Hello, roadrunner!
This is a tricky problem . . . with a number of "stages".
. . No wonder you had difficulty.
A solution is passing through a conical filter 24 in deep and 16 in across the top,
into a cylindrical vessel of diameter 12 in.
At what rate is the level of the solution in the cylinder rising if, when the depth of the solution in the filter is 12 in,
its level is falling at the rate 1 in/min? The answer given is 4/9 in/min.
The radius of the cylindrical vessel is \(\displaystyle 6\text{ in.}\)
The volume of solution in the cylindrical vessel is:
.\(\displaystyle V \:= \:\pi r^2h \:= \:\pi\cdot6^2h \:= \:36\pi h\text{ in}^3\)
. . Then:
.\(\displaystyle \frac{dV}{dt} \:= \:36\pi\left(\frac{dh}{dt}\right)\quad\Rightarrow\quad \frac{dh}{dt} \:= \:\frac{1}{36\pi}\left(\frac{dV}{dt}\right)\)
[1]
So if we knew \(\displaystyle \frac{dV}{dt}\) (how fast the solution is entering the cylinder), we'd be done now.
Now we must find \(\displaystyle \frac{dV}{dt}\), how fast the solution is <u>leaving</u> the conical filter.
Code:
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| \ | /
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24 | \:::|:r:/ Side view of conical filter
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| \:|:/
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The volume of solution in the cone is:
.\(\displaystyle V \:= \:\frac{1}{3}\pi r^2h\)
From the similar right triangles:
.\(\displaystyle \frac{r}{h} \ = \ \frac{8}{24}\quad\Rightarrow\quad r \ = \ \frac{h}{3}\)
Then we have:
.\(\displaystyle V \:= \:\frac{1}{3}\pi\left(\frac{h}{3}\right)^2h \:= \:\frac{\pi}{27}h^3\)
. . Hence:
.\(\displaystyle \frac{dV}{dt} \:= \:\frac{\pi}{9}h^2\left(\frac{dh}{dt}\right)\)
We are told that, when \(\displaystyle h = 12,\;\frac{dh}{dt} = -1\text{ in/min.}\)
. . Then:
.\(\displaystyle \frac{dV}{dt} \:= \:\frac{\pi}{9}(12^2)(-1) \;= \;-16\pi\;\text{in}^3\text{/min.}\)
There! . . . The solution is leaving the filter and entering the cylinder at \(\displaystyle 16\pi\text{ in}^3\text{/min.}\)
Substitute into
[1]:
.\(\displaystyle \frac{dh}{dt} \;= \;\frac{1}{36\pi}(16\pi) \;= \;\frac{4}{9}\text{ in/min.}\)
[Edit: Too fast for me, Galactus!]
