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Transcription de la vidéo
Let's say that the position of some particle as a function of time is given by this expression right over here. Negative d to the negative t power plus c to the fourth over c squared plus 1, where c and d are constants and both of them are greater than one. So what I want to do over the course of this video is see what can we infer based on this expression, this function definition, that we have here. And the first thing that I want you to think about is, what is the initial position? If I were to express the initial position, in terms of c's and d's, and try to simplify it. So I encourage you to pause the video and try to find an expression for the initial position. Well, the initial position is the position we're at when our time is equal to 0. So we essentially just want to find p of 0. And p of 0 is going to be equal to negative d to the negative 0. So I'll just write that down. Negative 0 plus c to the fourth over c squared plus 1. Well, d to the negative 0, that's the same thing as d to the 0, and since we know that d is non-zero, we know this is defined. Anything non-zero to the 0-th power is going to be 1. And the zero is actually under debate, what 0 to the 0-th power is. But we can safely say that this right over here is going to be equal to 1. And so the numerator here simplifies 2. This is equal to-- and I'll switch the order. c to the fourth minus 1 over c squared plus 1. And now this might jump out at you as a difference of squares. We could write this as c squared, squared, minus 1 squared over c squared plus 1. And that's the same thing as c squared plus 1 times c squared minus 1. All of that over c squared plus 1. And we have a c squared plus 1 in the numerator and that denominator so we can simplify. And so our initial position is going to be c squared minus 1. So that actually simplified out quite nicely. Now the next question I'm going to ask you is-- OK, we know that the initial position that time equals zero, the particle is going to be at c squared minus 1. But what happens after that? Does the position keep increasing? Does the position keep decreasing? Or does the position maybe increase and then decrease, or decrease and then increase as in keep swapping around? So I encourage you to pause the video now and think about what happens to the position. Does it keep increasing? Does it keep decreasing? Or does it do something else? Well, let's answer that question of what's happening to the position after our initial position. We really just have to focus on this term right over here. This d to the negative t. This is the only part that really t is driving. All these other things are staying the same as we go through time. So what happens to d to the negative t power, to this part of this first term, as t goes from 0 onwards. And to think about that, let's plot. Let's plot what the function d to the negative t looks like. d to the negative t would look like-- and we know that d is greater than 1. So when t is equal to 0-- So this is t right over here. And over here we're going to plot on this axis. We're going to plot d to the negative t. When t is equal to 0 this is going to be equal to 1. We've already seen that. Now what happens as t increases? Say t increases to 1. Well, now this is going to be d to the negative 1, which is the same thing as 1 over d. And we don't know the exact value for d, but we know, since d is greater than 1, 1 over d is going to be less than 1. So let's say this is 1 over d right over here. 1 over d. So it's going to be something like that. And then when t is at 2 we're going to be at 1 over d squared, which is going to be something like right over there. And you see at least this term what it's doing as t increases. As t increases, d to the negative t is strictly decreasing. And once again we know that because d is greater than 1. So this term right over here is strictly decreasing. So this is decreasing. That part of it. But we're not adding it. We're subtracting it. We're subtracting from the beginning. At first this starts off at 1. We subtract 1. And then we start subtracting smaller and smaller things than 1. So if this is decreasing, but we're subtracting it, we're subtracting smaller and smaller things, this whole thing, the negative of it, is going to be increasing. Another way to think about it. If you wanted to plot negative d to the negative t, it would look like this. It would just be the negative of this right over here. So it would look something like-- I'll do that yellow color-- this. So this whole term right over here, negative d to the negative t, is constantly increasing. And we know that all of these other things, well, these are just going to be fixed. So this entire expression is constantly increasing starting at t equals 0. And then t going into larger and larger positive values. Now the last thing I want to ask you is, what is the maximum value here? What is a value that this will never be able to get to? It might try to get close to it, but it's never going to quite get there. Well, we already know that it's increasing, but let's just think about what happens as t becomes really, really, really large numbers. Really you could think about it as t approaches infinity. Well, once again let's look at this d to the negative t term. You see the d to the negative t, as t gets larger and larger, is getting smaller and smaller. This term right over here is approaching 0 as t goes to infinity. Well, if this is approaching 0 that means we're subtracting 0. So this whole yellow thing, the negative d to the negative t, is increasing but it's increasing at a smaller and slower rate. So this is the negative d to the negative t. This is right over here. You see it increases but it never quite gets to the horizontal axis right over here. And so if we think about it as t approaches infinity, this whole thing just becomes 0 and our entire position is approaching but never quite gets to c to the fourth over c squared plus 1. So one way to think about it, it's approaching this but it never quite gets to a position of c to the fourth over c squared plus 1.