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December 30

cavitation

i need an equation to get the final speed of the water when a cavitation bubble collapses. p.s.: any equations about cavitation would be usefull. thanks, Jake1993811 (talk) 10:17, 30 December 2011 (UTC)[reply]

The final speed is zero. Bo Jacoby (talk) 11:50, 30 December 2011 (UTC).[reply]

It's not speed but the pressure Cavitation--Aspro (talk) 19:48, 30 December 2011 (UTC)[reply]

I believe the water is just accelerated in by the pressure so it would just depend on the pressure and the original size of the cavity. The cavity volume goes down as the cube of the size so the pressure must get pretty large at the end driven by the inertia. Dmcq (talk) 20:05, 30 December 2011 (UTC)[reply]

no ofence, but you people are completely not getting what i,m talking about. i know i was not to clear, and ill fix that tomorrow. but now, i,m to tired and enraged at your server admin to do so. Jake1993811 (talk) 04:13, 31 December 2011 (UTC)[reply]

Sorry we don't know is the result. You're best trying looking through google scholar or a book about it. Why should anybody here know much? They are singularities as far as any straightforward simple maths is concerned. There is no server admin to be enraged at offended or whatever etc. Dmcq (talk) 12:55, 31 December 2011 (UTC)[reply]

A cavitation bubble with volume V in water of pressure P and density ρ has potential energy PV. When the bubble collapses this potential energy is converted into kinetic energy Vρv²/2. So the order of magnitude of the velocity is v = √(P/ρ). A detailed solution to the equation of motion will provide this characteristic velocity with a dimensionless factor depending on the distance from the center of the bubble. Happy new year! Bo Jacoby (talk) 14:03, 31 December 2011 (UTC).[reply]

The water was at rest when the radius of the bubble was R₀. When the bubble has contracted to radius R<R₀, the inwards velocity of the surface of the bubble has increased to v>0. Then the velocity of the water at distance r>R from the center of the bubble is vR²/r² because of the incompressibility of the water. The volume of water having this velocity is 4πr²dr. The kinetic energy of the flow is

\int _{R}^{\infty }{\frac {1}{2}}\rho 4\pi r^{2}dr\left(v{\frac {R^{2}}{r^{2}}}\right)^{2}=2\pi \rho v^{2}R^{4}\int _{R}^{\infty }{\frac {dr}{r^{2}}}=2\pi \rho v^{2}R^{3}

Equating this to the work done

P{\frac {4}{3}}\pi (R_{0}^{3}-R^{3})

gives

v={\sqrt {\frac {P}{\rho }}}{\sqrt {\frac {2}{3}}}{\sqrt {{\frac {R_{0}^{3}}{R^{3}}}-1}}

For R→0 the assumption of incompressibility breaks down and an acoustic wave is emitted. Of course the velocity does not actually become infinite. Bo Jacoby (talk) 13:56, 1 January 2012 (UTC).[reply]

The mean velocity of the surface of the bubble is independent of the size of the bubble.

{\bar {v}}={\frac {R_{0}}{\int _{0}^{R_{0}}{\frac {dR}{v}}}}={\frac {R_{0}}{\int _{0}^{R_{0}}{\frac {dR}{{\sqrt {\frac {P}{\rho }}}{\sqrt {\frac {2}{3}}}{\sqrt {{\frac {R_{0}^{3}}{R^{3}}}-1}}}}}}={\frac {\sqrt {\frac {2}{3}}}{\int _{0}^{1}{\frac {dx}{\sqrt {{\frac {1}{x^{3}}}-1}}}}}{\sqrt {\frac {P}{\rho }}}=1.09328{\sqrt {\frac {P}{\rho }}}

The factor was computed by [1]. Bo Jacoby (talk) 16:40, 1 January 2012 (UTC).[reply]

Determinants

Back in high school I learned about matrices and determinants. I learned how to calculate them, but recently I realized my math teacher never provided word problems to illustrate how they are applied in real life. Now I cannot find any examples online; also, the article on determinants is of no help. What are some simple examples of how they are used to solve real life problems? — Preceding unsigned comment added by 75.36.223.223 (talk) 21:33, 30 December 2011 (UTC)[reply]

There are presumably very many ways, and others will definitely add to this, but one use is when you have a system of equations (like 4x + 2y = 5; 3x + 7y = 17) so you have to solve by inverting the matrix. For huge systems of equations, you can't do this by hand, so you get software to do it for you. The software has to calculate the inverse of the matrix, but it has to know ahead of time if this is possible. So it computes the determinant. If it is even close to zero, it spits out a warning, because its numerical computation will be prone to error. IBE (talk) 22:37, 30 December 2011 (UTC)[reply]

I'm not sure that's entirely true; generally something like LU decomposition would be used to find the inverse or the determinant, so it wouldn't really save anything to compute the determinant first and then compute the inverse. Perhaps a better application would be a formula for the tetrahedron bounded by four given points in space. Determinants don't really have many elementary applications, else they would have been studied long before they actually were, but the more math you study the more you see them popping up in unexpected places.--RDBury (talk) 03:17, 31 December 2011 (UTC)[reply]

