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University of North Texas



Problem 15 Medium Difficulty

Psychologists interested in learning theory study learning curves. A learning curve is the graph of a function $ P(t) $, the performance of someone learning a skill as a function of the training time $ t. $ The derivative $ dP/dt $ represents the rate at which performance improves.

(a) When do you think $ P $ increases most rapidly? What happens to $ dP/dt $ as $ t $ increase? Explain.

(b) If $ M $ is the maximum level of performance of which the learner is capable, explain why the differential equation

$ \frac {dP}{dt} = k(M - P) $ $ K $ a positive constant


(a) $P$ increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with learning a skill. As $t$ increases, we would expect $d P / d t$ to remain positive, but decrease. This is because as time progresses, the only points left to learn are the more difficult ones.
(b) $\frac{d P}{d t}=k(M-P)$ is always positive, so the level of performance $P$ is increasing.
As $P$ gets close to $M, \frac{d P}{d t}$ gets close to $0 ;$ that is, the performance levels off, as explained in part (a)
(c) See graph


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Video Transcript

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