anonymous
  • anonymous
Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
anonymous
  • anonymous
@dan815 look at the first comment, it continues the rest of the question :)
Michele_Laino
  • Michele_Laino
I think another way to solve the problem is to start from this differential equation: \[\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0\] and substituting the formula for each volume V, we get: \[\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}\]

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dan815
  • dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
TrojanPoem
  • TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
dan815
  • dan815
oh dang
TrojanPoem
  • TrojanPoem
V = 0 to get the time when it's empty .
anonymous
  • anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
Michele_Laino
  • Michele_Laino
so we have: \[\Large \frac{{d{h_1}}}{{dt}} = - \frac{{12}}{{\pi r_1^2}} = - \frac{{12}}{{16\pi }}\]
TrojanPoem
  • TrojanPoem
at t = 6 get the amount of water from v(t) you will get the water in the small one and and minus it from the total volume of samll you will get the water in the big one.
TrojanPoem
  • TrojanPoem
Yeah , samll -12 , big + 12
anonymous
  • anonymous
so t=6 is the time it takes for the small one to completely empty?
dan815
  • dan815
-16*
TrojanPoem
  • TrojanPoem
Nope, when the samll is empty V(t) = 0
anonymous
  • anonymous
oh, oh never mind, I see that now
TrojanPoem
  • TrojanPoem
@Michele_Laino , Your method is right, But the question specifies that it wants it with V(t)
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
So then how does it work throwing depth into the mix for part C?
dan815
  • dan815
okay so you know how the volume is changing in the big one, what can you say about the height then
TrojanPoem
  • TrojanPoem
I really don't understand it at all. ask Michele or dan.
TrojanPoem
  • TrojanPoem
The height is the height of water
anonymous
  • anonymous
Part C was supposed to read: C. Let z be the depth of the water in the large tank, which initially was empty. Compute dz/dt.
dan815
  • dan815
right, you know that the dv/dt = 12 cubic feet/sec
TrojanPoem
  • TrojanPoem
You can easily find it from the Volume = 2 pi r^2 h ( get dev for time) 2 pi r^2 are constants h is the height of water
TrojanPoem
  • TrojanPoem
dV/dt = 2pir^2 dh/dt solve for dh/dt
dan815
  • dan815
how much height is that for every 12 cubic feet?
TrojanPoem
  • TrojanPoem
We want rate of change of height ?
anonymous
  • anonymous
and for sure dh/dt=dz/dt??
dan815
  • dan815
the radius is 8 so 12 = pi * 8^2 * height height = ?
TrojanPoem
  • TrojanPoem
what is dz/dt ?
dan815
  • dan815
that tells you how much height you have changed for every 12 cubic feet of volume which is dz/dt
anonymous
  • anonymous
rate of change of depth? so it is dh/dt? and height would then be 12/(64pi)?
TrojanPoem
  • TrojanPoem
Yeah, but dh/dt for small is different from dh/dt for big so we have to solve for each separately
dan815
  • dan815
yeah
TrojanPoem
  • TrojanPoem
Yeah
dan815
  • dan815
dont they want you to just say the change in height for the big one
anonymous
  • anonymous
So we just solved for the small tank? But we used a radius of 8, which is from the large tank?
dan815
  • dan815
if its the small one do the same thing, you know the volume change over time is 12 cubic feet /sec so 12 cubic feet = pi*4^2 *height height = 12/16pi this means the hight is changing by 12/16pi every second
TrojanPoem
  • TrojanPoem
In the big tank dv/dt = 12 we solved already for the big one.
TrojanPoem
  • TrojanPoem
@dan815 , at the small one is will be -12/16pi the height is decreasing
dan815
  • dan815
right you can say that
anonymous
  • anonymous
but not dv/dt for the large one. dh/dt like the question asks?
TrojanPoem
  • TrojanPoem
12 = pi (8)^2 dz/dt ( that's the one your question asks for)
anonymous
  • anonymous
Oh! Thank you
TrojanPoem
  • TrojanPoem
yw
anonymous
  • anonymous
So, part D then addresses the fraction of the water that is in the tank at t=6... Is that v(6)/maximizedv(t)?
