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iop360 Group Title

A large container has the shape of a frustum of a cone with top radius 8m, bottom radius 2m, and height 6m. The container is being filled with water at the constant rate of 2.9m^3/min. At what rate is the level of water rising at the instant the water is 1m deep? Answer: 10.3cm/min

  • one year ago
  • one year ago

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  1. iop360 Group Title
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    \[\frac{ dV }{ dt } = +2.9, \frac{ dH }{ dt }=?\]

    • one year ago
  2. Chlorophyll Group Title
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    @iop360 Do you have the Volume formula?

    • one year ago
  3. iop360 Group Title
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    of a regular cone, yes

    • one year ago
  4. iop360 Group Title
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    \[V = \frac{ 1 }{ 3 }(\pi)r^2h\]

    • one year ago
  5. Chlorophyll Group Title
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    No this one, have 2 bases!

    • one year ago
  6. iop360 Group Title
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    i think what we are supposed to do though is make it into two cones though

    • one year ago
  7. iop360 Group Title
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    |dw:1353202550485:dw|

    • one year ago
  8. iop360 Group Title
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    R is top radius?

    • one year ago
  9. iop360 Group Title
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    b and B represent the different volumes?

    • one year ago
  10. iop360 Group Title
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    how did you derive this formula

    • one year ago
  11. iop360 Group Title
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    oh ok

    • one year ago
  12. iop360 Group Title
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    ill try it. thanks!

    • one year ago
  13. iop360 Group Title
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    hmm

    • one year ago
  14. iop360 Group Title
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    have you tried it?

    • one year ago
  15. iop360 Group Title
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    im getting a wrong answer

    • one year ago
  16. iop360 Group Title
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    i got .383...m/min

    • one year ago
  17. iop360 Group Title
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    answer is supposed to be 10.3 cm/min

    • one year ago
  18. iop360 Group Title
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    oh wait i think i made an error.. let me recalculate

    • one year ago
  19. iop360 Group Title
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    my expression for V is: \[V = [\frac{ 16 }{ 27 }(\pi)h^3 + \frac{ 4}{ 27 }(\pi)h^3 + \frac{ 1 }{ 9 }(\pi)h^3]\]

    • one year ago
  20. iop360 Group Title
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    what did you get?

    • one year ago
  21. iop360 Group Title
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    oh wait.. 1/27, not 1/9

    • one year ago
  22. iop360 Group Title
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    ok yeah thats what i get now

    • one year ago
  23. iop360 Group Title
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    taking its derivative, you get \[\frac{ 7 }{ 3 }(\pi)h^2\]

    • one year ago
  24. iop360 Group Title
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    dh/dt beside it of course

    • one year ago
  25. iop360 Group Title
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    (2.9)/(7pi/3) = dh/dt after you plug h =1, which does nothing

    • one year ago
  26. iop360 Group Title
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    0.395...m^3/min

    • one year ago
  27. iop360 Group Title
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    i think we did it right, not sure why its resulting in the wrong answer

    • one year ago
  28. iop360 Group Title
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    did you get 0.395...m/min

    • one year ago
  29. iop360 Group Title
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    doesnt that still result in 39.5cm/min then?

    • one year ago
  30. iop360 Group Title
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    try reading this http://www.askmehelpdesk.com/mathematics/related-rates-frustum-cone-352086.html

    • one year ago
  31. iop360 Group Title
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    it might be a solution, but i dont get it

    • one year ago
  32. iop360 Group Title
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    hmm is there anything extra in the formula you may have forgotten

    • one year ago
  33. iop360 Group Title
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    http://www.youtube.com/watch?v=1v1Pp-lJSKY this video ?

    • one year ago
  34. iop360 Group Title
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    http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html this link doesnt have a square root in the formula

    • one year ago
  35. iop360 Group Title
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    bleh, the formula without the sq.root doesnt work either...

