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PROFESSOR: So we go back to the integral. We think of k. We'll write it as k naught plus k tilde. And then we have psi of x0 equal 1 over square root of 2pi e to the ik naught x-- that part goes out-- integral dk tilde phi of k naught plus k tilde e to the ik tilde x dk. OK. So we're doing this integral. And now we're focusing on the integration near k naught, where the contribution is large. So we write k as k naught plus a little fluctuation. dk will be dk tilde. Wherever you see a k, you must put k naught plus k tilde. And that's it. And why do we have to worry? Well, we basically have now this peak over here, k naught. And we're going to be integrating k tilde, which is the fluctuation, all over the width of this profile. So the relevant region of integration for k tilde is the range from delta k over 2 to minus delta k over 2. So maybe I'll make this picture a little bigger. Here is k naught. And here we're going to be going and integrate ...

KRISTEN: So with that, I am going to turn it off to our first keynote speaker, Kris Clark. She also works at Lincoln Laboratory with me, but in a completely different field. She's going to talk about space cameras. KRIS CLARK: So as Kristen said-- I'm also Kristen, but I go by Kris-- I work at Lincoln Laboratory. And I want to say that this is an amazing program. So first, how many of you have ever done a program like this? Like an engineering kind of thing for girls. Come on, way up. I can't see that. OK, cool. Now, how many of you are freshmen? Sophomores? Juniors? Seniors? All right, no seniors. They're all busy doing college applications. All right, so let's get started. So today, I'm going to talk a little bit about me, just because we're kind of looking at what makes an engineer. It can be all over the map. So there is no one key picture of what an engineer looks like, or what a scientist looks like. Then we'll go into a little bit ab...

PROFESSOR: Welcome back. So in this session, we're going to look at unit step and impulse responses. So in this question, we ask you to find the unit impulse response to these two equations, x dot plus 2x equals f of t, and 2 x dot dot plus 27 x dot-- oops, it should be a 7-- plus 7 x dot plus 3x equals f of t. In the second part, you're asked to find the unit step response for the first equation. So here, the key points are really to remember what do we mean by unit impulse and unit step response. Which initial condition correspond to these responses? And what functions of f of t do you choose in each case? So why don't you pause the video and work through this problem? And I'll be right back. Welcome back. So let's look at the equation A. So the unit impulse response is simply-- I'm going to write this down, unit impulse response-- is simply the solution to the following problem, to our differential equation, x dot plus 2x that we're given, w...

GILBERT STRANG: OK. So today is unforced-- that means zero on the right-hand side, looking for null solutions-- damped-- that means there is a coefficient B in the first derivative. And what's the solution? This is really a basic, basic equation. In many applications, A would be the mass. In a spring, for example, A would be a mass. B is the damping, the friction. And C is the spring constant, the force that pulls the mass back. Or in electronics, B would be the resistance. It's giving some friction, giving some heat. So that's our equation. Just we have to be able to solve it. And we want to look for exponentials. A pure exponential is just right for a constant coefficient equation like that. So I'll substitute y equals e to the st. And what happens? Well, so there's a C e to the st. That's a Cy. The derivative brings down an s. Two derivatives bring down s squared. So I simply have As squared, Bs, and C all together, multiplying e to the st. And ...

PROFESSOR: So now, let me make the first of our technical asides, which you can skip over if you're not interested in the mathematical details or for those of you that have a higher-level, say, upper-level-undergraduate or even graduate-level understanding of transport phenomena and fluid mechanics. I'd like to show you some of the equations that are behind the results that I've been quoting in all the lectures. So in particular, let's derive the Wells curve. So part of that was a theory of drop settling. Here, I will quote a certain result, because the derivation would be a lot longer. Actually, for that, you could refer to my online class 10.50x, which is that if you have a droplet or a particle of a radius R and it is settling under gravity-- so it has a mass m, and the gravitational force is m g, where g is the gravitational acceleration-- then there is a flow of fluid around this object. And relative to the moving object, the flow is going the other w...

