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Problem 84 Hard Difficulty

Under certain circumstances a rumor spreads according to the equation
$$
p(t)=\frac{1}{1+a e^{-k t}}
$$
where $p(t)$ is the proportion of the population that has heard the rumor at time $t$ and $a$ and $k$ are positive constants. [In Section 9.4 we will see that this is a reasonable equation for $p(t) .]$
(a) Find $\lim _{t \rightarrow \infty} p(t)$
(b) Find the rate of spread of the rumor.
(c) Graph $p$ for the case $a=10, k=0.5$ with $t$ measured in hours. Use the graph to estimate how long it will take for $80 \%$ of the population to hear the rumor.

Answer

a. $\lim _{t \rightarrow \infty} p(t)=1$
b. $\frac{a k \cdot e^{-k t}}{\left(1+a e^{-k t}\right)^{2}}$
c. $p(t)=0.8$ when $t=7.38$ hours

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CA

Catherine A.

October 27, 2020

Finally, the answer I needed, thanks Heather Z.

SA

Sharieleen A.

October 27, 2020

This will help a lot with my midterm

Video Transcript

in this problem, we have this function which represents the proportion of the population that has heard a rumor as a function of time and what we want to do it with it first is find the limit as T approaches infinity. So as time is approaching infinity, what proportion of the population has heard the rumor? So let's imagine that we have infinity plugged in for tea. Then we have e raised to the negative K times infinity, and that would be e to the negative infinity. And that would be a number that is incredibly small. That's like one divided by E to the infinity, incredibly close to zero. So eight times that number is incredibly close to zero. So one plus something incredibly close to zero is very, very close to one one divided by one is very, very close to one. So we could say the limit is one. And what that means is that 100% of the population has heard the rumor when time is approaching infinity and that makes sense. Okay, in part B, we want to find the rate of spread of the rumor and the rate of spread would be the rate of change, and that would be the derivative. So what we're going to do is take the simplified version of the function where I wrote it here Instead of writing it as a quotient, I wrote it to the negative first power. And so we'll use the chain rule on that. So we bring down the negative one, and we raised the inside to the negative second power. And then we multiply by the derivative of the inside the derivative of a E to the negative K t would be a e to the negative Katie multiplied by negative K, the derivative of the exponents. Now we can simplify that derivative by changing it back to a quotient. Okay, now that we have that, what we want to dio is graph the original function, plugging in some specific values. So the value for a is 10 and the value for Caisse 100.5. And when we substitute them in, we get this function here. So we're going to put that into a calculator, and then what we want to do with it is use it to figure out how long it will take for 80% of the population to hear the rumor. So 80% would be represented by a pop by a proportion of 0.8. Okay, so let's grab a calculator and then we go toe y equals. And here we have our function. And for a window, I have chosen to try 0 to 20 on my y axis. I'll just go by five as a scale are sorry that 0 to 20 on the X axis and then negative 0.5 to 1.1 on my Y axis. So let's take a look at that graph. So that makes sense that as we get closer to the full population, reaching, uh, the point of having heard the rumor, it's just going to taper off in B flat. So the way we can find the time when it reaches 0.8 is to go back to y equals and go toe y two and put a 20.8 in there for why, to That gives us a horizontal line at a height of 0.8, and then we can find the point of intersection. So we go into the calculate menu, which is second trace. We go to number five Intersect. We put the cursor on the first curve. Press enter. Put the cursor on this. I concur. Press enter and put the cursor near the intersection. Point press, enter. And here we see at the bottom of the screen. The units would be hours in this case, so we have 7.4 hours roughly port to reach an 80%. Ah, rumor saturation point.