WEBVTT
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Hello. Welcome to today’s Biomathematics
lecture. We have been discussing differential
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equations. So, in the previous lecture, we
discussed the first part of differential equations;
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very simple, ordinary differential equations.
Today, we will have the second part on differential
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equations. So, the title of this lecture is
Differential equations part 2. We will go
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ead from where we stopped last time, in
the last lecture…So, where we discussed
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simple differential equations of the form…
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So, let me write here. The simple differential
equations we wrote, like, something like,
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d y by d t is some function f, either of y
or a t; something of this form. So, such differential
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equations, as we said, is called ordinary
differential equations and today, we will
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go and see...So, we already discussed a few
examples, few different cases, which is including
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bacterial growth. And today, we will go and
see, which are the, which are the contexts
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in Biology, where in Biology, you would need
to think of differential equations. As you
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all know, one of the important things in Biology,
as well as in Physics, in general, or in general
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in nature, is a relation between energy and
force. So, you might have heard of potential
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energy, like electrostatic potential energy,
when, where charges attract with each other;
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so, the force with which they attract and
the energy, they are related to each other.
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So, they are related through a very simple
differential equation in some sense. So, today,
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we will discuss, first, we will discuss the
relation between the force and energy.
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So, let us look at the relation, the relation
between force f and the energy E. So, this
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energy here, could be potential energy. So,
some places you might see this v. So, here,
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I am using the notation E for the energy;
typically, you might see at some or some other,
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in some books, you might see, it is v or many
other u; in some other places, you might see
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it as u; but essentially, what we are representing
is the energy, the interaction energy. We
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will, as we come to different examples, we
will see what E is. So, the difference, the
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relation between f and E, that is, if you
know the E, the way to calculate f is to calculate
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minus d by, d E by d R is f; that means, the
first derivative of energy d E by d R with
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a negative sign will give you the force. So,
if, this is like a differential equation.
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This is like a differential equation, d E
by d R is equal to minus f. So, one can integrate
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this equation and get energy, because if you
know the force, you can calculate the energy
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by this, by solving the simple, the ordinary
differential equation.
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So, let us think of some simple energies.
So, in Biology, like, there is all energy
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associated with proteins, their configurations,
so on and so forth. But here, we will take
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a very simple case. So, it turns out that,
one simple way of thinking about proteins
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is think about spring. So, spring is something
like, an extend and compress and all that;
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proteins can also be pulled out. So, it can
be extended in some sense; they get folded.
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So, they can come to a comeback state, from
where you can pull, pull out the proteins.
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So, essentially, the simplest way of thinking
about protein is to think about springs and
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charges; those are the two simple interactions,
that one can think of easily and they correspond
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in some sense with proteins. So, we will,
in this course, to understand the concept,
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we will think about the simplest example,
that is a spring.
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So, let us take the case of a spring and have
a look at this slide. The spring has force
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that is proportional to the distance. Now,
what is force and what is distance? So, what
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is the distance here? So, let us, let us try
and understand a bit. So, have a look at this
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drawing here. So, you have a drawing here.
Though, this distance…So, now, you, you
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have a spring, which you have some certain
distance here, which we call R. So, it turns
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out that, the more the distance, the more
the force is. So, you can always, all of you
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have taken a spring or go and take a spring
and try pulling it; try applying a force.
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The more you pull it, the more force you have
to apply; to hold it in a far distance, to
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hold it at a distance R, you have to apply
less force; to hold this two ends of this
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springs’ farther, you have to apply a larger
force. So, it turns out that, the force and
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the distance are proportional to each other.
So, that is why, the force is proportional
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to R, the distance, but it turns out that,
when you hold it, your experience, experiencing
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a force in the opposite direction. So, to
show this direction, we have minus sign; the
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spring force is proportional to minus R. So,
this is what, this is the basis of what is
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written here, in this slide, f is directly
proportional to minus R. So, the more the
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force, the more the distance, this between
two ends of this spring, if it is R, the more
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the distance, the more force you have to apply,
but the direction is in the opposite. So,
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there is a minus sign. So, we know that, f
is, you can remove this proportionality by,
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you can equate this with a constant k, where
k is, you can think of k as a spring constant,
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which depends on the property of this spring.
If the spring is made of some particular material,
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you will have particular k; if the spring
is made of some other material, you will have
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some other k. So, k essentially, depends on
the material property of the spring. So, we
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know that f is minus k R.
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Now, let us go back to the old relation, which
we, which we saw previously, and think about
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this. So, we saw this relation that, d E by
d R is minus f, that is what we said. This
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is the relation between energy and force.
Now, if we substitute f is equal to minus
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k R here, what you would get is, d E by d
R is equal to k R or d E is equal to k R d
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R; that is, you can take this d R on this
side; I can multiply both sides with d R in
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this part of equation and I get d E is equal
to k R d R. Now, we integrate both sides,
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integral d E is integral k R d R. So, what
you get is, E is equal to k R square by 2
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plus constant.
So, this is what the integral of, integral
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of d E is E; integral of k R is, k is k and
R is a constant, integral of R is R square
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by 2. So, basically, what you would get is,
half k R square plus constant. So, half k
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R square plus constant. So, what you get is,
energy is half k R square plus a constant.
