- #1

- 41

- 0

**This was the problem before the one I am asking about (backstory maybe?) - Maybe relevant to my current problem**

[tex]

\frac{dh}{dt}=-\frac{A_{h}}{A_{w}}\sqrt{2gh}

[/tex]

**Problem statement:**

It seems that, by placing marks on the tank corresponding to the water level at intervals of, say, one hour one should be able to create a reasonably accurate clock. The idea is distinctly not new. Such clocks, known later as clepsydra (from the Greek

*kleptein*, to steal, and

*hydro*, water) are believed to have appeared in China around 4000 BCE. The known versions all have the rather inelegant property that, as [itex]\frac{dh}{dt}[/itex]is not constant, the distances between successive hourly marks decreases with time. Your task here is to use the formula (??), with [itex]A_{w}=A_{w}(h)[/itex]

now assumed to depend on

*h*, to design a 12-hour clepsydra with the dimensions shown above and shaped like the surface obtained by roataing the curve [itex]x=g(y)[/itex] around the

*y*-axis. You are to determine a formula for the function

*g*describing the shape of your clepsydra, together with the radius of the circular bottom hole, in order that the water level [itex]h=h(t)[/itex] will fall at the

*constant*rate of 4 inches per hour (be careful with the units here).

**Here is my work:**

[tex]

\frac{dh}{dt}=\frac{4 in}{1 hr} = \frac{\frac{4}{12} ft}{60 s}

[/tex]

[tex]

\frac{dh}{dt}=-0.00\bar{55}

[/tex]

[tex]

h(t)=-0.00\bar{55}t+C_{0}

[/tex]

**find:**[itex]h(t)[/itex] for [itex]t=0, h=4[/itex]

[tex]

4=-0.00\bar{55}(0)+C_{0}

[/tex]

[tex]

C_{0}=4

[/tex]

[tex]

h(t) \rightarrow h(t)=0.00\bar{55}t+4

[/tex]

**find:**[itex]t[/itex] for [itex]h(t)=0[/itex]

[tex]

0=0.00\bar{55}t+4

[/tex]

[tex]t=\frac{4}{0.00\bar{55}}\rightarrow720\left(s\right)[/tex]

or

[tex]12\left(h\right)[/tex]

**Now, here is where I am stuck:**

I have tried and tried to find an equation for the line that should be [itex]x=g(y)[/itex]

I know that the eqn of a circle is involved.

It seems that the radius of the "circles" (stacked to form the volume of the paraboloid) should vary with time as well. ([itex]dx[/itex]?)

I even tried to deduce the eqn from the formula for a paraboloid ([itex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=z[/itex]), but obviously, other than being able to plot the eqn in Maple, this did me little good.

Someone throw me a bone, please?

Last edited: