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

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

|dw:1350805115720:dw|

- 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

|dw:1350805844589:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Ignore me too. I just realized this probably requires calculus stuff I'm not familiar with. :P

- anonymous

Maybe this is an optimization problem or something?

- anonymous

Let's see how this works out:
http://math.stackexchange.com/questions/217875/tilting-sealed-drums-for-fun

- 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

yes sir,,i do..

- anonymous

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

- anonymous

ohh \(\theta \) is also constant here as volume is fixed, hang on..

- calculusfunctions

Sorry @shubhamsrg I had to go away for a second but am back now. OK give me a second.

- shubhamsrg

take your time sir! :)

- anonymous

http://mathworld.wolfram.com/Quarter-TankProblem.html

- 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. â€“ Steve D
--
|dw:1350814466991:dw|

- 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

\[ \huge {\pi r^2 \over {2\pi \over \theta}} - {1\over2}rr \sin \theta = 0.09 \pi r^2 \]

- anonymous

The cylinder can be tilted any way or it is tilted in following way|dw:1350825580056:dw|

- shubhamsrg

res on its side right,,i'd assume the one in your figure..

- anonymous

Then, I dont think its a calculas problem.

- shubhamsrg

@mukushla ? :/

- anonymous

|dw:1350914228619:dw|

- 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

Isnt x=9 given???

- 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

And we have to prove |dw:1350915430127:dw|

- 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

Did u solve theta and pluged in its value ???

- 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%

Looking for something else?

Not the answer you are looking for? Search for more explanations.