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We always saw that, this constant is a part
of the constant of integration. Whenever you
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do indefinite integral, you have to have a
constant, which now depends, which can be
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fixed by boundary conditions, if you wish.
If you know the boundary conditions, the energy,
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the energy needed to pull from here to here,
if know the, if you say that, then, you can
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fix this constant. But otherwise, a constant,
you can always add a constant to an energy.
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It turns out that, what matters is the derivative
of energy, because in experiments, what you
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can measure is always force; it is difficult
to measure energy. So, whatever be this constant,
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whatever be the value of this constant that
we are dealing with, the derivative of this
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d E, when you calculate the d E by d R, the
constant will go to 0. So, you can put any
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constant you want in principle and still,
you will get the same force. So, now, let
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us quickly plot this and see.
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Let us get a feeling for this plot. So, let
us have a plot here. Let us plot energy versus
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R. So, you will get a parabola k R square.
I am not plotting this left hand side, because
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in this case, R is only positive for a spring.
So, this is, this is what half k R square.
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Now, even if you add a constant to this, the
shape of this curve will not change. So, let
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us have a look at this plot once more; even
if you add a constant to this, the curve will
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become something like this. So, essentially,
it will get shifted, but the shape, the shape
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of the curve will not change. So, this is,
this is half k R square and this is the half
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k R square plus a constant. So, let, let me
plot this little more carefully here. So,
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let me plot this little more carefully in
the next one.
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So, let us plot here. So, this is half k R
square, which is energy versus R. Now, we
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can add a constant to this and if you plot
that, you will get something like this. This
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is exactly the similar curve, parallel to
this. I have not drawn it very properly, but
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it is just like adding the constant everywhere;
the shape of the curve will not change. So,
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essentially, what you want to learn is that,
this is the force, which is the derivative
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of this curve at every point, if you calculate
the derivative at any point, if you calculate
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the derivative here, you should get the same
derivative; because, you are essentially adding
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a constant. So, basically, that, the force
will not change. Now, you can also do this
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by saying that, you can pull the spring from
this place to this place and you can ask the
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question, how much is the energy caused to
pull this spring from here to here. So, let
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us do that, let us do that, here. So, have
look at this slide here. So, if you look at
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the slide, what you have this equation is
integral d E is integral k R d R. So, now,
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let us write that here.
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So, let us write integral d E is integral
k R d R. Now, first you had this particular
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R, which is, let me call this R 1. Now, I
am pulling this spring to a greater distance,
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which is R 2. So, I am pulling from R 1 to
R 2. So, I have this spring, which is R 1.
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Then, I am pulling to a distance, length R
2. You can ask the question, how much energy
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it ((cost)) to pull from R 1 to R 2. So, then,
this integral goes from R 1, R 2. So, this
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is, energy is, this is energy, and here R
1 plus R 2. So, k R square by 2, now, in the
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limits R 1, R 2. So, if you have, this is
the integral, integration with the limits
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R 1 and R 2, so, this is done by, you can
write integral of k R d R is k R square by
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2 and you have to apply this limits. So, the
way of applying limits, as you learnt is that,
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k into…So, this 2 is also constant. So,
I can take this 2 out.
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So, basically, R 2 square minus R 1 square.
So, this is the energy. So, this is the energy
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needed to pull a spring from R 1 to R 2. If
this is R 1, this R 2, the energy needed to
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pull the spring from R 1 to R 2 is k by 2
into R 2 square minus R 1 square. So, this
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is the answer. So, this can be written as
k by 2 into, where k by 2, you can say, k
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by 2 R 2 square minus k by 2 R 1 square. So,
you can also write like this, k by 2 R 2 square
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minus k by 2 R 1 square. So, this is exactly
same as that. So, this is basically, the energy
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needed to pull the spring from R 1 to R 2.
So, this is essentially, the energy difference.
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If you look at this equation once more, little
more carefully…So, let us write it here,
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separately.
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So, let us write, let us write that, what
we got is energy is k by 2 R 2 square minus
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k by 2 R 1 square. So, this is nothing, but
energy to keep this...So, this is an one energy,
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which is E; let me call this E 2 and this,
may call this is E 1, where E 2 and E 1…So,
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this is R 1 and this is R 2. So, this is,
the energy of this spring is E 1 and this,
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energy of this spring is E 2. So, then, that,
the energy difference is basically, E 2 minus
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E 1. So, basically, what you find is that,
the change in energy, the energy needed to
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pull from here to here, is the energy difference
between this two and this thing, you can get
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by integrating this particular equation, this
particular differential equation. So, this
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is the simplest case that we learnt. Now,
let us go to a next case, for example. So,
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let us, which is basically, charges.
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So, as we said, we, if we learn about springs
and charges, we learn a lot about Biological
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system. So, let us take the simplest case
of a charge system, which is two positive
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charges at the distance R apart. So, this
are two charges plus q and a minus q and the
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distance between them is R. And we know that,
in such a case, what is force. The force is
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given by Coulomb’s Law and it is, f is q
square divided by 4 pi epsilon 0 epsilon r
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R square, where epsilon 0 and epsilon r, you
know as the permittivity of, at vacuum and
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relative permittivity. So, you know, you know
this constants from your school classes. So,
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where capital R here is the distance between
this two charges and this is the force and
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this is the famous Coulomb’s law. Now, known
this, how do we calculate energy of the system,
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energy this charged systems. So, the energy
of this charge system can be calculated the
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same way, by integrating the equation as you
see here.
