- #1
EtherealMonkey
- 41
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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?
[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: