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# Let $P(t)$ be the performance level of someone learning a skill as a function of the training time $t.$ The graph of $P$ is called a learning curve. In Exercise 9.1.15 we proposed the differential equation $\frac {dP}{dt} = k[M - P(t)]$as a reasonable model for learning, where $k$ is a positive constant. Solve it as a linear differential equation and use your solution to graph the learning curve.

## $$0 \leq P(0) \leq M, \text { so }-M \leq C \leq 0$$

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Differential Equations

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### Video Transcript

Okay, so we're given a question where p off t represent performance level or if you plot this gives a performance curve for learning. Any new skill on the differential equation that is proposed is given us DP by DT is k times a minus pt Here. We need to remember that case. Positive numbers. So case greater than equals toe. Well, it's not equals. It's greater than zero. Okay, so since it's a linear equation will change it toe a standard forms of DP by D t is request o k m minus K. P. Okay. Yeah, K and I'm both are constant value. So we'll write this as dp by d t minus capable. Take it here. So it becomes plus KP is it calls toe. Okay. Okay. So on comparing this with d y by D. X plus p Wyche was toa que we get P s k. Remember this p and this piece not Sims will name it s p one. So p one is K and Q is came. Okay, So what we get here, so are integrating factor becomes a raise to integration off KDD and since case constant so it will become it is too, Katie. Okay, so r p solution will be p times it is two k t equals to integration. Off Q times just came. Integration off it is to Katie. DT. Thank you. Okay, so you're came is constant, so it will come out. An integration of it is to Katie's. It is to Katie. Okay. Plus C can cancel this k So we have our solution now. We don't know. It s to Katie here, so we'll divide by it is to Katie, so we'll have p equals to it is to Katie. Divided by it is to Katie's one. So we left with m plus c divided by it is to Katie, will go in numerator and become Erez to minus Katie by in this room. So we have our value off p. But we need value off. See? Okay. So to calculate value off see, at T equals to zero value off p will be between zero. And m, therefore, P can be written is value between zero and m. She's equals toe m plus C times. It is toe Kate, times zito, and this gives us zero. Tow em is equals toe m plus C since this value is equals to one. Okay, so M plus C has reigned. Zero toe m, Therefore see will have ranged between minus m two zero. Okay, so this is a solution now to plot it, Let's assume is you m as one and for different and see will be between then minus 1 to 0. So we'll see what the values off see gives what kind off learning curve on the positive value K. Let's take you this Do. Okay, so now our function becomes peas one plus if c is minus one. So let's say minus one Here is two minus two t and when we plot it So this is T and this is a learning co. So for P equals to one minus. It is two minus two t. We have a graph which passes through video, and it then saturates over time. Okay. Similarly, if you go what a lower values off. See, let's say four Sequels to minus 40.8. What we get is are learning still saturates the same value, but it starts with lesser known. Okay, so see is what sees representing. She's representing how much information but how much learning person already has already? Yes. Okay, So if you change, See, it just shifted in the graphs because it representing how much learning a particular person already has. And based on that is getting. And since we're selecting K as to the cause, comes and meet it too. Okay, so whatever value you take, it well can watch at two. And after that, they all have same kind off learning. Okay, so that's autograph. Thank you.

Mumbai university

#### Topics

Differential Equations

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