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So, the equation here is, d E by d R is minus
f. So, d E by d R is minus f and the f we
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know. So, let us find out what happens. If
you substitute this f here, you get q square
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by 4 pi epsilon 0 epsilon r R square, that
is what you substitute for f; this is what
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you get. So, essentially, you can take this
d R this way, this side, or multiply both
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sides by d R and you get d E is q square by
4 pi epsilon 0 epsilon r R square d R. So,
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this is basically, rewriting this equation
and now, we can integrate both sides. So,
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q is a constant; 4 is a constant; pi, epsilon
0, epsilon r, all of them are constants. So,
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you can take out all of this and you can have
just integral of 1 over R square d R. So,
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what is integral of 1 over R square d R? We
know that, integral of R power, x power minus
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n or 1 over x power n is, if you have integral
of x power minus n, it is x power minus n
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plus 1 divided by minus n plus 1. We know
this.
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So, let us have a look at it. So, if you do
this integral of R power minus 2 d R, that
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is what you have to calculate.
18:24.600 --> 18:31.600
And, we know that, integral R power minus
2 d R is R power minus 2 plus 1 divided by
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minus 2 plus 1 plus a constant. And, this
is R power minus 1, with the minus sign. This
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is what integral R power minus 2 d R. So,
essentially, you will get R power minus 1
18:52.620 --> 18:59.620
is 1 over R with a minus sign plus a constant.
So, have a look at this, here in the slide.
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You get a minus sign extra. This minus sign
goes away and q square 4 pi epsilon 0 epsilon
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r remains the same and R power minus 2 becomes
1 over R plus a constant. So, this is basically,
19:13.580 --> 19:20.580
you know that, this is nothing, but the potential
energy of two charges. So, the potential energy,
19:23.470 --> 19:28.899
when you have two charges, the potential energy
of this system was q square by 4 pi epsilon
19:28.899 --> 19:35.899
0 epsilon r by R. So, this the potential energy.
Then, do not confuse this, this E is here,
19:38.000 --> 19:44.120
E is not electric field; this is potential
energy. I just use a symbol E for energy.
19:44.120 --> 19:51.120
So, some, some places u; in the text book,
you might see this as v. Typically, this is
19:51.690 --> 19:58.690
what is used, v. So, you will see that, this
is, this is typically written as v. So, some
19:59.000 --> 20:02.049
places, you will see that potential energy…
20:02.049 --> 20:09.049
So, let me write, potential energy of two
charge systems is q square by 4 pi epsilon
20:14.070 --> 20:21.070
0 epsilon r R square plus a constant and that
constant, can be anything; that does not change
20:21.789 --> 20:27.610
the force. Whatever be the constant, again,
the force remains the same, because, the derivative
20:27.610 --> 20:34.610
of energy is force and derivative of any constant
is 0. So, again, you have the force. Again,
20:38.340 --> 20:45.340
here also, you can do this integral in the
limit. You can do, what we had here is, if
20:47.769 --> 20:54.769
you look at this slide, what you have here
is, integral minus q square by 4 pi epsilon
20:57.240 --> 21:04.240
0 epsilon r, integral 1 over R d R. So, you
can do this from R 1 to R 2 again; this is
21:11.250 --> 21:17.470
the same. What are we asking here? We are
asking, if, we have charges this distance
21:17.470 --> 21:22.169
and if it move, if you move this charge from
this distance to this distance, how much energy
21:22.169 --> 21:29.169
it cost, to move this charge. So, again, you
can do this. So, if you do this integral…
21:30.970 --> 21:37.820
So, let us have a look at what you end up
here. Basically, what you have to do is, q
21:37.820 --> 21:44.820
square by 4 pi epsilon 0 epsilon r is a constant.
R 1 to R 2. 1 over R square d R, this is what
21:50.889 --> 21:57.889
we have to do. So, in the previous one, we
had integral 1 over R square. So, this is,
21:58.980 --> 22:05.980
this is the energy we have. So, integral of
this, in the limits R 1 to R 2 is nothing,
22:06.039 --> 22:11.870
but…So, this is equal to, if you do this,
this constants, this constant will exactly
22:11.870 --> 22:18.870
remain the same, q square by 4 pi epsilon
0 epsilon r and integral 1 over R square,
22:21.110 --> 22:27.409
as we saw, is 1 over R with a minus sign.
So, we had a minus sign here.
22:27.409 --> 22:34.409
So, that minus sign goes with the plus, it
becomes plus here, in the limits R 1, R 2.
