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shubhamsrg

  • 2 years ago

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 .

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  1. muntoo
    • 2 years ago
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    |dw:1350805115720:dw|

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

  3. nincompoop
    • 2 years ago
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    |dw:1350805844589:dw|

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

  5. muntoo
    • 2 years ago
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    Maybe this is an optimization problem or something?

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

  7. calculusfunctions
    • 2 years ago
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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?

  8. shubhamsrg
    • 2 years ago
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    yes sir,,i do..

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

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

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

  12. shubhamsrg
    • 2 years ago
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    take your time sir! :)

  13. estudier
    • 2 years ago
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    http://mathworld.wolfram.com/Quarter-TankProblem.html

  14. muntoo
    • 2 years ago
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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|

  15. sara12345
    • 2 years ago
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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

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

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

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

  19. sauravshakya
    • 2 years ago
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    Then, I dont think its a calculas problem.

  20. shubhamsrg
    • 2 years ago
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    @mukushla ? :/

  21. sara12345
    • 2 years ago
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    |dw:1350914228619:dw|

  22. sara12345
    • 2 years ago
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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

  23. sauravshakya
    • 2 years ago
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    Isnt x=9 given???

  24. sara12345
    • 2 years ago
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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

  25. sauravshakya
    • 2 years ago
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    And we have to prove |dw:1350915430127:dw|

  26. sara12345
    • 2 years ago
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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.

  27. sauravshakya
    • 2 years ago
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    Did u solve theta and pluged in its value ???

  28. sara12345
    • 2 years ago
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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%

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