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A model for the spread of a rumor is given by the equation

$$ p(t) = \dfrac{1}{1 + ae^{-kt}} $$

where $ p(t) $ is the proportion of the population that knows the rumor at the time $ t $ and $ a $ and $ k $ are positive constants.

(a) When will half the population have heard the rumor?

(b) When is the rate of spread of the rumor greatest?

(c) Sketch the graph of $ p $.

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Missouri State University

Oregon State University

Harvey Mudd College

Idaho State University

were given that a model p A T is equal to one over one plus a e to the negative. Katie will tell us the proportion of the population that will no other rumor at a Thai tea and where A and K are positive. Constance, what we want to do is first find out when half the population will have heard of this rumor. And then we also want to determined win is the spread of the rumor at its greatest. And then after that, we want to use that to sketch our function. So this first part here this means we want it to be one have, since pft will tell us the proportion that will know so we could go ahead and write. 1/2 is equal to P A. T, which is one plus a e to the negative k t all over one. Then we can go ahead and start trying to salt pretty so we can reciprocate each side of this and doing that will give us two is equal to one plus a e to the negative k t. Then we can go ahead and subtract one and then divide by a which is going to give us that one over A is equal to e to the negative, Katie. Then we can take the natural log on each side to cancel out our exponential E and then divide by negative K so we'll end up with T is equal to the natural log of one over a over negative. Okay, Now something we might recall about natural logs is if we have an exponents, we could go ahead and move that out front, and we just so happen to have that. Well, one of her A is really a to the negative first power so we could move that exponents out front. So we get negative natural log of a all over negative k Will. These negatives here will comes up and we'll get that The natural log of a over kay will be when the population, or half of the population will. Now, Now this brings up a point of if a is between zero and one, then we get natural. Gabe's going or yeah, that's a lot of it is going to be negative and then says K is positive we'll have a negative number. So what extra restriction outside of ages. Being positive that we're going to need to add is that a is greater than one, because otherwise this first question doesn't really make any sits for us. So this will actually play a role in Part B as well. So it's good that we come to this conclusion now, as opposed to later now to find when the spread is greatest for a rumor. Well, if we were to just take the derivative of P well, this year tells us the rate. And then if we want to find when the rate is at its largest, that means we need to look for the critical values of the second derivative or the derivative of our derivative. So this is the derivative that we're going to want to find to be able to find our critical bounties once we set this equal to zero. So this is what we want to determine. So let's go ahead and first take the derivative of P. And then we can also take the derivative of P A T. So let's go ahead and do this on a new page. So we had that p of tea was equal to one plus a e to the negative, Katie all over one. So remember, this can actually be rewritten as one plus a e to the negative. Katie. All to the negative First power. So to take this derivative, we're going to go ahead and use general as well as powerful. So the first derivative it's gonna be be of tea. So we're going to use power ruled negative 11 plus a e to the negative K t. Now, this is gonna be to the negative second power since we subtract one off, and then we need to take the derivative of what was on the inside here. So D by DT of one plus a e negative k t. And now let's go ahead and start regretting all this. So this should be negative. One over one, plus a e to the negative, Katie and the derivative of only having green. There should be negative a k e to the negative k t. All right, so let's go ahead and rewrite this a little bit more. So first these negatives counts out. And then I almost dropped my square right here. Sense. Remember? This was negative. Too far power. And now something that might help us is if we've rewrite this as won over E to the Katie, this part right here and expand out arv denominator, I don't just make taking the next derivative a little bit easier. So let's do that. So we're going to have a K all over e to the Katie. And now when we do one plus a e to the negative Katie squared. Well, that's gonna be one plus two a to the negative Katie, plus a squared E to the negative two k. And this is just by the perfect square formula. Now let's go ahead and distribute the C to the K t. And doing that is going to get, well, e to the k t plus two a plus a squared e to the negative, Katie. And this is all going to be over a kit. And now I wanted to write it like this so we don't have to use the quotient, rule or anything else, because before, if we would have left it like this here, we would have had either used product. Rourke sociable. Now we can just use essentially what we did before chain rule as well as power rule. So let's go ahead and find he double prime empty. So again, remember, this is actually to the negative first power. So we're going to have a negative A k Yeah, to the K T plus two a plus. Or actually, just go ahead and write it like we did before because I might run out of space. Otherwise. So e to the Katie plus two way plus a squared e to the negative Katie. And this will all be squared. And then we have to take the derivative on the inside. Here, let me do that. Blooms. Remember, we're taking the derivative of this because it's general. So the derivative of E to the Katie is going to be K e. Typically 80 the derivative of to a will just be zero and then the derivative of a squared E to the negative. Katie should be negative. A squared k e to the negative k. But now to find our Eric See, before we do that, let's go ahead and clean this up a little bit because, well, we will need to do a little bit of algebra for stuff. So