22:36.750 --> 22:42.639
So, you have to apply a limit to this 1 over
R. So, essentially, what you would get is,
22:42.639 --> 22:49.639
q square by 4 pi epsilon 0 epsilon r into
1 over R 2, 1 over R 2 minus 1 over R 1. This
22:56.029 --> 22:59.679
is what essentially, the potential energy
difference, the energy difference between
22:59.679 --> 23:06.679
two points; or, in other words, the energy
needed to bring this charge from R 1 to R
23:07.490 --> 23:12.090
2. If you want to move this charge, let us
say, this proteins keep moving; when they
23:12.090 --> 23:17.960
change, when they their configuration, the
charges get, has to move from one place to
23:17.960 --> 23:24.590
another place. So, that much energy has to
be put in to move one charge from one place
23:24.590 --> 23:31.590
R 1 to another place R 2. So, this is basically,
what we have seen here is, the energy needed
23:32.080 --> 23:39.080
for a charge to move from R 1 to R 2. Again,
we got this by solving this simple differential
23:40.769 --> 23:44.110
equation.
So, this is all a part of calculus, where
23:44.110 --> 23:50.730
you can apply all this ideas, simple ideas
of integration or differentiation, to understand
23:50.730 --> 23:57.730
about forces, bacterial growth, energies and
motion. So, when we talk about motion, the
24:00.840 --> 24:06.440
next question is, how does this differential
equation governs motion. So, that is what,
24:06.440 --> 24:10.149
something we ill, we will, we will come and
see later in this talk.
24:10.149 --> 24:15.629
But there is another simple differential equation,
which we have to understand. So, one another
24:15.629 --> 24:22.509
thing that is often comes in Biology, is concentration.
So, concentration varies with, within the
24:22.509 --> 24:29.220
cell or concentration can vary within the,
within a particular regime, region. So, when
24:29.220 --> 24:34.830
this concentration varies, a diffusion can
happen; that you know. So, our aim is to try
24:34.830 --> 24:41.830
and understand this diffusion, from this perspective
of differential equations. So, there is something
24:42.620 --> 24:47.269
called diffusion equation; but at this moment,
immediately, we will not go directly to the
24:47.269 --> 24:53.279
diffusion equation, which is a bit complicated
for beginners. So, first, we will try and
24:53.279 --> 24:58.909
understand, how the concentration gradient
or the difference in concentration, how do
24:58.909 --> 25:01.370
we represent that, through a differential
equation.
25:01.370 --> 25:08.370
So, let us have a look at this particular,
this slide. So, let us imagine such a cell
25:08.379 --> 25:13.740
having uniform concentration of something.
So, this is the green thing here, which is,
25:13.740 --> 25:20.070
let us say, some protein; it is almost uniform
all over. So, what does, if we want to make
25:20.070 --> 25:26.929
this statement mathematically, that this is
uniform all over, the way to make the statement
25:26.929 --> 25:33.929
is exact following; that is what written here.
We can say that, d C by d X is a, is 0; that
25:34.870 --> 25:40.100
is, the c does not change which x. So, here
the x is this distance.
25:40.100 --> 25:47.100
So, so, let me, let me draw here. So, we had…Have
a look at this drawing here. So, we had, let
25:47.990 --> 25:54.029
me call, this x as this distance, for convenience,
here. So, this, starting from here, the x
25:54.029 --> 26:01.029
goes in this particular way. So, all over
here, the concentration is pretty much the
26:01.090 --> 26:08.090
same; let us say, uniform concentration everywhere.
In that case, what you want to say is that,
26:09.340 --> 26:16.039
the concentration does not change with x.
So, if you say C as a function of x does not
26:16.039 --> 26:23.039
change, that means, d C by d x, the change
in C, when you change x is 0 ; that means,
26:25.690 --> 26:32.690
the concentration does not change. This only
means that, C is a constant. How do you know
26:34.490 --> 26:41.490
that, because you can integrate this equation;
I can multiply both sides with d x. So, you
26:41.519 --> 26:46.240
will get d C equal to 0, because, if you multiply
0 with anything is 0.
26:46.240 --> 26:53.240
Now, you can integrate this and you can integrate
this. So, 0 is anyway a constant, as there
26:56.480 --> 27:03.350
is the constant of integration. So, you will
get C is equal to a constant. So, that is
27:03.350 --> 27:10.350
how you integrate by this equation. So, essentially,
what we did here is that, we conveyed a message
27:13.720 --> 27:20.720
that, the concentration is uniform across
this cell or this particular region that we
27:22.269 --> 27:28.580
saw and this is mathematically explained,
this is mathematically expressed this way
27:28.580 --> 27:35.580
that, d C by d x is 0 . Now, let us, let us
look the next case, where it is not uniform,
27:36.500 --> 27:39.919
but it is slowly varying. How do we express
that?
27:39.919 --> 27:46.450
So, have a look at the next one. So, from
here to here, the concentration is slowly
27:46.450 --> 27:51.700
increasing. Here, the concentration is less;
here, the concentration is more. So, the concentration
27:51.700 --> 27:58.700
is slowly increasing. So, how do we say that,
the concentration is slowly increasing? So,
27:59.740 --> 28:06.740
the way to say this is that, d C by d X is
a constant; so, that means, the change in
28:07.100 --> 28:11.169
the concentration, when you change x is a
constant. This is a simplest thing, constant
28:11.169 --> 28:18.169
change in concentration. And, you can integrate
this equation, as we always do; multiply both
28:20.740 --> 28:27.740
sides with d X. So, d C can be written as
k d X and you can integrate both sides, like
28:28.149 --> 28:35.149
we do. So, C is a constant; integral of d
C is C; k is a constant; integral of d X is
28:36.320 --> 28:41.360
just X. So, k X plus a constant. So, that
means…So, uniform, it is a concentration
28:41.360 --> 28:48.360
gradient. So, there is a concentration gradient.
