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GILBERT STRANG: OK. This video is about the third of the great trio of partial differential equations. Laplace's equation was number one. That's called an elliptic equation. The heat equation was number two. That's called a parabolic equation. Now we reach the wave equation. That's number three, and it's called a hyperbolic equation. So somehow the three equations remind us of ellipses, parabolas, and hyperbolas. They have different types of solutions. Laplace's equation, you solve it inside a circle or inside some closed region. The heat equation and the wave equation, time enters, and you're going forward in time. The heat equation is first order in time, du dt. And the wave equation, the full-scale wave equation, is second order in time. That stands for the second derivative, d second u dt squared. And it matches the second derivative in space with a velocity coefficient c squared. I'm in one-dimensional space. If I were in three dimensi...

In this video for 3D geometry, we will be talking about two topics. One is the angle bisector of two planes and one is line of intersection of two planes. Both of the topics are important especially the line of intersection of two planes, and I will be spending some time on that. Angle bisector of two planes is a very easy topic, something you should remember while preparing for JEE. I will just give you the formula and I won’t be doing a problem but I will spending sometime on this (Line of Intersection of two planes), because this is a relatively important topic. So let us say we have a pair of planes. And you have to find the angle bisector. This is the angle bisector — plane which is the angle bisector of two planes. This angle is the same as this angle. You have been given equations for P1 and P2. You have find equation of angle bisector. If you think about this — easiest way to think about planes is using a notebook. I always recommend you do that if you are a...

Hi everyone In this video, we will be talking about two topics of 3D chapter for mathematics of JEE preparation. The topics are: doing problems on finding perpendicular distance of a point from line and we will be finding equations of a line perpendicular to two lines given. These are small topics but as in other maths topics, you should learn the approach to solve problems like these. Let us start by finding perpendicular distance of a point from a line. We have a line given. Let us saw we have point A, and that is a_vector, and we have been given b_vector, which is the parallel vector. And we have been given a point P. Let us call it p_vector. And we have been asked to find the perpendicular distance of the point from the line. You may recall these kind of problems in 2D in straight lines. But now its 3D or 3-Dimensinal Geometry. So how will we solve this problem? One of the approaches to start this to think that there is a point C here, let us call it c_vector. And the...

TINA CHEN: In this video, we will be exploring the effect of bond energies on vacancy diffusion using a Mathematica simulation. Even the most carefully produced materials have defects. These defects often affect the physical properties of the material. One type of defect is a vacancy in which an atom or molecule is missing from a point in the lattice. Vacancies can diffuse or move around. In this video, we will investigate diffusion of a single vacancy in a binary AB alloy. To simplify the simulation process, this AB alloy will have a two-dimensional, square lattice. In order to simulate the movement of a vacancy, we must consider, on the atomic level, the interaction between A and B. Specifically, we will look at the bond energies. Here, we have our AB alloy square lattice with randomly placed atoms A and B. If we have a vacancy here, then the vacancy can potentially jump to one of four places. We look at the energy required to jump to each of the four places. If the vac...

In this video, we're going to take a look at how the social contract issues have been addressed around the world. It turns out there's quite a bit of experience with this, sometimes with success for a limited period of time and sometimes without as much success. So let's see how other countries address these issues. Let's start with Australia. Australia has a long history. In the 1980s, a new prime minister came along by the name of Bob Hawke. And he came into office negotiating a new labor agreement called the accord, with the labor movement, where he was saying to labor, if you limit your wage increases to the price increases that are happening in the country, we will then find some offsetting things that we can do in social welfare. So they negotiated a national health insurance. They negotiated a national pension program. They put in place a variety of other worker adjustment programs. And for business, they allowed over time much more flexibility in p...

