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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 $h=3$?

$$ V =dfrac{pi}{3}r^2 h $$

Note that there are two terms when you differentiate a product.

$$ dfrac{dV}{dt} =dfrac{pi}{3} (2rdfrac{dr}{dt} h + r^2 dfrac{dh}{dt})tag1$$

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) $

$$ dfrac{dV}{dt}=frac{pi}{3}(2htdfrac{dh}{dt} th+h^2t^2dfrac{dh}{dt}) =pi h^2 t^2dfrac{d h}{dt} =pi r_{max}^2 dfrac{dh}{dt}= pi dfrac{dh}{dt};,, dh/dt= frac{1.2}{pi} tag3 $$

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.

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}pileft(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 $dh/dt$.

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