A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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...
anonymous
 one year ago
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...

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 look at the first comment, it continues the rest of the question :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I 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: \[\Large \frac{{d{h_2}}}{{dt}} =  {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3You were right, volume as function of time is V(t) = pi (r)^2h  12 t

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3V = 0 to get the time when it's empty .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0so we have: \[\Large \frac{{d{h_1}}}{{dt}} =  \frac{{12}}{{\pi r_1^2}} =  \frac{{12}}{{16\pi }}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3at 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
 one year ago
Best ResponseYou've already chosen the best response.3Yeah , samll 12 , big + 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so t=6 is the time it takes for the small one to completely empty?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3Nope, when the samll is empty V(t) = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, oh never mind, I see that now

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3@Michele_Laino , Your method is right, But the question specifies that it wants it with V(t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then how does it work throwing depth into the mix for part C?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay so you know how the volume is changing in the big one, what can you say about the height then

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3I really don't understand it at all. ask Michele or dan.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3The height is the height of water

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Part 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
 one year ago
Best ResponseYou've already chosen the best response.2right, you know that the dv/dt = 12 cubic feet/sec

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3You 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
 one year ago
Best ResponseYou've already chosen the best response.3dV/dt = 2pir^2 dh/dt solve for dh/dt

dan815
 one year ago
Best ResponseYou've already chosen the best response.2how much height is that for every 12 cubic feet?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3We want rate of change of height ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and for sure dh/dt=dz/dt??

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the radius is 8 so 12 = pi * 8^2 * height height = ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2that tells you how much height you have changed for every 12 cubic feet of volume which is dz/dt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0rate of change of depth? so it is dh/dt? and height would then be 12/(64pi)?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, but dh/dt for small is different from dh/dt for big so we have to solve for each separately

dan815
 one year ago
Best ResponseYou've already chosen the best response.2dont they want you to just say the change in height for the big one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we just solved for the small tank? But we used a radius of 8, which is from the large tank?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.3In the big tank dv/dt = 12 we solved already for the big one.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3@dan815 , at the small one is will be 12/16pi the height is decreasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but not dv/dt for the large one. dh/dt like the question asks?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.312 = pi (8)^2 dz/dt ( that's the one your question asks for)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, part D then addresses the fraction of the water that is in the tank at t=6... Is that v(6)/maximizedv(t)?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3You mean the percentage of water ? at t = 6 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.3Fraction = v(6)/ v(0) if I didn't misunderstand

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3It will be like a/b from the small one is in the big one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That 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
 one year ago
Best ResponseYou've already chosen the best response.3If 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2why are you doing v(6)/v(0)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3Current water / total water hmm ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2v(0) for the big tank is = remember

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3I am doing v(0) for the samll tank

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3v(t) = pi (4)^2  12 t

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3It's like current water/ volume of small

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i thought D was asking for large tank

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3It'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
 one year ago
Best ResponseYou've already chosen the best response.0It is asking for large tank so would it look like: pi (8)^2 + 12 (6) /pi (4)^2  12 (0) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which 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
 one year ago
Best ResponseYou've already chosen the best response.3I was thinking of get 12 * 6/ 4^2 pi

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.312 * 6 ( amount lost) / total volume

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3pi (8)^2 + 12t < means we are adding more than the volume of the large

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3Check with me number D dan

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.2its 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
 one year ago
Best ResponseYou've already chosen the best response.3so they want fraction in the big , find we get amount in big / volume of big

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But they only want the fraction of the water, which would be amount in big/ volume of small

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3Ey selective reading fails dude _

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so do v(t)= pi*4^2*612t 1 (V(6)/V(0))

dan815
 one year ago
Best ResponseYou've already chosen the best response.2V(6)/V(0) gives you the fraction in the small tank so 1 that is the fraction in the big tank

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.3I agree with dan815 ++

dan815
 one year ago
Best ResponseYou've already chosen the best response.2gimme medal + monay plz

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what 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
 one year ago
Best ResponseYou've already chosen the best response.3Let's say v(6)/ v(0) is 2/3 of the volume in the water so in the big is 1/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh! Gotcha. Thank you both so much!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.