Hello, Dug!
4. Two inverted conical tanks have their vertices connected by a pipe.
One tank, initially full of water, has a height of 6 feet and a base radius of 3.5 feet.
The other tank, initially empty, has a height of 9 feet and a base radius of 4 feet.
If the water is allowed to flow through the connecting pipe, find the level to which
the water will rise in the empty tank. (Neglect the water in the pipe.)
The water will be at the same height $\displaystyle h$ in both tanks.
Code:
4
*---------+---------*
3.5 :\ | /
*-------+-------* : \ | /
:\ | / : \ | /
: \ | r1 / : \ | r2 /
: \----+----/ 9 \----+----/
6 \ | / : \ | /
: \ |h / : \ |h /
: \ | / : \ | /
: \|/ : \|/
- *-------------------*
The original volume of water is: .$\displaystyle V \:=\:\tfrac{\pi}{3}r^2h \:=\:\tfrac{\pi}{3}(\tfrac{7}{2})^2(6) \:=\:\frac{49}{2}\pi\text{ ft}^3$
The volume of water in the small tank is: .$\displaystyle V_1 \:=\:\tfrac{\pi}{3}(r_1)^2h$
. . where: .$\displaystyle \frac{r_1}{h} \,=\,\frac{\frac{7}{2}}{6} \quad\Rightarrow\quad r_1 \:=\:\tfrac{7h}{12}$
Hence: .$\displaystyle V_1 \;=\;\tfrac{\pi}{3}(\tfrac{7h}{12})^2h \;=\;\frac{49\pi}{432}h^3$
The volume of water in the large tank is: .$\displaystyle V_2 \;=\;\tfrac{\pi}{3}(r_2)^2h$
. . where: .$\displaystyle \frac{r_2}{h} \,=\,\frac{4}{9} \quad\Rightarrow\quad r_2 \:=\:\frac{4h}{9}$
Hence: .$\displaystyle V_2 \;=\;\tfrac{\pi}{3}(\tfrac{4h}9})^2h \;=\;\frac{16\pi}{243}h^3$
Since $\displaystyle V_1 + V_2 \:=\:V$, we have: .$\displaystyle \frac{49\pi}{432}h^3 + \frac{16\pi}{243}h^3 \;=\;\frac{49\pi}{2}$
. . $\displaystyle \frac{697\pi}{3888}h^3 \:=\:\frac{49\pi}{2} \quad\Rightarrow\quad h^3 \:=\:\frac{95,\!256}{697} \:=\:136.6657102$
$\displaystyle \text{Therefore: }\;h \;=\;\sqrt[3]{136.6657102} \;=\;5.150940353 $