The determinant of a matrix is equal to the product of its eigenvalues. Widener (talk) 04:26, 31 December 2011 (UTC)[reply]

You know that the difference between two fractions is computed like this

{\frac {a}{b}}-{\frac {c}{d}}={\frac {ad-bc}{bd}}

The numerator is the determinant

{\begin{vmatrix}a&c\\b&d\end{vmatrix}}

If the determinant is zero then of course

{\frac {a}{b}}={\frac {c}{d}}

and then the tupples (a,b) and (c,d) are proportional or linearly dependent. Bo Jacoby (talk) 09:01, 31 December 2011 (UTC).[reply]

One "real life" application of determinants (as noted above by RDBury) is to calculate areas and volumes. The area of a parallelogram with vertices at (0,0), (a,b), (c,d) and (a+c, b+d) is |ad - bc| i.e. the absolute value of the determinant

{\begin{vmatrix}a&b\\c&d\end{vmatrix}}

Similarly, the determinant of a 3x3 matrix is related to the volume of the parallelpiped formed from its rows (or from its colums). If the determinant is 0, then in the 2x2 case this means that (a,b) and (c,d) are proportional (as Bo pointed out) and the "parallelogram" is in fact a straight line through the origin. In the 3x3 case, a determinant of 0 means that the "parallelpiped" is reduced to a plane through the origin. Gandalf61 (talk) 12:04, 31 December 2011 (UTC)[reply]

Another useful application is in vector (or cross) products see cross_product

\mathbf {a} \times \mathbf {b} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\\end{vmatrix}}.

83.100.189.252 (talk) 15:47, 31 December 2011 (UTC) Yet another application is to find the eigenvalues of a matrix, see Eigenvalue_algorithm, although as above it's a more of a technique or algorithm. The best 'real-world' application is probably finding areas/volumes as described above - at least for 'worded' maths problems. — Preceding unsigned comment added by 83.100.189.252 (talk) 15:52, 31 December 2011 (UTC)[reply]

This is a mnemonic for remembering the cross product formula; it is not truly an application of the determinant. Matrices cannot have vectors as entries. Widener (talk) 16:32, 31 December 2011 (UTC)[reply]

The OP says that the determinant article is not good, yet it has its own applications section. I would direct the OP to the latter half of the article. — Fly by Night (talk) 23:55, 31 December 2011 (UTC)[reply]

It seems like that would be hard to follow for someone who apparently hasn't touched the subject since high school, and the OP has clearly tried to look it up first. It also might not appear comprehensive - one could be forgiven for thinking there should be more than just those listed, as I myself had assumed. IBE (talk) 07:32, 2 January 2012 (UTC)[reply]

To me, the best intuitive way to understand the determinant is with volumes. A few people touched on this, but I think it's worth emphasizing. An n by n matrix is a linear transformation, mapping the points in n-dimensional space to other points in n-dimensional space. The determinant is the scaling factor for volume under this transformation. (When I say "volume" here I mean n-dimensional volume, so for n=3 this is traditional volume, for n=2 this is area, but we can also generalize the notion to larger n.) So if I have some some solid with a volume of 1 (the shape of it doesn't matter), and then apply a matrix with determinant c, the resulting solid will have a volume of c. One place this idea shows up is in multivariable calculus, when you do a change of variables on an integral using the Jacobian. Rckrone (talk) 23:08, 2 January 2012 (UTC)[reply]

I would give an iconoclast answer to the initial question. Determinants, by them selves, have no other real application than to prove that some computations in linear algebra have a unique result or that some objects defined from linear algebra are well defined. In fact, my favorite definition of the determinant is that it is the product of the diagonal elements after a Gaussian elimination. This is sufficient for almost all applications, but the theory of determinants remains needed to prove that this definition of the determinant is independent on the choices made during the Gaussian elimination. Note that this unusual definition is the one which is the closer with the practice of the computation: If you have a computer which is able to do a billion of arithmetic operations in a second, you will need much more than the age of the earth to compute the 60! terms of the expansion of a determinant 60x60. With a clever use of recursive expansion with respect to a row of the same determinant with respect to a row you may reduce the computation to around 2⁶⁰ operations, and you will get the answer in around 30 years. With Gaussian elimination you will get the answer in less than a second on your laptop. This explains why determinants, by them selves, are not used directly in the applications. 92.128.119.13 (talk) 22:36, 7 January 2012 (UTC) (Replacing 50 by 60 to correct a minor mathematical mistake. D.Lazard (talk) 11:31, 8 January 2012 (UTC))[reply]

In line with the volume interpretation Rckrone rightly emphasizes, an appendix to Emil Artin's book on Galois Theory, IIRC, has a beautiful synthetic "geometric algebra" treatment of linear algebra & determinants thought of as volumes. Don't think the OP could do better to understand determinants than reading it.John Z (talk) 09:33, 9 January 2012 (UTC)[reply]