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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...

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

  2. anonymous
    • one year ago
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    @dan815 look at the first comment, it continues the rest of the question :)

  3. Michele_Laino
    • one year ago
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    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
    • one year ago
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    am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec

  5. TrojanPoem
    • one year ago
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    You were right, volume as function of time is V(t) = pi (r)^2h - 12 t

  6. dan815
    • one year ago
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    oh dang

  7. TrojanPoem
    • one year ago
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    V = 0 to get the time when it's empty .

  8. anonymous
    • one year ago
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    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
    • one year ago
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    so we have: \[\Large \frac{{d{h_1}}}{{dt}} = - \frac{{12}}{{\pi r_1^2}} = - \frac{{12}}{{16\pi }}\]

  10. TrojanPoem
    • one year ago
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    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
    • one year ago
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    Yeah , samll -12 , big + 12

  12. anonymous
    • one year ago
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    so t=6 is the time it takes for the small one to completely empty?

  13. dan815
    • one year ago
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    -16*

  14. TrojanPoem
    • one year ago
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    Nope, when the samll is empty V(t) = 0

  15. anonymous
    • one year ago
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    oh, oh never mind, I see that now

  16. TrojanPoem
    • one year ago
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    @Michele_Laino , Your method is right, But the question specifies that it wants it with V(t)

  17. Michele_Laino
    • one year ago
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    ok!

  18. anonymous
    • one year ago
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    So then how does it work throwing depth into the mix for part C?

  19. dan815
    • one year ago
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    okay so you know how the volume is changing in the big one, what can you say about the height then

  20. TrojanPoem
    • one year ago
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    I really don't understand it at all. ask Michele or dan.

  21. TrojanPoem
    • one year ago
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    The height is the height of water

  22. anonymous
    • one year ago
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    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
    • one year ago
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    right, you know that the dv/dt = 12 cubic feet/sec

  24. TrojanPoem
    • one year ago
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    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
    • one year ago
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    dV/dt = 2pir^2 dh/dt solve for dh/dt

  26. dan815
    • one year ago
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    how much height is that for every 12 cubic feet?

  27. TrojanPoem
    • one year ago
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    We want rate of change of height ?

  28. anonymous
    • one year ago
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    and for sure dh/dt=dz/dt??

  29. dan815
    • one year ago
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    the radius is 8 so 12 = pi * 8^2 * height height = ?

  30. TrojanPoem
    • one year ago
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    what is dz/dt ?

  31. dan815
    • one year ago
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    that tells you how much height you have changed for every 12 cubic feet of volume which is dz/dt

  32. anonymous
    • one year ago
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    rate of change of depth? so it is dh/dt? and height would then be 12/(64pi)?

  33. TrojanPoem
    • one year ago
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    Yeah, but dh/dt for small is different from dh/dt for big so we have to solve for each separately

  34. dan815
    • one year ago
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    yeah

  35. TrojanPoem
    • one year ago
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    Yeah

  36. dan815
    • one year ago
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    dont they want you to just say the change in height for the big one

  37. anonymous
    • one year ago
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    So we just solved for the small tank? But we used a radius of 8, which is from the large tank?

  38. dan815
    • one year ago
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    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
    • one year ago
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    In the big tank dv/dt = 12 we solved already for the big one.

  40. TrojanPoem
    • one year ago
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    @dan815 , at the small one is will be -12/16pi the height is decreasing

  41. dan815
    • one year ago
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    right you can say that

  42. anonymous
    • one year ago
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    but not dv/dt for the large one. dh/dt like the question asks?

  43. TrojanPoem
    • one year ago
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    12 = pi (8)^2 dz/dt ( that's the one your question asks for)

  44. anonymous
    • one year ago
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    Oh! Thank you

  45. TrojanPoem
    • one year ago
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    yw

  46. anonymous
    • one year ago
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    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
    • one year ago
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    yep , v(6)

  48. TrojanPoem
    • one year ago
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    You mean the percentage of water ? at t = 6 ?

  49. anonymous
    • one year ago
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    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
    • one year ago
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    Fraction = v(6)/ v(0) if I didn't misunderstand

  51. TrojanPoem
    • one year ago
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    It will be like a/b from the small one is in the big one.

  52. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    asks for*

  55. dan815
    • one year ago
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    why are you doing v(6)/v(0)

  56. TrojanPoem
    • one year ago
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    Current water / total water hmm ?

  57. dan815
    • one year ago
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    v(0) for the big tank is = remember

  58. dan815
    • one year ago
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    v(0) for big tank = 0

  59. TrojanPoem
    • one year ago
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    I am doing v(0) for the samll tank

  60. TrojanPoem
    • one year ago
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    v(t) = pi (4)^2 - 12 t

  61. TrojanPoem
    • one year ago
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    It's like current water/ volume of small

  62. dan815
    • one year ago
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    i thought D was asking for large tank

  63. TrojanPoem
    • one year ago
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    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
    • one year ago
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    It is asking for large tank so would it look like: pi (8)^2 + 12 (6) /pi (4)^2 - 12 (0) ?

  65. anonymous
    • one year ago
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    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
    • one year ago
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    I was thinking of get 12 * 6/ 4^2 pi

  67. anonymous
    • one year ago
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    why?

  68. TrojanPoem
    • one year ago
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    12 * 6 ( amount lost) / total volume

  69. TrojanPoem
    • one year ago
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    @dan815 , come.

  70. dan815
    • one year ago
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    ah i gotcha!

  71. TrojanPoem
    • one year ago
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    pi (8)^2 + 12t < means we are adding more than the volume of the large

  72. TrojanPoem
    • one year ago
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    Check with me number D dan

  73. dan815
    • one year ago
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    you are doing it right

  74. dan815
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    xD

  77. TrojanPoem
    • one year ago
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    so they want fraction in the big , find we get amount in big / volume of big

  78. anonymous
    • one year ago
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    But they only want the fraction of the water, which would be amount in big/ volume of small

  79. dan815
    • one year ago
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    yes water

  80. TrojanPoem
    • one year ago
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    Ey selective reading fails dude -_-

  81. dan815
    • one year ago
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    so do v(t)= pi*4^2*6-12t 1- (V(6)/V(0))

  82. dan815
    • one year ago
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    V(6)/V(0) gives you the fraction in the small tank so 1- that is the fraction in the big tank

  83. TrojanPoem
    • one year ago
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    I agree with dan815 ++

  84. dan815
    • one year ago
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    gimme medal + monay plz

  85. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Oh! Gotcha. Thank you both so much!

  88. TrojanPoem
    • one year ago
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    yw

  89. dan815
    • one year ago
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    sure thing :)

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