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. Angle A and height above the water h The exposed wetted area is the full circle less the inner (dry) circle less the submerged segment. In terms of r, the radius of the disc, and the parameters angle A, and height h, the exposed area is A h Area = πr 2 − πh 2 −
r2
(2A − sin2A) 2
(1) I tried putting all the variables in terms of h (i.e. removing the references to angle A), but that led to some messy algebra, and I never was able to get out of that thicket. Next I tried putting things in terms of A, using the relation €
h = r cos A (2) to get this version of the equation for the exposed wetted area: Area = πr 2 − πr 2 cos2 A − Ar 2 +
€
r2
sin2A
2
r2
sin2A
2
⎛
⎞
1
2
2
Area = r ⎜ π sin A − A + sin2A⎟
⎝
⎠
2
Area = πr 2 sin 2 A − Ar 2 +
(3) Now we take the derivative of Area with respect to A, and set it equal to zero: € dArea
= r 2 (π 2sin Acos A −1+ cos2A) = 0
dA
π sin2A + cos2A = 1
(4) where we used the identity 2sinA cosA = sin2A. € To solve (4), I found these Weierstrass substitutions (see https://en.wikipedia.org/wiki/Weierstrass_substitution ) t = tan A
2t
1+ t 2
1− t2
cos2A =
1+ t 2
sin2A =
(5) (6) Using (5) in (4), we get €
€
π
2t
1− t2
+
= 1 1+ t 2 1+ t 2
Multiplying both sides of (6) by 1+t2, and simplifying we find t 2 − πt = 0
t = 0, t = π
(7) (8) From the definition of t given by (5), we solve for angle A: €
A = 0,
A = tan −1 π ≈ 72.34º
The solution A=0 leaves no wetted area (the disc is above the waterline), so we have only one solution. To put the answer in terms of h, we note that € tan A = π →cos A =
1
1+ π 2
(9) (10) and (9) in (2) gives us €
€
h=
r
1+ π 2