So, in this point…So, this thing is called
28:50.429 --> 28:55.330
a concentration gradient. The d C by d X,
the derivative of concentration. But, when
28:55.330 --> 29:00.100
you talk about the real gradients, there is
something more we have to understand; we have
29:00.100 --> 29:06.740
to incorporate the idea of the vector, the
direction. So, whenever we talk about the
29:06.740 --> 29:13.220
force, and whenever we talk about gradients,
we have to include the direction, somehow,
29:13.220 --> 29:18.679
in this, they all, there is a gradient in
a particular direction. As we go from here
29:18.679 --> 29:24.450
to here, there is a gradient. We have to bring
this idea of gradient here. In, we will discuss
29:24.450 --> 29:30.899
that, how do we bring about this idea of direction
here. We have, we have to incorporate some
29:30.899 --> 29:37.690
ideas from vectors. So, the vectors that we
discussed, we will, we will see how do we
29:37.690 --> 29:44.690
incorporate the idea from vectors, to port,
to stay, to say here, to incorporate the direction
29:47.370 --> 29:51.980
in this gradient. The, essentially, we want
to say that, the concentration is increasing
29:51.980 --> 29:57.850
in this particular direction; in which direction
is concentration changing.
29:57.850 --> 30:03.600
So, we have to incorporate this idea of a
vectors here, to exactly, mathematically convey
30:03.600 --> 30:10.600
this idea. So, we will discuss, how do we
do that, in another lecture; but at this moment,
30:14.779 --> 30:20.700
we basically, just want to understand this
differential equation d C by d X is a constant,
30:20.700 --> 30:27.279
and the integral of that is C is equal to
k X plus constant and that is what the integral
30:27.279 --> 30:34.279
is. So, we know this much. Now, we will go
to a different things. So, we said, concentration
30:35.309 --> 30:38.080
is one of the important things in Biology.
Another thing is, the very important thing
30:38.080 --> 30:43.779
is the force, but there is force associated
with motion, motion. So, we saw force associated,
30:43.779 --> 30:49.419
force and energy. Now, we, what we will see
is that, force and motion; because things
30:49.419 --> 30:55.769
move around; molecules move around; and, when
molecules move around, in Biology, you have
30:55.769 --> 30:59.889
to add some force.
And, how does this force and motion are related.
30:59.889 --> 31:06.379
Because, basically, for example, you know
that, protein, which is completely straight,
31:06.379 --> 31:12.580
they fold up into a folded configuration.
So, the protein which is unfold, which is
31:12.580 --> 31:17.629
not in this fold, which is in some random
configuration, they all come together and
31:17.629 --> 31:24.629
fold. So, there is some force, leading force,
making this protein fold. In other words,
31:26.869 --> 31:33.869
the protein molecules are moving in a particular
direction, so that, they end up in a folded
31:33.899 --> 31:39.379
configuration. So, how does this force and
motion, how are they related? And, they are
31:39.379 --> 31:44.490
related through a differential equation and
this is what we will see today. So, let us
31:44.490 --> 31:51.490
have a look at this. So, before discussing
this, we should understand one thing about
31:55.419 --> 31:57.320
differential equations.
31:57.320 --> 32:03.619
So, there is something called second order
differential equation. So, what we have discussed
32:03.619 --> 32:08.730
so far is called the first order differential
equation. So, let me just say, what I mean
32:08.730 --> 32:11.049
by first order differential equation once
more.
32:11.049 --> 32:18.049
So, we said things of the form d X by d t
is something. So, here, there is, we only
32:24.639 --> 32:31.639
use the first derivative. So, such things
are called first order, ordinary differential
32:34.480 --> 32:40.649
equations. We will see what are, we will see
some other kind of differential equations.
32:40.649 --> 32:47.649
So, but, whatever we saw so far are first
order, ordinary differential equations. Now,
32:52.090 --> 32:59.090
we will go and look at something called second
order differential equations. So, have a look
32:59.600 --> 33:05.429
at this slide. If we have d square X by d
t square is something…So, there is a d,
33:05.429 --> 33:12.429
this is second derivative. So, such things
are called ordinary differential equations.
33:12.629 --> 33:16.970
So, now, we will immediately come and see
what are this examples, what is the example
33:16.970 --> 33:22.769
of this, because, when we say about force
and motion, we have to deal with second order
33:22.769 --> 33:29.519
differential equation, which is one level
up from what we discussed so far. So, now,
33:29.519 --> 33:36.519
how do we learn about force and motion? So,
we all know that, something that connects
33:36.730 --> 33:41.909
the force and motion is this famous equation
called Newton’s equations, where Isaac Newton
33:41.909 --> 33:48.909
proposed this equations in, way back in 60,
late 60s, like, in 17th century and early
33:51.090 --> 33:58.090
18th century. So, in late 1600s and 1700s,
Newton discussed about this equations. So,
34:01.320 --> 34:06.759
the equation that connects the force and motion
are called Newton’s equations.