PROFESSOR: So, as an aside for more advanced students, let's try to fill in some mathematical details to provide a theory to support or interpret the Lin-Marr hypothesis of disinfection kinetics having to do with the concentration of solutes during drying and their effect on deactivating viruses. So, to put it in mathematical terms, if we have a certain number of viruses Nv in a droplet, then we'll postulate that d Nv dt is minus lambda v0, the deactivation rate per solute virion collision, times the volume fraction of disinfecting solutes we'll call phi d, which is time dependent, having to do with the size of the droplet, times Nv. The volume fraction of disinfecting solutes we'll write as alpha D, a constant, times phi s, which is the total volume fraction of solutes present. And that might be, for example, the fraction of solutes that are sodium chloride or some other salt that might be causing the damage to the virus, as opposed to the mucins or other...

PROFESSOR: So as a more technical aside, let's analyze more carefully the problem of release of viral load from a drop by process of diffusion. So here, again, I sketch a droplet, which would typically be an aerosol droplet in the size range of, let's say, microns. And the virion of interest has a size that is much smaller than that on the order of, let's say, 100 nanometers. And this white path is showing how such a virion would go from its initial position R, let's say, as a radio position, to the boundary. Now, the general problem of finding the expected first passage time from the point inside a domain to a boundary is a classical problem in the theory of stochastic processes and random walks. And it has the following representation. So the mean are expected first passage time from a point to an absorbing surface solves the following problem. Is the Laplacian of that time with a minus sign is 1 over D where D is the diffusivity. And so this is the DV t...

PROFESSOR: So besides our physical expectation that a virus such as SARS-CoV-2, coronavirus, could be transmitted through respiratory droplets, especially aerosol droplets through the airborne route of transmission, there is substantial evidence-- both epidemiological evidence and some physical evidence-- to support this hypothesis. So first, let's go through some of the epidemiological evidence. This is a very small fraction of what is available to date. So one of the first incidents that gave a sense there might be airborne transmission was a religious event that took place at the Tiantong Temple in Ningbo China, where there were hundreds of people in attendance. But in particular, there were two buses that brought worshippers in one-hour bus rides to this location. And on one of the buses was what was known to be the first infected person with COVID-19, entering this region after having had contact with others from Wuhan China, the initial source of the outbreak. A...

PROFESSOR: So let's think of a specific virus, the coronavirus, including the case of interest today, which is the SARS-CoV-2 novel coronavirus. So this virus comes in the form of virion, which is the capsid containing RNA, which is then going to infect a cell. And the size of that virion is around 120 nanometer diameter, and it's nearly a perfect sphere. Now, the size here of 120 nanometers compared to bacteria is about 1,000 times smaller than [some] bacteria. So this actually has a very big implication in terms of how a virion can be spread from one organism to another. So the bacteria, if you recall, are the size of several microns. That's the scale of large droplets, which sediment -- or larger than that, they would begin to sediment out of the air. And we can think about transmission through coughing. Also, besides the fact that the virus is much smaller, it cannot swim. So bacteria have various means of locomotion -- cilia, flagella, et cetera, whereas ...

PROFESSOR: So let's look at a little more detail at the equilibrium size of respiratory droplets that are emitted during breathing, or coughing, or speaking. And the key idea is that these droplets are not pure liquid. As explained in the wells curve, pure droplets that are small enough will shrink completely and evaporate. However, these droplets contain a significant amount of solutes. And those solutes, in the case of mucus coming from your lungs or from your vocal cords, your nasal pharynx, are full of proteins and other macromolecules, carbohydrates. And also there are always in bodily fluids plenty of dissolved salts such as sodium and chloride or calcium or potassium ions. Also, in saliva, many of these species are present, although it's not quite as thick of a liquid. And of course, virions as well will find themselves in here, and it also constitutes solutes. So the idea is that we don't just have a pure liquid. So there is some initial volume fractio...