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 $h=3$?

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
