables in the equation relating $dV/dt$ and $dh/dt$.
A: Much simpler is to note that when $h=3$ cm the radius is $1$ cm.  The area of the water is then $\pi$ cm$^2$ so the rate of rise is $\frac {1.2}\pi$ cm/sec