TrojanPoem
  • TrojanPoem
yep , v(6)
TrojanPoem
  • TrojanPoem
You mean the percentage of water ? at t = 6 ?
anonymous
  • anonymous
But not just v(6), right? Because that would only be a number and I need the fraction? So I would have to divide v(6) by the maximized volume found by v(t), yes?
TrojanPoem
  • TrojanPoem
Fraction = v(6)/ v(0) if I didn't misunderstand
TrojanPoem
  • TrojanPoem
It will be like a/b from the small one is in the big one.
anonymous
  • anonymous
That makes sense, except that t=0 is empty in the large tank? I understand that the water is only what fills up the small thank, though... I'm honestly really confused by this part
TrojanPoem
  • TrojanPoem
If I didn't misunderstand the question for ex the question wants how much from the water is currently is in the big like 2/3 or 1/3 so we current amount/ total but how to get the total ? put t = 0 when no water has drained from the small one.
TrojanPoem
  • TrojanPoem
asks for*
dan815
  • dan815
why are you doing v(6)/v(0)
TrojanPoem
  • TrojanPoem
Current water / total water hmm ?
dan815
  • dan815
v(0) for the big tank is = remember
dan815
  • dan815
v(0) for big tank = 0
TrojanPoem
  • TrojanPoem
I am doing v(0) for the samll tank
TrojanPoem
  • TrojanPoem
v(t) = pi (4)^2 - 12 t
TrojanPoem
  • TrojanPoem
It's like current water/ volume of small
dan815
  • dan815
i thought D was asking for large tank
TrojanPoem
  • TrojanPoem
It's asking for the fraction of water in the large so we have to get the total water and the current in the big am I mistaken ?
anonymous
  • anonymous
It is asking for large tank so would it look like: pi (8)^2 + 12 (6) /pi (4)^2 - 12 (0) ?
anonymous
  • anonymous
Which was my hope of putting v(t) into large tank terms for v(6) and dividing it by v(t) for the small tank at v(0)? hopefully?
TrojanPoem
  • TrojanPoem
I was thinking of get 12 * 6/ 4^2 pi
anonymous
  • anonymous
why?
TrojanPoem
  • TrojanPoem
12 * 6 ( amount lost) / total volume
TrojanPoem
  • TrojanPoem
@dan815 , come.
dan815
  • dan815
ah i gotcha!
TrojanPoem
  • TrojanPoem
pi (8)^2 + 12t < means we are adding more than the volume of the large
TrojanPoem
  • TrojanPoem
Check with me number D dan
dan815
  • dan815
you are doing it right
dan815
  • dan815
i have bad habit of not reading question lol, i just read what fraction and stopped lol, and assumed they wanted the fraction of the height of the big tank
dan815
  • dan815
its called selective reading xD u only read words here and there, u tend to do it when u gotta read so many questions over and over
TrojanPoem
  • TrojanPoem
xD
TrojanPoem
  • TrojanPoem
so they want fraction in the big , find we get amount in big / volume of big
anonymous
  • anonymous
But they only want the fraction of the water, which would be amount in big/ volume of small
dan815
  • dan815
yes water
TrojanPoem
  • TrojanPoem
Ey selective reading fails dude -_-
dan815
  • dan815
so do v(t)= pi*4^2*6-12t 1- (V(6)/V(0))
dan815
  • dan815
V(6)/V(0) gives you the fraction in the small tank so 1- that is the fraction in the big tank
TrojanPoem
  • TrojanPoem
I agree with dan815 ++
dan815
  • dan815
gimme medal + monay plz
anonymous
  • anonymous
what I'm not understanding is that it can't be the same v(t) because one is referring to the big tank and one is referring to the small tank? I'm confused as to how the 1- fixes that?
TrojanPoem
  • TrojanPoem
Let's say v(6)/ v(0) is 2/3 of the volume in the water so in the big is 1/3
anonymous
  • anonymous
Oh! Gotcha. Thank you both so much!
TrojanPoem
  • TrojanPoem
yw
dan815
  • dan815
sure thing :)

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