34:06.759 --> 34:13.759
So, let us see an example. So, we can think
of an example, which is d square x by d t
34:15.389 --> 34:22.389
square is some, some constant f by m. So,
this f is a force.
34:22.679 --> 34:29.679
So, Newton’s equation is essentially, as
you all know, is that, d square X by d t square
34:34.310 --> 34:41.310
is m. So, this is called Newton’s equation.
So, you all know that, f is equal to m a.
34:43.850 --> 34:50.850
This is the Newton’s equation that, as all
of you know. Now, a, acceleration can be written
34:50.890 --> 34:57.890
as d v by d t; that is, a is acceleration
and acceleration is change in velocity by
34:58.850 --> 35:05.850
change in time, d v by d t; and, v can be
written as, v can be written as d X by d t,
35:11.400 --> 35:18.400
that is change in position divided by change
in time. So, if we, d X by d t is v and d
35:20.540 --> 35:27.540
by d t of this is acceleration...So, this
whole thing can be written as d square x by
35:34.130 --> 35:41.130
d t square. So, this is essentially, d square
X by d t square, this whole thing. This whole
35:42.480 --> 35:49.480
thing is written as d square X by d t square.
So, essentially, acceleration can be written
35:51.340 --> 35:58.340
as d square X by d t square and acceleration
is force by mass. So, this is basically, Newton’s
35:58.480 --> 36:04.290
equation. And, the simplest Newton’s equation
you can think of is something falling under
36:04.290 --> 36:10.470
gravity. So, let us say, this pen is, if I
leave the pen, it will fall under gravity.
36:10.470 --> 36:17.470
So, the falling of this pen or any object
that is falling under gravity, can be described
36:18.830 --> 36:25.830
by this equation. If you say f is equal to
m g, if I say f is equal to m g, I get this
36:31.210 --> 36:37.050
equation, which is d square X by d t square
equal to g, which is a simple equation, which
36:37.050 --> 36:43.140
is basically, equation of, Newton’s equation
for an object falling under gravity, where
36:43.140 --> 36:50.140
g is acceleration due to gravity, which is
the constant, which is 9.8 meters square per
36:51.920 --> 36:58.920
second. So, g is a universal constant for,
in Earth, which is 9.8 meter per second square.
37:02.380 --> 37:09.380
And, once you know this g, we can solve this
differential equation d square X by d t square,
37:10.430 --> 37:17.430
the second order differential equation and
get a solution for X, where X is the position
37:18.130 --> 37:22.690
of the object falling. So, let us try and
understand little more, what is this equation
37:22.690 --> 37:26.010
actually mean. So, let us try and understand
little more.
37:26.010 --> 37:31.340
Let us say an object…Let us have a look
at here. Let us say, an object is falling
37:31.340 --> 37:38.230
from here. And so, this is the object, which
is falling. It started falling from this particular
37:38.230 --> 37:45.230
position. So, let X is the distance from the,
from the initial position, at any point. So,
37:48.410 --> 37:55.410
sorry, I, it, consider X as the distance from…So,
X can be thought of as a distance from any
37:58.400 --> 38:04.760
point. We will come and discuss, what is X,
basically. X is basically, the distance of
38:04.760 --> 38:09.590
this object falling. You can measure either
the distance from the ground up, or from a,
38:09.590 --> 38:15.050
some fixed point. So, we can, we can decide,
where you want to take this distance.
38:15.050 --> 38:22.050
But, X is essentially, this distance and how
does this X, this distance vary as the function
38:22.380 --> 38:27.070
of time. So, that is what, this equation is
described. How does this distance, for example,
38:27.070 --> 38:31.720
the distance from the ground, how does it
vary as the function of time; or, distance
38:31.720 --> 38:36.920
from a fixed point, how does this distance
vary as the function of time, when this object
38:36.920 --> 38:43.920
is falling. So, that is, this equation describes.
So, again, let us have a look at this equation,
38:44.040 --> 38:50.280
d square X by d t square. How does this X,
which is the distance from the Earth, for
38:50.280 --> 38:56.560
this particular object that is falling, is
changing with respect to time, that is what
38:56.560 --> 39:02.560
the question we are asking. So, it turns out
that, this can be integrated the way we did
39:02.560 --> 39:04.010
for all other cases.
39:04.010 --> 39:10.110
So, let us have a look at this next slide.
So, this slide says, d square X by d t square
39:10.110 --> 39:17.110
can be written as d by d t of d X by d t,
as we saw. Once we write this in a particular
39:18.200 --> 39:24.980
way, this is basically, equal to g. This is
our equation. Now, d x by d t, as we said,
39:24.980 --> 39:31.980
is change in position by change in time. It
is nothing, but velocity v. So, v is d x by
39:32.300 --> 39:39.300
d t. And, this is, d v by d t is equal to
g. So, this equation becomes d v by d t equal
39:40.390 --> 39:46.150
to c, g and as we…This is a simple, ordinary
differential equation that we have been seeing
39:46.150 --> 39:51.050
all in this class, in the previous class.
And, such an equation has an, immediately,
39:51.050 --> 39:55.680
we can see that, such an equation has a solution,
which is v is equal to g t plus constant.
