## shubhamsrg 3 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 .

1. muntoo

|dw:1350805115720:dw|

2. muntoo

Ugh. OK, that says $$w = \textrm{water_volume} = \pi r^2 \cdot 9x$$ Surface area of round thingamagic is $s = 2 \pi r \cdot 100x$

3. nincompoop

|dw:1350805844589:dw|

4. muntoo

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

5. muntoo

Maybe this is an optimization problem or something?

6. muntoo

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

7. 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?

8. shubhamsrg

yes sir,,i do..

9. sara12345

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

10. sara12345

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

11. calculusfunctions

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

12. shubhamsrg

13. estudier
14. muntoo

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|

15. sara12345

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

16. muntoo

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

17. sauravshakya

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

18. shubhamsrg

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

19. sauravshakya

Then, I dont think its a calculas problem.

20. shubhamsrg

@mukushla ? :/

21. sara12345

|dw:1350914228619:dw|

22. sara12345

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

23. sauravshakya

Isnt x=9 given???

24. sara12345

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

25. sauravshakya

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

26. sara12345

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.

27. sauravshakya

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

28. sara12345

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%