## shubhamsrg Group Title 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

1. muntoo Group Title

|dw:1350805115720:dw|

2. muntoo Group Title

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 Group Title

|dw:1350805844589:dw|

4. muntoo Group Title

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

5. muntoo Group Title

Maybe this is an optimization problem or something?

6. muntoo Group Title

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

7. calculusfunctions Group Title

@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 Group Title

yes sir,,i do..

9. sara12345 Group Title

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

10. sara12345 Group Title

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

11. calculusfunctions Group Title

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

12. shubhamsrg Group Title

13. estudier Group Title
14. muntoo Group Title

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 Group Title

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 Group Title

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

17. sauravshakya Group Title

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

18. shubhamsrg Group Title

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

19. sauravshakya Group Title

Then, I dont think its a calculas problem.

20. shubhamsrg Group Title

@mukushla ? :/

21. sara12345 Group Title

|dw:1350914228619:dw|

22. sara12345 Group Title

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 Group Title

Isnt x=9 given???

24. sara12345 Group Title

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 Group Title

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

26. sara12345 Group Title

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 Group Title

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

28. sara12345 Group Title

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%