    • one year ago
  36. iop360 Group Title
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    !!! i think i got it

    • one year ago
  37. iop360 Group Title
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    forgot the pi from the above forumla

    • one year ago
  38. iop360 Group Title
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    http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html

    • one year ago
  39. iop360 Group Title
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    i got 10.1 though...calculator didnt take exact values

    • one year ago
  40. iop360 Group Title
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    \[V = \frac{ (\pi)h }{ 3 }[R^2 + Rr + r^2]\]

    • one year ago
  41. iop360 Group Title
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    im going to calculate it again..

    • one year ago
  42. iop360 Group Title
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    im getting 10.05... cm/min that is slightly off

    • one year ago
  43. iop360 Group Title
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    hm ok

    • one year ago
  44. iop360 Group Title
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    thanks

    • one year ago
  45. iop360 Group Title
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    im going to try it with different values

    • one year ago
  46. iop360 Group Title
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    oh wait...i used base for r in that last formula

    • one year ago
  47. iop360 Group Title
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    uh oh

    • one year ago
  48. iop360 Group Title
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    ill try it with r/R

    • one year ago
  49. iop360 Group Title
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    annnnnnd were back to 21pi/27 h^3

    • one year ago
  50. iop360 Group Title
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    yeah, i accidently used the B/b for that forumla instead of R/r

    • one year ago
  51. iop360 Group Title
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    and yet i got close to a right answer

    • one year ago
  52. iop360 Group Title
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    do we sub in h = 1 or h =6? h = 1 is right isnt it?

    • one year ago
  53. Chlorophyll Group Title
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    Yes, we calculate the instant rate when h = 1

    • one year ago
  54. iop360 Group Title
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    the other guy reading this question: are you doing this question?

    • one year ago
  55. Callisto Group Title
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    *learning* ._.!

    • one year ago
  56. iop360 Group Title
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    i could switch the values for the question and try it again if you want, haha

    • one year ago
  57. iop360 Group Title
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    ok i will

    • one year ago
  58. iop360 Group Title
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    a large container has the shape of a frustum of a cone with top radius 10m, bottom radius 6m, and height 4m. The container is being filled with water at the constant rate of 2.9m^3/min. At what rate is the level of water rising at the instant rate the water is 2m deep?

    • one year ago
  59. iop360 Group Title
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    answer: 1.4cm/min

    • one year ago
  60. iop360 Group Title
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    alright

    • one year ago
  61. iop360 Group Title
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    i get 1.9 ..hmmm

    • one year ago
  62. iop360 Group Title
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    it could be a glitch with the thing, maybe.

    • one year ago
  63. iop360 Group Title
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    ok, im pretty sure were doing it right. imma close this down

    • one year ago
  64. iop360 Group Title
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    thanks for the help

    • one year ago
  65. Chlorophyll Group Title
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    Check with your teacher about the formula! I'll sure will message you if I find out something interesting :)

    • one year ago
  66. iop360 Group Title
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    thanks!

    • one year ago
  67. Chlorophyll Group Title
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    I'm so sorry that I'm underestimate this Related Rate frustum of Cone! Here's the correct logic: V = π/ 3 ( R² + Rr + r² ) H With r = 2 and R is Radius variable and H is the Height variable -> V = π/ 3 ( R² + 2R+ 4 ) H ( 1 ) From the ratio of right triangle: ( R - 2 ) / ( 8 - 2 ) = H/ 6 --> R = H + 2 (2) Plug (2) into (1): => V = π/ 3 [ ( H +2) + 2( H +2) + 4 ] H = π/ 3 [ H³ + 6H² + 12H ] so V' = π/ 3 [ 3H² + 12H + 12 ] H' At H = 1: V' = 27 π/ 3 * H' 2.9 = 9 π * H' Thus H' = 2.9 / 9π = .1025 m/ min = 10.3 cm/ min

    • one year ago
  68. iop360 Group Title
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    thanks!

    • one year ago
  69. iop360 Group Title
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    @Chlorophyll one question though.. what do you do exactly during the ratios of the triangle part? eg. say my top radius was 9, bottom was 8, and height was 5 how do you set it up exactly?

    • one year ago
  70. Chlorophyll Group Title
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    Ratio R and H: ( R - 8 )/ ( 9-8) = H/5 -> R = ( H/5 ) + 8

    • one year ago
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