shubhamsrg
  • 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 .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1350805115720:dw|
anonymous
  • anonymous
Ugh. OK, that says \(w = \textrm{water_volume} = \pi r^2 \cdot 9x\) Surface area of round thingamagic is \[s = 2 \pi r \cdot 100x \]
nincompoop
  • nincompoop
|dw:1350805844589:dw|

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anonymous
  • anonymous
Ignore me too. I just realized this probably requires calculus stuff I'm not familiar with. :P
anonymous
  • anonymous
Maybe this is an optimization problem or something?
anonymous
  • anonymous
Let's see how this works out: http://math.stackexchange.com/questions/217875/tilting-sealed-drums-for-fun
calculusfunctions
  • calculusfunctions
@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?
shubhamsrg
  • shubhamsrg
yes sir,,i do..
anonymous
  • anonymous
im getting this minimize : .09 r/2 [1 - \(\frac{sin\theta}{\theta}\) ]
anonymous
  • anonymous
ohh \(\theta \) is also constant here as volume is fixed, hang on..
calculusfunctions
  • calculusfunctions
Sorry @shubhamsrg I had to go away for a second but am back now. OK give me a second.
shubhamsrg
  • shubhamsrg
take your time sir! :)
anonymous
  • anonymous
http://mathworld.wolfram.com/Quarter-TankProblem.html
anonymous
  • anonymous
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|
anonymous
  • anonymous
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
anonymous
  • anonymous
\[ \huge {\pi r^2 \over {2\pi \over \theta}} - {1\over2}rr \sin \theta = 0.09 \pi r^2 \]
anonymous
  • anonymous
The cylinder can be tilted any way or it is tilted in following way|dw:1350825580056:dw|
shubhamsrg
  • shubhamsrg
res on its side right,,i'd assume the one in your figure..
anonymous
  • anonymous
Then, I dont think its a calculas problem.
shubhamsrg
  • shubhamsrg
@mukushla ? :/
anonymous
  • anonymous
|dw:1350914228619:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
Isnt x=9 given???
anonymous
  • anonymous
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
anonymous
  • anonymous
And we have to prove |dw:1350915430127:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
Did u solve theta and pluged in its value ???
anonymous
  • anonymous
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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