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Math Question (1 Viewer)
- Thread starter chet
- Start date
chet
Footballguy
Megaton
Footballguy
Galileo
Footballguy
If 104 cm^3 of water covers the ball, then the volume of water + volume of ball equals the volume of a cylinder with a height equal to 2xradius of the ball. This leads to the following...
104 + 4/3 pi r^3 = pi R^2 h
r - radius of ball
R - radius of cylinder
h - height of cylinder = 2r
So... 104 + 4/3 pi r^3 = pi (64)(2r)
104 + 4.19 r^3 = 402.12 r
4.19r^3 - 402.12r + 104 = 0
There's 3 solutions to this equation. One of them is negative (-9.9 cm)...toss it out. There are 2 positive solutions r = 0.26 cm or 9.7 cm. The 9.7 cm solution would mean the ball is larger than the cylinder can accommodate. Thus the ball must be 0.26 cm in radius.
ETA...obviously, I introduced some slight rounding differences into my work. I used pi = 3.14159 in my calculation, but then rounded those results as coefficients in my equation. So, if you work backwards using the 0.26 cm result, it doesn't land exactly on 104 cm^3...but close enough for government work.
104 + 4/3 pi r^3 = pi R^2 h
r - radius of ball
R - radius of cylinder
h - height of cylinder = 2r
So... 104 + 4/3 pi r^3 = pi (64)(2r)
104 + 4.19 r^3 = 402.12 r
4.19r^3 - 402.12r + 104 = 0
There's 3 solutions to this equation. One of them is negative (-9.9 cm)...toss it out. There are 2 positive solutions r = 0.26 cm or 9.7 cm. The 9.7 cm solution would mean the ball is larger than the cylinder can accommodate. Thus the ball must be 0.26 cm in radius.
ETA...obviously, I introduced some slight rounding differences into my work. I used pi = 3.14159 in my calculation, but then rounded those results as coefficients in my equation. So, if you work backwards using the 0.26 cm result, it doesn't land exactly on 104 cm^3...but close enough for government work.
Last edited by a moderator:
GregR
Footballguy
Not enough information to tell.
We also need to know if the ball is more or less dense than the water.
More edits:
No actually, that isn't right either. We also need to know if the ball is waterproof.
If it is waterproof and heavier than water, we can find the answer. If it is not waterproof we can find the answer with an error of the thickness of the ball. If it's less dense than water and watertight, then it cannot be determined with the information provided, though we could stick a bound on it.
We also need to know if the ball is more or less dense than the water.
More edits:
No actually, that isn't right either. We also need to know if the ball is waterproof.
If it is waterproof and heavier than water, we can find the answer. If it is not waterproof we can find the answer with an error of the thickness of the ball. If it's less dense than water and watertight, then it cannot be determined with the information provided, though we could stick a bound on it.
Last edited by a moderator:
jvdesigns2002
Footballguy
I thought for sure the math question would somehow involve a humble brag or five about massive wealth. I was wrong. I also have nothing to contribute to the question presented- but am impressed with the level of math knowledge in here.
Galileo
Footballguy
the moops
Footballguy
You sound like students I used to have who would find any little way to avoid answering what the obvious intent of the question was
chet
Footballguy
chet
Footballguy
Galileo
Footballguy
Frostillicus
Footballguy
The Viknigs won.
Megaton
Footballguy
You did not properly account for the surface tension of water which would require the water level to be slightly higher than the ball before it would just cover the ball. Normally negligible, but relative to an object around a half inch high, it will be significant.
Megaton
Footballguy
Nobody is. They are saying the height of the cylinder (at least the height filled with water) is the same as the diameter of the ball because the height of the water in the cylinder just covers the ball.
Ignoramus
Footballguy
Dan Lambskin
Footballguy
Assume spherical chickens
chet
Footballguy
Thx
What's the best way to get the solutions to the equation? i.e. y = 0 r = -9.9, 0.26 and 9.7
I messed around with a couple of function calculators--is that the way to gp?
Galileo
Footballguy
That's what I did...skipped the grunt work. This is the one I used. The nice part about it as it shows you the worked out solution and not just the answers.
chet
Footballguy
Volume of cylinder = volume of ball + 104 cm^3.
Volume of cylinder is rcylinder^2*pi*rball
Volume of ball = 4/3*pi*rball^3
8cm*8cm*pi*r=64cm^2*pi*rball
Some basic algebra and you get 4/3*pi*rball^3-64*pi*rball+104=0
Put that into your ti89 and solve.
Galileo
Footballguy
chet
Footballguy
Actually, I don't think you're accounting for the volume of the ball and because the volume of the ball (~0.074cm^3) is relatively small compared to the water (104cm^3), you get about the same answer.
GAlmgren
Footballguy
You are right, I was thinking of the 104 as the combined volume of the sphere and water..
chet
Footballguy
Galileo
Footballguy
Hooper31
Footballguy
I was being ironical. I teach math for a living. "Nobody likes math" was an alias account that someone (Tanner?) created many years ago when he was poking me with a stick through my cage. I was once upon a time a mod around these parts.
I had a free moment or ten this morning and solved it as well. My solution is similar to Galileo's. Using a graphing calculator I got one possible solution at about 0.25880736 by numerically solving the following cubic: (4/3)r^3-128(Pi)r+104=0
Hooper31
Footballguy
Tom Servo
Nittany Beavers
My electrical engineering degree is also worthless after 32 years and a worn out TI-55 II.
Mjolnirs
Footballguy
Hooper31
Footballguy
Megaton
Footballguy
Hooper31
Footballguy
Actually 5i^2=-5, but (5i)^2=-25. It's a common mistake. Don't feel bad for being wrong. Feel brave for hanging yourself out there with the risk of being ridiculed. Atta kid.
