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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 .
Ugh. OK, that says \(w = \textrm{water_volume} = \pi r^2 \cdot 9x\) Surface area of round thingamagic is \[s = 2 \pi r \cdot 100x \]
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Ignore me too. I just realized this probably requires calculus stuff I'm not familiar with. :P
Maybe this is an optimization problem or something?
Let's see how this works out: http://math.stackexchange.com/questions/217875/tilting-sealed-drums-for-fun
@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?
im getting this minimize : .09 r/2 [1 - \(\frac{sin\theta}{\theta}\) ]
ohh \(\theta \) is also constant here as volume is fixed, hang on..
Sorry @shubhamsrg I had to go away for a second but am back now. OK give me a second.
take your time sir! :)
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|
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
\[ \huge {\pi r^2 \over {2\pi \over \theta}} - {1\over2}rr \sin \theta = 0.09 \pi r^2 \]
The cylinder can be tilted any way or it is tilted in following way|dw:1350825580056:dw|
res on its side right,,i'd assume the one in your figure..
Then, I dont think its a calculas problem.
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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
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
And we have to prove |dw:1350915430127:dw|
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.
Did u solve theta and pluged in its value ???
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%