A tank of water has a base a circle of radius 2 meters and vertical sides. If water leaves the tank at a rate of 4 liters per minute, how fast is the water level falling in centimeters per hour? [1 liter is 1000 cubic centimeters]

Question

anonymous · Answer

Its a tough problem, I tried it

anonymous · Answer

I'm on it.. typing and trying to explain. patience please =]

anonymous · Answer

no problem haha

anonymous · Answer

Ok let's see
$$
\text{Base area} = \pi \cdot (2_{m})^{2} = \pi \cdot (200_{cm})^{2} = \pi \cdot 40\,000_{cm^2} \
$$
Now, we know the tank is a cylinder (circle base... vertical sides) and therefore we know that the water inside are in shape of cylinder with same base, only the height changes.
We know that 4 liters are $4000_{cm^3}$
That means that if we lost 4 liters of the water inside, the shape of the water lost $4000{cm^3}$ of its volume.
The volume is basically $base \cdot height$ and therefore we can find the height loss in the volume:
$$
h =\text{Height loss in 1 minute}=\frac{4\;000_{cm^3}}{\pi40\,000_{cm^2}} = \bigg(\frac{1}{10\pi}\bigg)_{cm}
$$
Means, every 4 liters the volume of the water in the tank gets $h$ less of its height.
We know that we lose 4 liters in a minute.. so in hour we lose like in 60 minutes so:
$$
\text{Height loss in 1 hour} = 60\cdot h =  60 \cdot\bigg(\frac{1}{10\pi}\bigg)_{cm} = \bigg(\frac{6}{\pi}\bigg)_{cm}
$$
Hopefully no dumb mistakes.. I go get a drink, then I'll come back and recheck that (again)

anonymous · Answer

wow amazing work, do you know how to solve half-life problems?

anonymous · Answer

Well, honestly.. I failed before getting there.
But I can read about it a bit and then try to help.

anonymous · Answer

It is known that the mass, m(t), of a radio-active substance decreases as it decays, and that the equation governing this is m'(t)=-0.02 m(t) where t is in years. What is the half-life?