Wikipedia:Reference desk/Archives/Mathematics/2008 August 27

= August 27 =

Laplace Transforms
I am having trouble using Laplace Transform to solve a system of 2 ODEs involving a step function.

$$(y_1)'=y_1+6u(t-2)exp(4t)$$

where u represent the Heaviside step function

$$(y_2)'=y_1+2y_2$$

I arrived at

$$Y_2=(6*exp(8-2s))/(s-1)(s-2)(s-4) - 1/(s-2)$$

I then use partial fractions to break the first term up.

but when I take the inverse transform of this I end up with an answer which I know is wrong.

I'm not clear on how the step function works with this type of inverse.

Do you leave the 's' (from the second shifting theorem) in when you are evaluating the partial fractions?

Or do you ignore it, and add a Heaviside step function to your answer? I can't find any clear explanation in my textbook.

Please help if you know how to do these things.

P.S. My apologies if the formatting does not come out properly but I can never seem to make it work. —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 01:09, 27 August 2008 (UTC)

10000000000000000000000000
what is 1000000000000000=100000000000000000= —Preceding unsigned comment added by Kelly4.taylor (talk • contribs) 10:10, 27 August 2008 (UTC)
 * Nonsense. --hydnjo talk 12:24, 27 August 2008 (UTC)
 * Maybe the poster would be interested in Names of large numbers. PrimeHunter (talk) 17:43, 27 August 2008 (UTC)
 * I doubt it. -hydnjo talk 01:04, 28 August 2008 (UTC)

matrices
could you please explain me any practical usage of matrix?, with examples —Preceding unsigned comment added by 59.160.206.225 (talk) 12:11, 27 August 2008 (UTC)


 * Rotation matrix is an important one. -- Coneslayer (talk) 12:31, 27 August 2008 (UTC)


 * The simplest example is that matrices can be used to represent a system of linear equations, so rather than having lots of equations of the form $$y_n = a_n x_n + b_n z_n + \cdots$$, you have a single matrix equation of the form $$\vec{y} = A \vec{x}$$, which can then be solved using matrix manipulation. That's a boring example,though.
 * There are many many examples. Matrices are useful for representing rotation (see rotation matrix). They are useful for nonlinear applications, too because they can help you generalize derivatives. That's the matrix of partial derivatives, the Jacobian.
 * If you are just starting to learn about these things and are like me, you'll get confused by wrote things like row reduction and Gaussian elimination. There's a lot more to linear algebra than those things. Personally, I don't think it's usually taught in a useful order. —Ben FrantzDale (talk) 12:39, 27 August 2008 (UTC)


 * Google's PageRank; finite element method structural analysis programs and almost all other physical modeling and simulation programs; raytrace and collision algorithms used in animation and games; regression analysis etc. etc. etc.  As above, systems of linear equations are much easier to manipulate using matrix notation.  Good luck!  Saintrain (talk) 17:36, 27 August 2008 (UTC)


 * In computer graphics, we use matrices for all sorts of things. For 3D graphics, we typically use a 4x4 "homogeneous" matrix that allows us to represent rotation, translation, scaling, shearing, perspective and a bunch of other effects (like converting an object into it's own shadow when cast onto an arbitary plane) - all in a single matrix.  We may also use matrices to represent colour changes too.  The hardware that makes up that big chip on your graphics card has a considerable amount of circuitry devoted to multiplying, adding, inverting 4x4 matrices.  It's really neat that we can take the matrix representing an object's rotation, another representing it's position in the world, another representing the position and rotation of the camera, another representing the lens that camera is using - and another representing the position of the game's graphics window on your desktop.  Multiplying that mess of 4x4's together produces a single matrix that we can use to transform the vertices of the polyhedra that represent that object from however they were modelled into a 2D representation on the screen. SteveBaker (talk) 15:03, 30 August 2008 (UTC)

Tank
I've just done an exam question. Could someone please check it for me? If I've gone wrong, please let me know but don't tell me where, how or why. I want to do as much of this by myself as possible.

i) At time t=0, a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time 't' is 'y'. Show that there is a constant b<1 such that $$y=b^t$$.

ii)Suppose instead that the tank contains one unit of water at t=0 but that in addition to water flowing out as described, water is added at a rate a>0. Show that

$$\frac{dy}{dt} -y \ln{b} = a,$$

and hence find y in terms of a, b and t. ''

i)
 * $$\frac{dy}{dt} = -ky$$
 * $$\int \frac{1}{y} dy = \int -k dt$$
 * $$lny=-kt+c\,\!$$
 * $$y=e^{-kt+c}\,\!$$
 * $$y=Ce^{-kt}\,\!$$ where $$C=e^c\,\!$$


 * at t=0, y=1


 * $$1=Ce^0\,\!$$
 * $$1=C\,\!$$
 * $$y=e^{-kt}\,\!$$

let $$-k=\ln b$$


 * $$y=e^{t \ln b}\,\!$$
 * $$y=b^t\,\!$$

ii)

using the same constant k as above,
 * $$\frac{dy}{dt} = -ky+a$$
 * $$\frac{dy}{dt} = y \ln b +a$$
 * $$\frac{dy}{dt}- y \ln b = a$$

as required.


 * $$\int \frac{1}{y \ln b +a} dy = \int 1 dt$$
 * $$ \frac{1}{\ln b}(\ln|y \ln b +a|) = (t + c)$$

at t=0, y=1


 * $$ \frac{1}{\ln b}(\ln| \ln b +a|) = c$$
 * $$ \frac{1}{\ln b}(\ln|y \ln b +a|) = (t + \frac{1}{\ln b}(\ln| \ln b +a|))$$
 * $$ \ln|y \ln b +a| = t\ln b + (\ln| \ln b +a|)\,\!$$
 * $$ y \ln b +a = e^{t\ln b + (\ln| \ln b +a|)}\,\!$$
 * $$ y \ln b +a = e^{t\ln b}e^{\ln| \ln b +a|}\,\!$$
 * $$ y \ln b +a = {(e^{\ln b})}^t e^{\ln| \ln b +a|}\,\!$$
 * $$ y \ln b +a = b^t e^{\ln| \ln b +a|}\,\!$$
 * $$ y \ln b +a = b^t (\ln b +a)\,\!$$
 * $$ y \ln b = b^t (\ln b +a)-a\,\!$$
 * $$ y =\frac{b^t (\ln b +a)-a}{\ln b}\,\!$$

Thanks 92.5.221.82 (talk) 16:49, 27 August 2008 (UTC)


 * At a very quick look, the first part seems fine, although you may want to explicitly demonstrate why b<1. I haven't looked at the second part, but remember you can always check whether a function is a solution to an equation (differential or otherwise) by substituting it back in. Confusing Manifestation (Say hi!) 23:35, 27 August 2008 (UTC)
 * In the second part, you've dropped the absolute value signs without justification, but I don't think this has stopped you getting the right answer. More generally, how well this is marked will depend on the exam mark scheme; it might require more explanation of correct reasoning. Algebraist 23:43, 27 August 2008 (UTC)