39:55.680 --> 40:00.980
And, the constant, as we said previously,
is basically, the velocity at time is equal
40:00.980 --> 40:07.870
to 0. So, essentially, you can write, v of
t is equal to g t plus v of 0. If you have
40:07.870 --> 40:12.680
confusion, let me do this little more carefully.
Let me do this little more carefully. How
40:12.680 --> 40:14.160
did we do this?
40:14.160 --> 40:21.160
So, d v by d t was g; this was our equation.
So, now, you can integrate this by d v is
40:24.080 --> 40:31.080
g d t and integrate both sides. So, basically,
v is equal to, v as a function of time is
40:35.290 --> 40:42.290
equal to g t plus a constant. Now, how do
we calculate this constant? Put time equal
40:46.380 --> 40:53.380
to 0. At time is equal to 0, what you have,
v of 0 is equal to v of, v at time equal to
40:56.110 --> 41:01.970
0, is this constant. So, essentially, what
we have, you substitute this thing back in
41:01.970 --> 41:08.970
the above equation. Then, you get v of t is
equal to g t plus v of 0. So, this is essentially,
41:14.050 --> 41:20.370
what I have written here. This is exactly
what I have written here, v of t is g t plus
41:20.370 --> 41:27.370
v of 0. So, you know, as this is falling,
the velocity seems to be increasing. So, if
41:27.440 --> 41:33.050
you plot this g as a constant, it is like,
y is equal to m x plus c. This is a straight
41:33.050 --> 41:40.050
line. So, that means, as an object is falling,
its speed is increasing; its velocity is increasing.
41:41.990 --> 41:46.240
So, that is what it, that is what it says;
that is what the first part of the solution
41:46.240 --> 41:53.240
tells you that, as an object is falling under
gravity, its speed is increasing; that is
41:54.840 --> 42:01.840
what, the first part of this equation says,
v t is equal to g t plus v 0. Now, this is
42:04.290 --> 42:10.680
only the fist integration we did and we got
just the velocity and to do the one integration,
42:10.680 --> 42:15.080
we need to understand one constant of integration,
v of 0.
42:15.080 --> 42:20.450
Now, let us do the second integration. The
second integration is…So, we have, as of
42:20.450 --> 42:27.450
now, we have v of t is g t plus v 0. Now,
we know that, v is nothing, but d x by d t.
42:31.140 --> 42:37.880
If we substitute v of t is d x by d t here,
we have another differential equation now.
42:37.880 --> 42:43.330
Now, we can integrate this differential equation
by, as we…This is again, is the first order
42:43.330 --> 42:49.440
differential equation. You can multiply both
sides with d t. So, what you will end up is,
42:49.440 --> 42:56.440
integral d x is integral g t plus v 0 d t;
that is what you get. So, you know that, integral
42:59.240 --> 43:06.240
g t is g t square by 2 and integral v 0, which
is a constant, is v t, v 0 t. You have and
43:07.070 --> 43:12.690
you have another constant, which is, we call
x 0. Essentially, it turns out that, x 0 is
43:12.690 --> 43:19.690
the position at time equal to 0. This is something
which you can simply see, just as we typically
43:20.590 --> 43:23.800
do. So, let us have a look at here, how did
we do that.
43:23.800 --> 43:30.800
So, we had this equation, which is d X by
d t is equal to g t plus v 0. And, we multiply
43:36.350 --> 43:43.350
both sides by d t and integrate it. So, integral
g t plus v 0 d t and the first part of this
43:46.360 --> 43:53.360
is integral g of t d t plus integral v 0 d
t. This is equal to, integral g t d t is g
44:01.730 --> 44:08.730
t square by 2 plus a constant and v 0 is integral
v 0 t plus some other constant. So, this two
44:11.990 --> 44:18.990
constants together, let me call this constant
C, as a constant C. Now, this C…So, basically,
44:22.930 --> 44:29.830
what you get is x of t. Now, when you put
t equal to 0, this term goes to 0; this term
44:29.830 --> 44:36.830
also goes to 0. So, C is nothing, but x equal
to 0. So, C is nothing, but x at t equal to
44:37.390 --> 44:44.390
0. So, you substitute this back here; so,
you get this particular thing, which is x
44:48.990 --> 44:52.870
of t is g t square plus v 0 t plus x 0.
44:52.870 --> 44:59.570
So, this is essentially, what does it say
is that, if something is falling under gravity,
44:59.570 --> 45:05.010
how does the distance…So, distance is, let
us say, you, I start, from this particular
45:05.010 --> 45:12.010
point, I drop this pen and the distance will
increase. The distance from my hand to this
45:12.260 --> 45:19.260
pen will increase as the distance, as the
time goes. So, then, what you get is essentially,
45:24.500 --> 45:28.970
x of t, which is, this equation is basically,
gives you the position as the function of
45:28.970 --> 45:33.780
time. As the, after…So, you can ask this
question by looking at this equation. So,
45:33.780 --> 45:36.290
let us have a look at this equation once more.
