Ok here is a question I was working on today but I cannot seem to visualize it properly.
A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that the center of the disk should be positioned at a height r / sqrt(1 + pi^2) above the surface of the liquid.
What is the "wetted" area of the disk? It cannot be the part of the disk that is submerged because then the maximum area would be the entire disk submerged.
Stacey Warren - Expert brainly.com
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It may just be a poorly worded question...
That's true. I found a solution here (http://answers.yahoo.com/question/index;_ylt=AjP6YX4F6.ERQSVCCXH8zhHty6IX;_ylv=3?qid=20091128095615AA1AuqI&show=7#profile-info-91HhPVl0aa) but that did not really help much. This particular question was posted in a chapter on integration yet it is an optimization problem. I am not sure how integration can be used here.
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So this disk is actually an annulus? Is this bit of information assumed in the question?
Oh wait..scratch that. I see.
vertical plane was the give away....up/down. just trying to work out the integration set up
the integration involved is to determine the area of the submerged disc section of the circle
although, to find a max/min you will have to set a Area equation and take the derivative.
Yes, I agree. I do not believe integration is required to solve this problem. Thank you so much for the drawing! I see now the "wetted" portion of the disk is the portion exposed to the air ready to evaporate. I understand now how the person from Yahoo Answers set this question up. I will modify your drawing and add the geometry involved|dw:1356733401864:dw|
I supposed now my question is how do we write the area of the triangle AOB and the area of the inner circle in terms of theta (angle AOB)? If we can do this the rest is a matter of optimization. Attached is my version of the region that needs to be maximized.
see attached, hopefully its not too messy.
I've broken the drawing into 4 parts each can be solved relatively easily. summed they form the area of the 'wetted' area.
1. Derivative will find angle of maximum wetness
2. integration may be used to find area but you can sub your solved angle into that.