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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
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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muntooBest ResponseYou've already chosen the best response.0
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

nincompoopBest ResponseYou've already chosen the best response.0
dw:1350805844589:dw
 one year ago

muntooBest ResponseYou've already chosen the best response.0
Ignore me too. I just realized this probably requires calculus stuff I'm not familiar with. :P
 one year ago

muntooBest ResponseYou've already chosen the best response.0
Maybe this is an optimization problem or something?
 one year ago

muntooBest ResponseYou've already chosen the best response.0
Let's see how this works out: http://math.stackexchange.com/questions/217875/tiltingsealeddrumsforfun
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
@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

sara12345Best ResponseYou've already chosen the best response.1
im getting this minimize : .09 r/2 [1  \(\frac{sin\theta}{\theta}\) ]
 one year ago

sara12345Best ResponseYou've already chosen the best response.1
ohh \(\theta \) is also constant here as volume is fixed, hang on..
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
Sorry @shubhamsrg I had to go away for a second but am back now. OK give me a second.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
take your time sir! :)
 one year ago

estudierBest ResponseYou've already chosen the best response.2
http://mathworld.wolfram.com/QuarterTankProblem.html
 one year ago

muntooBest ResponseYou've already chosen the best response.0
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

sara12345Best ResponseYou've already chosen the best response.1
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

muntooBest ResponseYou've already chosen the best response.0
\[ \huge {\pi r^2 \over {2\pi \over \theta}}  {1\over2}rr \sin \theta = 0.09 \pi r^2 \]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
The cylinder can be tilted any way or it is tilted in following waydw:1350825580056:dw
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
res on its side right,,i'd assume the one in your figure..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Then, I dont think its a calculas problem.
 one year ago

sara12345Best ResponseYou've already chosen the best response.1
dw:1350914228619:dw
 one year ago

sara12345Best ResponseYou've already chosen the best response.1
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

sauravshakyaBest ResponseYou've already chosen the best response.0
Isnt x=9 given???
 one year ago

sara12345Best ResponseYou've already chosen the best response.1
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

sauravshakyaBest ResponseYou've already chosen the best response.0
And we have to prove dw:1350915430127:dw
 one year ago

sara12345Best ResponseYou've already chosen the best response.1
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

sauravshakyaBest ResponseYou've already chosen the best response.0
Did u solve theta and pluged in its value ???
 one year ago

sara12345Best ResponseYou've already chosen the best response.1
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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