WEBVTT
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So P is giving the performance of someone learning,
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eh? Topic. So it would make sense that
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at the very start you pick up a lot of
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the material really quickly. So when so in terms
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of time, we can think about this as he
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is increasing most rapidly when t is small. So
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one of our feast together and as the change of
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the performance over time Well, since it's increasing most
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rapidly at beginning, we would expect that it to
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slow down at the end So we can say that
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DP Bye. Auntie should go to zero. As
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t goes on, so goes to infinity. And
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now if we're told that him is some maximal level
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that anyone should be able to perform at if we
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want to see if this differential equation actually does give
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a good representation for the learning model, you might
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want to first start by just solving for what he
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is. So let's go ahead and do that.
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So the first thing I'm going to do is you
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might look at this and knows that this is a
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separable equation. So I'm going to go ahead and
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try to get all my peas on the same side
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as my derivative here since I want to group all
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the same terms together. So I don't see any
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tease anywhere. But I do see these p right
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here. So I'm going to go ahead and divide
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by n minus p. And when I do that
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, I end up with one over and minus p
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d p over dt and they'LL just be equal to
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K. Okay, now the next thing I want
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to do is move my DT over. And when
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I do that, I end up with one or
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i'll move the dp appear so d he only arrest
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us, James. So BP over and minus P
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is equal to K Dean team. So what?
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I'm integrating this M Since it's just some maximum level
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, this will be a constant as well as K
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. We're told it's just a constant. So if
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I integrate each side of this So the right hand
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side we'll just be kay t plus some constant seem
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one. And if you look at this here,
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you may knows that it looks very close to a
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natural logs derivative because I have he appear, then
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I have p e in the denominator. So I'm
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going to write natural log of and minus he.
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But if you were to take the driven about this
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, you would end up with it being negative.
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So also need to throw on a negative here.
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And you can see that if you just do a
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u substitution in for this instead, if you let
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you equal in minus p, you can also get
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this here, right? So I have this now
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I want to solve for what P is. So
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the first thing I'm gonna do is move the negative
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over I'LL get natural Aga and minus P is equal
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to negative Okay T minus C one and I'LL go
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ahead and move this up here now What I need
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to do is exponentially ate each side So put each
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side too, eh? Base of so the left
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hand side will counsel Oh, I'll get it and
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minus p is equal to he to the mega knew
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k t minus C one And then since I'm solving
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for P, I would add pee over this and
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I would end up with P is equal to m
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minus into the negative Kay t minus scene one All
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right, So now let's maybe rewrite just a little
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bit so you can look a little bit prettier for
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us. So remember that when I have expert exponents
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being added or subtracted, it's the same thing as
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the base of those experiments being multiplied. So I
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could write this as negative e to the negative C
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one times e to the negative. Kay t i'LL
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just go ahead and call this e to the negative
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C one since it also is just a constant I'LL
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call it, See two So that is me m
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minus c too. And then recall that when I
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have a negative experiment I could just write this as
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one over whatever I have here. So I get
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I mean to the k t. So if you're
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looking at this, you may notice that as time
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goes on that this portion here will go to zero
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. So as time goes on, so t goes
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to infinity. If I take the limit, em
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, since it's a constant, would still just go
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to him. But this here, so see to
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was a constant and e k to the tea will
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get very large. So I've see two over some
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very large number. So this here would go to
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zero. So it first makes sense that this could
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be a model because we would expect that as time
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were to go on forever, we should reach our
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maximal level of performance. And the other thing that
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I should be worried about What I'm looking at office's
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This value here will always be negative. So,
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hee, this expression right here never actually outputs anything
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bigger than one. So it would make sense to
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say that well, before I start, I should
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always be below this right. And if I were
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to sketch a graph for this, what this is
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really telling me is that this is a horizontal assam
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toe because I'm looking at the behaviour. Has disfunction
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tends towards infinity? So at am I have a
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horizontal ask him to, and I can just go
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ahead and choose any value for sea to here since
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it just once a possible solution, which just means
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I can't choose c too to be whatever I want
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. So I'll just go ahead and make this so
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that, uh, C two is equal to one
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. And when I do that, I end up
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with and minus one. And I'm just going to
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do that for convenience. And then you could see
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that this will tend towards it like that. So
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this is one possible solution. If you were to
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just picks a random number but it would make a
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lot more sense for us to actually start at zero
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. So if C to is actually em So maybe
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I should write that this curve here is see to
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illegal to one. If I have see, two
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is equal to em. Instead, I'll get to
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start at zero and then ten towards my horizontal acid
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Toby So si tu is equal time.