Question 114.
What will be the dimensions of an inverted hollow cone that will hold equal volumes of water....?
..........above and below the solid sphere of 10cm. radius placed inside the cone?
(Surface of the water is tangent to the sphere.)
This excellent question was posted by Questor and correct solutions were first posted by rozeta53 and Scythian. All three are my valuable contacts in YA! and their contributions in YA! is an asset to the community.
rozeta53's solution was adjudged the best by the asker and the answer posted below is just an elaboration of rozeta53's answer. Detailed steps are provided for the benefit of the students to whom this blog is directed.
Answer 114.
The radius "r" and height "h" of the cone in terms of radius "R" of the sphere and semi-vertical angle "θ" of the cone are given by
r = h tanθ and
h = R(1 + cosecθ)
=> Volume of cone
= (π/3) r^2 h
= (π/3) tan^2 θ * R^3 * (1 + cosecθ)^3
Volume of Cone - volume of sphere,
V = (πR^3/3) tan^2 θ * (1 + cosecθ)^3 - (4/3) πR^3 ... ( 1 )
Taking the center of the sphere as the origin, equation of circular section of the sphere can be written as
x^2 + y^2 = R^2 =>x^2 = R^2 - y^2
The equation of the slanting side of the cone can be written as
y = xcotθ - Rcosecθ => x^2= (ytanθ + Rsecθ)^2
Volume of water above the sphere is obtained by rotating the area between the slanting side of the cone and the circular section of the sphere between the limits -Rsinθ to R.
=> Volume of water above the sphere
= π ∫ (-Rsinθ to R) [(ytanθ + Rsecθ)^2 - (R^2 - y^2)] dy
= π ∫ (-Rsinθ to R) [(ysecθ + Rtanθ)^2 dy
= (π/3secθ) [ysecθ + Rtanθ]^3 ... [from -Rsinθ to R]
= (πR^3/3secθ) [secθ + tanθ]^3
= (πR^3/3cos^2 θ) [1 + sinθ]^3 ... ( 2 )
From( 1 ) and ( 2 ),
(πR^3/3) tan^2 θ * (1 + cosecθ)^3 - (4/3) πR^3
= 2 * (πR^3/3cos^2 θ) (1 + sinθ)^3
=> (1/sinθ cos^2 θ) * (1 + sinθ)^3 - 4 = (2/cos^2 θ)(1+ sinθ)^3
=>(1 +sinθ)^3 * (1 - 2sinθ) = 4sinθ cos^2 θ
=>(1 +sinθ)^3 * (1 - 2sinθ) = 4sinθ (1 - sin^2 θ)
=>(1 +sinθ)^2 * (1 - 2sinθ) = 4sinθ (1 - sinθ)
=> (1 + 2sinθ + sin^2 θ) (1 - 2sinθ) = 4sinθ - 4sin^2 θ
=> 2sin^3 θ - sin^2 θ + 4sinθ - 1 = 0
Let sinθ = x. Then,
2x^3 - x^2 + 4x - 1 = 0
=> x^2 - 4x + 4 = 3 + 2x^3
=> (x - 2)^2 = 3 + 2x^3
To solve this cubic equation, we note that θ < 45°
=> x = sinθ < 0.707
=> 2x^3 < 0.707
To begin with, neglecting the value of 2x^3 compared to 3,
x = 2 - √3 = 0.26795
=> 2x^3 = 0.038476
Next better solution for x is
x = 2 - √(3 + 0.038476) = 0.25688
=> 2x^3 = 0.033901
Final acceptable value of x will be
x = 2 - √(3 + 0.033901) = 0.25819
Accepting this as sinθ = 0.25819,
h = R(1 + cosecθ) = 10(1 + 1/0.25819) = 48.73 cm. and
r = h tanθ = 48.73 * 0.26725 = 13.02 cm.
Wolfram Alpha gives the exact values as
h = 48.75 cm and r = 13.02 cm.
which are quite close to the above values.
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Wednesday, February 3, 2010
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Wow! This is great. We have been pampered with easy and touch-go methods of solving algebraic equations now. But this is resorting to the fundamental and vital concepts of Mathematics which are the backbone of the real solution. Commendable efforts on your part Mr Daftary, you are a great mentor any worthy student should have.
ReplyDeleteThanks for the appreciation. Since my participation in YA! I have observed that math enthusiasts from developing countries are resorting to software which has yet to percolate in India. Sometimes I wonder if this will ultimately produce better mathematicians or it will result in our losing control over the fundamentals!
ReplyDeleteI am afraid that I might become slave to the machine. I will use the wolfram facility very sparingly : only when it is absolutely necessary,
ReplyDeletelest I might lose whatever little grasp I have of
fundamentals. It looks like a great tool for professionals but I don't think we should encourage our students to 'solve'problems using wolfram. For verifying the answers ? Yes. But for finding the answers ? Strictly , No !