PROFESSOR: The previous video was about positive definite matrices. This video is also linear algebra, a very interesting way to break up a matrix called the singular value decomposition. And everybody says SVD for singular value decomposition. And what is that factoring? What are the three pieces of the SVD? So this is the fact is every matrix, rectangular, every matrix factors into-- these are the three pieces. U sigma V transpose. People use those letters for the three factors. The factor U is an orthogonal matrix, an orthogonal matrix. The factor sigma in the middle is a diagonal matrix. The factor V transpose on the right is also an orthogonal matrix. So I have orthogonal, diagonal, orthogonal, or physically, rotation, stretching, rotation. Now we have seen three factors for a matrix, V, lambda, V inverse. What's the difference? What's the difference between this SVD, this, and the V, lambda, V transpose, V inverse, V lambda, V inverse for diagonalizing other...

PROFESSOR: OK. This video is about derivatives. Two rules for finding new derivatives. If we know the derivative of a function f-- say we've found that-- and we know the derivative of g-- we've found that-- then there are functions that we can build out of those. And two important and straightforward ones are the product, f of x times g of x, and the quotient, the ratio f of x over g of x. So those are the two rules we need. If we know df dx and we know dg dx, what's the derivative of the product? Well, it is not df dx times dg dx. And let me reduce the suspense by writing down what it is. It's the first one times the derivative of the second, we know that, plus another term, the second one times the derivative of the first. OK. So that's the rule to learn. Two terms, you see the pattern. And maybe I ought to use it, give you some examples, see what it's good for, and also some idea of where it comes from. And then go on to the quotient rule, which...

Hi. In this video, we're going to compute some useful quantities for the exponential random variable. So we're given that x is exponential with rate lambda. PDF looks like this, and the formula is here. First question, part a, what's the CDF? So let's go right in. The CDF of x is the probability that X is less than or equal to little x. Let's look at some cases here. What if little x is less than 0? Well, x random variable only takes on these non-negative values. And so the probability that X is less than or equal to some negative number is going to be 0. On the other hand, if x is greater than or equal to 0, we do actually have to integrate here. So to do that, we take the integral from minus infinity to x of fx of t-- the dummy variable here used is t. Notice that again, fx of t is going to be 0 for negative values, so we take the integral here from 0. And now we plug in for fx of t. That's minus lambda t dt. And recall that the integral of u t...

In this last video, we illustrate how to use the techniques we have recently learned in order to answer some questions about the following classical problem-- consider a gambler putting a bet of $1 in a game that has a pay off of one dollar if she wins. We assume that this is a fair game, so the probability of winning $1 on each play of the game is one-half. And so the probability of losing the bet is also one-half. Suppose that she starts with i dollars and continues to play the game until either she reaches a goal of n dollars, or she has no money left, whatever comes first. Let us consider a first question, which is the following-- what is the probability that she ends up with having her goal of n dollars? Now, how to go about solving this problem? Can we think of a Markov chain representation for this problem? But in that case, what would be good choices for the definition of the states? Let us think. At any point in time, the only relevant information is the amount o...

In this video, we continue our exploration of quantities of interest associated with the short-term behavior of Markov chain. This time, we suppose that we have a Markov chain composed of a single recurrent class, such as this one. Here, we have our recurrent class. And these are transient states. We also assume that we're interested in a specific recurrent state-- let's say 9. So this is my state s. And we will assume that the Markov chain starts from a given initial state, state i, and assume that i is 1. Now the question we ask is, given that you started in 1, how long is it going to take to reach 9 for the first time? We know this is a recurrent class, so we know that the Markov chain eventually will come to that class and circulate here. So you know that the Markov chain will get to that state 9. You are interested in knowing when it does so for the first time. Of course, we don't know for sure. This is random. It is a random variable. And let's try t...

In this video, let us look at a second quantity of interest that has to do with absorbing states. Now that we know how to calculate the probability of getting to a given absorbing state, we would like to know how long it would take to get to it. Let us first deal with that question when we have only one absorbing state. Let us consider the following Markov chain, which is a little simpler than the one that we had in the previous video. We have transient states. One, two, and three are transient states. And we have one recurrent state, four. And that recurrent state four is an absorbing state, because once you get to it you stay there forever. So in this simple example, the absorption probability to four is trivially one. No matter where you start, with probability one you're guaranteed that eventually you will reach four. But the question of interest is to know how long would it take to get to four. In other words, how many transitions would you have to do until you r...