Recent content by needhelp81

Hey Chaos, made a little progress with the problem. So I figured what I have to do it take both the above equations and combine pi(R^2-X^2)-1/2R^2(theta-sin(theta)) (sorry don't know how to do the eqations like you =) and First, Solve for angle theta. The angle theta will be the inverse...

now if anybody knows how to tie in the area underneath the water with the formula for the area that's wet..

Awesome! its starting to click =) now finding that function should be fun. Thank you again!

Firstly, Thanks for your help and quick response...very much appreciated! So, if I'm understanding correctly, I'll find the function of the wet area above the waterline by combining the previous to functions of A..then, take its first derivative?

maybe my mind is tired but I'm confused, should I be thinking about using derivatives to optimize the wet area above the waterline?

Yeah I've got that far, But would you use both the formula's A=pi(R^2-r^2) and A=piR^2? I'm having a difficult time trying to figure out what formulas to use.. and I'm almost sure I'll be taking the deriv of which ever it is.

Hey all, I'm struggling to even start this problem: An evaporative cooler design uses a rotating wheel that is placed upright (vertical) and is partially submerged in water. The center of the circle is above the waterline. Let R be the radius of the wheel and x be the distance from the...