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shubhamsrg

a sealed cylindrical drum of radius r is filled with 9% of water. Now if the drum is tilted to rest on its side, show that the fraction of the curved surface area (not counting the flat sides) that will be under is water is less than 1/12 .

  • one year ago
  • one year ago

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  1. muntoo
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    |dw:1350805115720:dw|

    • one year ago
  2. muntoo
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    Ugh. OK, that says \(w = \textrm{water_volume} = \pi r^2 \cdot 9x\) Surface area of round thingamagic is \[s = 2 \pi r \cdot 100x \]

    • one year ago
  3. nincompoop
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    |dw:1350805844589:dw|

    • one year ago
  4. muntoo
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    Ignore me too. I just realized this probably requires calculus stuff I'm not familiar with. :P

    • one year ago
  5. muntoo
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    Maybe this is an optimization problem or something?

    • one year ago
  6. muntoo
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    Let's see how this works out: http://math.stackexchange.com/questions/217875/tilting-sealed-drums-for-fun

    • one year ago
  7. calculusfunctions
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    @shubhamsrg the curved surface area of a cylinder is also known as the lateral area and is equal to the product of the circumference of the base and the height of the cylinder. In other words, the lateral area is given by\[2\pi rh\]. Did you already know this?

    • one year ago
  8. shubhamsrg
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    yes sir,,i do..

    • one year ago
  9. sara12345
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    im getting this minimize : .09 r/2 [1 - \(\frac{sin\theta}{\theta}\) ]

    • one year ago
  10. sara12345
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    ohh \(\theta \) is also constant here as volume is fixed, hang on..

    • one year ago
  11. calculusfunctions
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    Sorry @shubhamsrg I had to go away for a second but am back now. OK give me a second.

    • one year ago
  12. shubhamsrg
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    take your time sir! :)

    • one year ago
  13. estudier
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    http://mathworld.wolfram.com/Quarter-TankProblem.html

    • one year ago
  14. muntoo
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    From http://math.stackexchange.com/q/217875/2736 : The height and radius of the cylinder are irrelevant. Think of a unit circle in the plane. You have a chord, parallel to the y axis, that cuts your circle. The cut piece accounts for 9% of the area of the circle, and you're interested in the percentage of circumference between the two endpoints of the chord. – Steve D -- |dw:1350814466991:dw|

    • one year ago
  15. sara12345
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    that gave me the area ratio as, \(\frac{\theta}{2\pi} = .09 + \frac{sin \theta}{2\pi }\) when \(\theta = 1.565\) , gives the ratio of 1/4

    • one year ago
  16. muntoo
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    \[ \huge {\pi r^2 \over {2\pi \over \theta}} - {1\over2}rr \sin \theta = 0.09 \pi r^2 \]

    • one year ago
  17. sauravshakya
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    The cylinder can be tilted any way or it is tilted in following way|dw:1350825580056:dw|

    • one year ago
  18. shubhamsrg
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    res on its side right,,i'd assume the one in your figure..

    • one year ago
  19. sauravshakya
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    Then, I dont think its a calculas problem.

    • one year ago
  20. shubhamsrg
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    @mukushla ? :/

    • one year ago
  21. sara12345
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    |dw:1350914228619:dw|

    • one year ago
  22. sara12345
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    Volume of water = (Area of Sector - Area of triangle) x h = (\(\pi r^2\frac{\theta}{2\pi} - \frac{1}{2}r^2\sin \theta\)) x h = \(\frac{r^2}{2}(\theta - \sin \theta\)) x h say, this volume equals x% of total volume => \(\frac{r^2}{2}(\theta - \sin \theta\)) x h = \(\frac{x}{100} \ \pi r^2 h\) \(\frac{1}{2}(\theta - \sin \theta\)) = \(\frac{x}{100} \ \pi \) \(x = \frac{50}{\pi}(\theta - \sin \theta\)) \) plugin \(\theta = 2\pi/12 = \pi/6\) and see what \(x\) % you get

    • one year ago
  23. sauravshakya
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    Isnt x=9 given???

    • one year ago
  24. sara12345
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    only when \(\theta = \pi/2\), we get x = 9.1 % => the water covers less than 1/4 th of the curved surface area when volume = 9% of tot volume

    • one year ago
  25. sauravshakya
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    And we have to prove |dw:1350915430127:dw|

    • one year ago
  26. sara12345
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    9% volume and 1/12 surface area doesnt look correct. for water to cover less than 1/12th of curved surface area it needs to have very high surface tension ;P wid my working im getting 9% volume covers 1/4 surface area.

    • one year ago
  27. sauravshakya
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    Did u solve theta and pluged in its value ???

    • one year ago
  28. sara12345
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    i didnt solve \(\theta\), since the problem says \(\theta\) needs to be less than \(\pi/6\), ive just plugged in \(\theta=\pi/6\) and computed volume of water which comes nowhere near to 9%

    • one year ago
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