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Vous préparez un test ? Révisez avec ces 6 leçons sur Les fonctions exponentielles.

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# Une fonction exponentielle

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.