At which depth will the glass, which is it not a right cylinder, be volumetrically half-filled? Assume that the thickness of the base is negligible and the thickness of the walls is constant.
I think I would first have to find the derivative of perhaps the circular area or something like that... Out of the box answers are welcome:))
The glass.
I think I would first have to find the derivative of perhaps the circular area or something like that... Out of the box answers are welcome:))
The glass.
1 Answer
The split should occur at a height of approx
Explanation:
No Calculus required:
The mathematical name for the given "bucket" , or "glass" shape is a frustum, we can readily calculate the volume using the formula:
V = pi/3(R^2 + rR + r^2)h \ \ \ (See Notes)
We are required to split the volume of the given frustum of lower radius
Steve M
Suppose the volumeatic split occurs at a height
Using this information, and utilising the frustum formula, we can write the identical volume of each frustum as:
pi/3(3.5^2 + 3.5R + R^2)h = pi/3(R^2 + 3.9R + 3.9^2)(10-h)
And by similar
h : (R-3.5) = 10: (3.9-3.5)
:. h/(R-3.5) = 10/(0.4)
:. 0.4h = 10(R-3.5)
:. 0.4h = 10R-35
:. 10R = 0.4h +35
Then we solve the two equations simultaneously, which results in the solution:
h ~~ 5.26948 R~~3.71078 (5dp)
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Notes:
We can readily derive the frustum volume by considering the difference in the volumes of standard cones#
http://jwilson.coe.uga.edu/
Using the standard volume of a cone, V=1/3pir^2h; we have:
Volume of Smaller cone:
V_1 = 1/3pir^2H
Volume of Larger cone:
V_2 = 1/3piR^2(H+h)
So the volume of the frustum is:
V = V_2-V_1
\ \ \ = 1/3piR^2(H+h) - 1/3pir^2H
\ \ \ = 1/3pi(R^2H+R^2h-r^2H)
\ \ \ = 1/3pi(H(R^2-r^2)+R^2h)
\ \ \ = 1/3pi(H(R+r)(R-r)+R^2h)
And by similar
R : H+h = r : H
:. R/(H+h) =r/H
:. RH = r(H+h)
:. RH = rH+rh
:. H(R-r) = rh
:. H = (rh)/(R-r)
Substituting into out volume formula we get:
V = 1/3pi((rh)/(R-r)(R+r)(R-r)+R^2h)
\ \ \ = 1/3pih((r)(R+r)+R^2)
\ \ \ = 1/3pi(rR+r^2+R^2) QED