45:36.290 --> 45:43.290
So, basically, x of t is g t square plus v
0 t plus x 0. If you know x 0, and if we know
45:48.420 --> 45:55.420
v 0, and we know g, which is the constant
of, as gravity is a constant, acceleration
45:55.990 --> 46:02.990
due to the gravity is a constant. If we know
this much, for any time t, we can substitute
46:03.320 --> 46:10.320
in this equation and get x of t. So, we can
get x after 3 seconds, is equal to g into
46:13.380 --> 46:20.380
3 square plus v 0 into 3 plus x 0. Now, if
we know that, x 0 is 0, the, if we said that,
46:25.390 --> 46:30.680
x 0 is your height, let us say, you are putting
from your height, which is, let us say, your
46:30.680 --> 46:36.680
height is like 1 and a half meters. So, let
us say, you are dropping this pen from 1.5
46:36.680 --> 46:43.680
meters, I can get x after 3 seconds is g,
which we know is like, close to 10 meter square
46:47.850 --> 46:54.850
per second into 3 square plus v 0 and I am
dropping it at 0 velocity; at the beginning,
46:57.800 --> 47:04.490
there is no velocity at all. So, v 0 is 0.
So, let us assume, in this equation v 0 is
47:04.490 --> 47:11.490
0. So, this is 0 into 3 is 0, plus x 0, I,
initial position is 1.5 meters. So, if you
47:13.320 --> 47:20.320
do this, what I get is, 10 into 9, 10 into
3 square is 9. So, it is 90 plus 1.5, 91.5.
47:24.040 --> 47:31.040
So, this is, there is a slight absurd thing
here, which is…This is assuming that, the
47:32.450 --> 47:39.450
force is there forever, but we have, we have
Earth after 1.5 meters. So, we have to take
47:39.860 --> 47:46.860
this numbers properly. So, same way that,
you put from, like, from reasonably good height
47:47.670 --> 47:52.920
and within, before 3 second, you would have
hit the ground in this particular case; because,
47:52.920 --> 47:58.550
the answer you get essentially, is 91.5 meters.
This equation says that, if you have such
47:58.550 --> 48:05.550
a force, in 3 second, if there is nothing
to stop, it would have gone 91.5 meters. It
48:05.810 --> 48:11.150
is a slightly absurd thing, because, you are
only putting 1.5 meters height and then, it
48:11.150 --> 48:18.150
cannot go beyond that, because we have Earth.
But if you take this from very, if you drop
48:18.210 --> 48:24.090
this pen from very, from a very height point,
you can see that, it will fall down and this
48:24.090 --> 48:28.620
equation will be obeyed.
So, take or do this exercise by taking x 0
48:28.620 --> 48:35.620
as, let us say, you are putting it from a
top of a building and let us say, how far
48:36.200 --> 48:42.220
this would go, if you had taken all this constants
properly. So, let, we will, we will do this
48:42.220 --> 48:47.860
exercise in the next class, but have a, go
home and then, also think about what you would
48:47.860 --> 48:52.930
get by putting the right numbers in this equation.
But as of now, you understood, how do we solve
48:52.930 --> 48:54.440
this differential equation.
48:54.440 --> 49:01.390
So, one point, important point to remember
is, again, I am reiterating in the next slide,
49:01.390 --> 49:08.390
which is that, we have two constants of integration,
v 0 and x 0. So, to solve any second order
49:10.010 --> 49:15.480
differential equation, any equation of the
form d square x by d t square or d square
49:15.480 --> 49:21.660
y by d x square or d square something by d
something square, any second order equation,
49:21.660 --> 49:27.880
if you want to solve it, you need to know
two constants. To solve a first order equation,
49:27.880 --> 49:32.430
you need to know one constant. To solve a
second order equation, you need to know two
49:32.430 --> 49:37.820
constants. Here, it turns out that, the two
constants, in the case of equation of motion,
49:37.820 --> 49:44.600
Newton’s equation, they are the initial
position and the initial velocity. But in
49:44.600 --> 49:49.760
some other context, it will be something else.
But you need to know two constants, basically,
49:49.760 --> 49:56.760
to understand, to solve a second order differential
equation. So, basically, essentially, knowing
49:58.630 --> 50:01.040
this, let us summarize what we learnt today.
50:01.040 --> 50:08.040
So, we learnt the first and second order differential
equation. the second part of this differential
50:09.610 --> 50:15.190
equation is, what we discussed today. And,
we said that, how the relation between energy
50:15.190 --> 50:22.190
and force can be expressed through a differential
equation; and, how Newton’s equation is
50:22.590 --> 50:27.780
essentially, a second order differential equation.
So, we also learnt a bit about concentration,
50:27.780 --> 50:33.780
but in the, in the coming lectures, we will
discuss more about concentration and understand.
50:33.780 --> 50:40.780
We will go and understand diffusion; but to
understand diffusion, we have to discuss,
50:40.980 --> 50:47.980
how does the concentration vary in a concentration
gradient, for which we have to use ideas from
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vectors. So, we will discuss that, and discuss
this in one of the future lectures. So, essentially,
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we will go slowly towards diffusion, because
diffusion is an interesting phenomenon in
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Biology, which we want to understand using
Mathematics. And for that, whatever tools
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we needed, we will develop as we go along
and we will see those differential equations.
51:12.860 --> 51:17.210
In the coming classes, we will also solve
many other types of differential equations
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and learn a few more different techniques
to solve differential equations. So, with
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this, I will stop today’s class. Thank you.