I noticed that we can factor out a e to the negative. Katie at a k from these. And I'm also going to distribute this negative into here. So factoring out that Kay would give us a k squared and we'd have e to the negative. Katie. So all over e to the Katie, plus two a plus a sward e to the negative, Katie. And this is all still squared. And then over here, we're going to have que because we factor that k out so e to the two k t. And then we're going to have minus a squared. But remember, we distributed that negative in. So this should be plus at let me do this in blue. So this should be plus, And now this should be minus now to find our possible maximum of our first derivative. Remember, we set the second derivative equal to zero. And so first we know that the denominator will never be able to equal zero. So we really don't care about that part. And we also know that this part here can also never be equal zero. Because each of negative Katie can ever be zero in eight and care both larger than zero by assumption. So that means the only thing in this that can actually be negative is this part here. So we really just need to set this equal to zero to solve for teeth. So we have negative e to Katie lost. A square is equal to zero, so we can add heat to the Katy over and then take natural log on each side. So that's going to give us that to Katie is eager to the natural log of a squirt. Then we can divide each side by two. K. So we get tea is into the natural log of a squared over to K. And remember, earlier we talked about how we can move that out front by one of the law rules. So we get to natural log of a over to K the Jews Council out when we get natural log of a over. Okay, so we're not quite done with this because we'll still need to use the soul. I'm gonna say first derivative test, Um, in quotes saying is we have to use it on the second derivative, all reference of our function to see if the first derivative has a maximum. So Let's go ahead and zoom out a little bit. I love it much. So now what we want to do is find a value that's going to be smaller in this and one that's larger so we can make a number line. So are critical. Value here is natural of a okay, and then this here is going to be Remember our values for P of T r P double prime of teeth. So let's go ahead and on the left plug in zero. So let's see what P double prime of zero is going to be a P double prime of zero. Well, we can just plug everything in to what we have right here and doing that. Everything should simplify down and give us the following. So a K squared all over one plus two a plus, a squared, all squared and then plug it into the other part will have negative one plus a squared. Right? So all of this here, we know, is going to be bigger than zero since a and care both bigger than zero. And we also know this part here is going to be larger than zero. Because remember that extra assumption that we made of a being bigger than one. Because otherwise remember, natural log of a K wouldn't make sense. So over here we know that to the left of this this function is going to be positive, since something larger than zero something largest zero multiplied together gives us something larger than zero. Now let's go ahead and plug in the number larger than this one. And let's just go ahead and do just natural log of a so match log of a and so plugging all. Listen, let's see what we end up getting. So it should be a K squared A to the K all over. Um, I should That should be negative, k right there and then in the denominator, it's going to give us a to the K plus two a plus. Hey, to the negative k. So should be a square A to the negative k. And then all this here is going to be squared and let me screwed all this over a little bit. And then when we put this in over here will end up with so negative a to the two K plus a squared. And so now, once again we know that this part here is just gonna be strictly larger than zero sense A and Caroll positive. And then over here que is also going to be larger than one. So then this here has to be larger than zero officer or a less than zero because we have minus a tow something larger than two plus a squared. So this should be less than zero. So two things multiply together. Positive negative will give us negative. So that means the function is increasing or the first derivative. It's increasing it to this point and then decreasing after. So that shows that natural log of a K is going to be a max. Some go ahead and zoom in a little bit. So we have t is equal to match a log of a over K. And this is by the first derivative test on the first Resident. Now the last part. They want for us to sketch the graph. So let's just go ahead and do that really quickly. So some things that we might need to know before we sketches or possibly our intercepts. So let's find out what just p of zero is going to be. So plug it in here. Zero will end up getting one over one, plus a So just do, like 1 1/2 And then somewhere around here is gonna be, like one over one, plus a depending on how large a is. Remember, since we assumed this larger than one, it should be below that point. Then we know it. 1/2. It'll be natural log of a k so we can put that point there. And now we need to figure out what the in behavior of this thing is. And if we were to go ahead and take the limit as T approaches infinity of one over one plus a e to the negative. Katie the negative k t. So the only thing that has a T is this part here, and we know that e to the negative. Katie as T approaches Infinity Ghost zero. So we'd have a time zero, which is zero plus one. This one. So this limit here is one. So we know we're going to have a horizontal ask himto at Why is it you want? So the graph of this function is going to look something kind of like this so it has to curve up to this point and then it's gonna have to fly and out so we can get closer to this. And actually, if we go and look at our, um, chart over here, remember that the second derivative will tell us that this function should be con cave up here since it's positive. And over here it should be negative or con cape down to the right of our natural aga a over K And if we look at this to the left of it definitely looks can't give up and to the right of it definitely looks con cave down.

University of North Texas