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shubhamsrg
 4 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 .
shubhamsrg
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350805115720:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ugh. 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350805844589:dw

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Maybe this is an optimization problem or something?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let's see how this works out: http://math.stackexchange.com/questions/217875/tiltingsealeddrumsforfun

calculusfunctions
 4 years ago
Best 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im getting this minimize : .09 r/2 [1  \(\frac{sin\theta}{\theta}\) ]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh \(\theta \) is also constant here as volume is fixed, hang on..

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

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0take your time sir! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0From 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
 4 years ago
Best ResponseYou've already chosen the best response.0that 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \huge {\pi r^2 \over {2\pi \over \theta}}  {1\over2}rr \sin \theta = 0.09 \pi r^2 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The cylinder can be tilted any way or it is tilted in following waydw:1350825580056:dw

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then, I dont think its a calculas problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350914228619:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Volume 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
 4 years ago
Best ResponseYou've already chosen the best response.0only 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
 4 years ago
Best ResponseYou've already chosen the best response.0And we have to prove dw:1350915430127:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.09% 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
 4 years ago
Best ResponseYou've already chosen the best response.0Did u solve theta and pluged in its value ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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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