Question

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...

Answer 1

...empty. Compute dz/dt.
D. What fraction of the total amount of water is in the large tank at time t = 6?

Answer 2

@dan815 look at the first comment, it continues the rest of the question :)

Answer 3

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:
\[\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}\]

Answer 4

am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec

Answer 5

You were right, volume as function of time is V(t) = pi (r)^2h - 12 t

Answer 6

oh dang

Answer 7

V = 0 to get the time when it's empty .

Answer 8

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)?

Answer 9

so we have:
\[\Large \frac{{d{h_1}}}{{dt}} = - \frac{{12}}{{\pi r_1^2}} = - \frac{{12}}{{16\pi }}\]

Answer 10

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.

Answer 11

Yeah , samll -12 , big + 12

Answer 12

so t=6 is the time it takes for the small one to completely empty?

Answer 13

-16*

Answer 14

Nope, when the samll is empty V(t) = 0

Answer 15

oh, oh never mind, I see that now

Answer 16

@Michele_Laino , Your method is right, But the question specifies that it wants it with V(t)

Answer 17

ok!

Answer 18

So then how does it work throwing depth into the mix for part C?

Answer 19

okay so you know how the volume is changing in the big one, what can you say about the height then

Answer 20

I really don't understand it at all. ask Michele or dan.

Answer 21

The height is the height of water

Answer 22

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.

Answer 23

right,
you know that the dv/dt = 12 cubic feet/sec

Answer 24

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

Answer 25

dV/dt = 2pir^2 dh/dt solve for dh/dt

Answer 26

how much height is that for every 12 cubic feet?

Answer 27

We want rate of change of height ?

Answer 28

and for sure dh/dt=dz/dt??

Answer 29

the radius is 8 so

12 = pi * 8^2 * height
height = ?

Answer 30

what is dz/dt ?

Answer 31

that tells you how much height you have changed for every 12 cubic feet of volume which is dz/dt

Answer 32

rate of change of depth? so it is dh/dt? and height would then be 12/(64pi)?

Answer 33

Yeah, but dh/dt for small is different from dh/dt for big so we have to solve for each separately

Answer 34

yeah

Answer 35

Yeah

Answer 36

dont they want you to just say the change in height for the big one

Answer 37

So we just solved for the small tank? But we used a radius of 8, which is from the large tank?

Answer 38

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

Answer 39

In the big tank dv/dt = 12 we solved already for the big one.

Answer 40

@dan815 , at the small one is will be -12/16pi the height is decreasing

Answer 41

right you can say that

Answer 42

but not dv/dt for the large one. dh/dt like the question asks?

Answer 43

12 = pi (8)^2 dz/dt ( that's the one your question asks for)

Answer 44

Oh! Thank you

Answer 45

yw

Answer 46

So, part D then addresses the fraction of the water that is in the tank at t=6... Is that v(6)/maximizedv(t)?

Answer 47

yep , v(6)

Answer 48

You mean the percentage of water ? at t = 6 ?

Answer 49

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?

Answer 50

Fraction = v(6)/ v(0) if I didn't misunderstand

Answer 51

It will be like a/b from the small one is in the big one.

Answer 52

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

Answer 53

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.

Answer 54

asks for*

Answer 55

why are you doing v(6)/v(0)

Answer 56

Current water / total water hmm ?

Answer 57

v(0) for the big tank is = remember

Answer 58

v(0) for big tank = 0

Answer 59

I am doing v(0) for the samll tank

Answer 60

v(t) = pi (4)^2 - 12 t

Answer 61

It's like current water/ volume of small

Answer 62

i thought D was asking for large tank

Answer 63

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 ?

Answer 64

It is asking for large tank
so would it look like:
pi (8)^2 + 12 (6) /pi (4)^2 - 12 (0) ?

Answer 65

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?

Answer 66

I was thinking of get 12 * 6/ 4^2 pi

Answer 67

why?

Answer 68

12 * 6 ( amount lost) / total volume

Answer 69

@dan815 , come.

Answer 70

ah i gotcha!

Answer 71

pi (8)^2 + 12t < means we are adding more than the volume of the large

Answer 72

Check with me number D dan

Answer 73

you are doing it right

Answer 74

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

Answer 75

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

Answer 76

xD

Answer 77

so they want fraction in the big , find we get amount in big / volume of big

Answer 78

But they only want the fraction of the water, which would be amount in big/ volume of small

Answer 79

yes water

Answer 80

Ey selective reading fails dude -_-

Answer 81

so do
v(t)= pi*4^2*6-12t

1- (V(6)/V(0))

Answer 82

V(6)/V(0) gives you the fraction in the small tank so 1- that is the fraction in the big tank

Answer 83

I agree with dan815 ++

Answer 84

gimme medal + monay plz

Answer 85

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?

Answer 86

Let's say v(6)/ v(0) is 2/3 of the volume in the water so in the big is 1/3

Answer 87

Oh! Gotcha. Thank you both so much!

Answer 88

yw

Answer 89

sure thing :)