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Missouri State University’s Advanced Problem May 2011
Vincent Pantaloni (Orléans, France)
Problem : A humidifier has a circular disk of radius r that slowly rotates. It is partially submerged in a tank of water (the disk is perpendicular to the surface of the water). In order to have the greatest humidifying effect, the total area of the wetted part of the disk that is above the water should be maximized. How high above the surface of the water should the center of the disk be to attain this maximum ?
Source : James Stewart Solution :
Here’s a picture of what is happening. We want to find AB=x, such that the blue area is as great as possible. Notice that if we find θ, we findxbeacuse x=rcos
θ 2
. Let’s call Athe blue area. A is the area of the whole ring minus the white circular seg- mentW.
A=π(r2−x2)− W (1) W is the area of the circular sectorADC minus the area of the triangleADC, so withb=DC we have :
W=πr2× θ 2π−1
2b×x= r2 2
θ−b
r ×x r
Ab
Cercle
b
B d
b
C W
A
b
D
θ= 144.7◦ x
b
E
b
F
b
G
b
H
b
If you consider the right triangle ABC you will see that x
r = cosθ 2 and b
r = 2× b/2
r = 2 sinθ 2. So b
r×x
r = 2 sinθ 2cosθ
2 = sinθ. That gives us a nice formula for W : W=r2
2 (θ−sinθ) (2)
Sincex=rcosθ2 the blue areaAis a function ofθ. Using (1) and (2) we get : A(θ) =π
r2−r2cos2θ 2
−r2
2 (θ−sinθ)
= r2 2
2π+ sinθ−2πcos2θ 2 −θ
= r2
2 (2π+f(θ))
We are now looking for the value ofθ∈[0;π]for whichf(θ) = sinθ−2πcos2θ2−θreaches it’s maximum.
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Let’s differentiatef and have some fun with trigonometry : f′(θ) = cosθ−2π×2 cosθ
2 ×
−1 2sinθ
2
−1
= cosθ+π×2 cosθ 2×sinθ
2−1
= cosθ+πsinθ−1
=p π2+ 1
1
√π2+ 1cosθ+ π
√π2+ 1sinθ
−1
=p
π2+ 1×cos
θ−Acos 1
√π2+ 1
−1
Now solving f′(θ)>0forθ∈[0;π]gives : f′(θ)>0 ⇐⇒ p
π2+ 1×cos
θ−Acos 1
√π2+ 1
−1>0
⇐⇒ cos
θ−Acos 1
√π2+ 1
> 1
√π2+ 1
⇐⇒ θ−Acos 1
√π2+ 1 6Acos 1
√π2+ 1 (BecauseAcosis a decreasing function.)
⇐⇒ θ62Acos 1
√π2+ 1
This shows that f andAare max when
θ= 2Acos 1
√π2+ 1
This angle is approximately equal to 144.7˚as shown on the figure of the previous page. To answer the question we need x=rcos
θ 2
.
x= r
√π2+ 1 ≈0.303×r
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