## 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...

1. anonymous

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

2. anonymous

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

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

4. dan815

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

5. TrojanPoem

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

6. dan815

oh dang

7. TrojanPoem

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

8. 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)?

9. Michele_Laino

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

10. 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.

11. TrojanPoem

Yeah , samll -12 , big + 12

12. anonymous

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

13. dan815

-16*

14. TrojanPoem

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

15. anonymous

oh, oh never mind, I see that now

16. TrojanPoem

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

17. Michele_Laino

ok!

18. anonymous

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

19. dan815

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

20. TrojanPoem

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

21. TrojanPoem

The height is the height of water

22. 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.

23. dan815

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

24. 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

25. TrojanPoem

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

26. dan815

how much height is that for every 12 cubic feet?

27. TrojanPoem

We want rate of change of height ?

28. anonymous

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

29. dan815

the radius is 8 so 12 = pi * 8^2 * height height = ?

30. TrojanPoem

what is dz/dt ?

31. dan815

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

32. anonymous

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

33. TrojanPoem

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

34. dan815

yeah

35. TrojanPoem

Yeah

36. dan815

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

37. anonymous

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

38. 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

39. TrojanPoem

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

40. TrojanPoem

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

41. dan815

right you can say that

42. anonymous

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

43. TrojanPoem

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

44. anonymous

Oh! Thank you

45. TrojanPoem

yw

46. 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)?

47. TrojanPoem

yep , v(6)

48. TrojanPoem

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

49. 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?

50. TrojanPoem

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

51. TrojanPoem

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

52. 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

53. 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.

54. TrojanPoem

55. dan815

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

56. TrojanPoem

Current water / total water hmm ?

57. dan815

v(0) for the big tank is = remember

58. dan815

v(0) for big tank = 0

59. TrojanPoem

I am doing v(0) for the samll tank

60. TrojanPoem

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

61. TrojanPoem

It's like current water/ volume of small

62. dan815

i thought D was asking for large tank

63. 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 ?

64. anonymous

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

65. 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?

66. TrojanPoem

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

67. anonymous

why?

68. TrojanPoem

12 * 6 ( amount lost) / total volume

69. TrojanPoem

@dan815 , come.

70. dan815

ah i gotcha!

71. TrojanPoem

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

72. TrojanPoem

Check with me number D dan

73. dan815

you are doing it right

74. 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

75. 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

76. TrojanPoem

xD

77. TrojanPoem

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

78. anonymous

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

79. dan815

yes water

80. TrojanPoem

Ey selective reading fails dude -_-

81. dan815

so do v(t)= pi*4^2*6-12t 1- (V(6)/V(0))

82. dan815

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

83. TrojanPoem

I agree with dan815 ++

84. dan815

gimme medal + monay plz

85. 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?

86. TrojanPoem

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

87. anonymous

Oh! Gotcha. Thank you both so much!

88. TrojanPoem

yw

89. dan815

sure thing :)