A news item is broadcast by mass media to a potential...

A news item is broadcast by mass media to a potential audience of 50,000 people. After $t$ days, $$ f(t)=50,000\left(1-e^{-0.3 t}\right) $$ people will have heard the news. The graph of this function is shown in Fig. 8. (a) How many people will have heard the news after 10 days? (b) At what rate is the news spreading initially? (c) When will 22,500 people have heard the news? (GRAPHS CANNOT COPY) (d) Approximately when will the news spread at the rate of 2500 people per day? (e) Use equations (3) and (4) to determine the differential equation satisfied by $f(t)$ (f) At what rate will the news spread when half the potential audience has heard the news?

A news item is broadcast by mass media to a potential audience of 50,000 people. After $t$ days,
$$
f(t)=50,000\left(1-e^{-0.3 t}\right)
$$
people will have heard the news. The graph of this function is shown in Fig. 8.
(a) How many people will have heard the news after 10 days?
(b) At what rate is the news spreading initially?
(c) When will 22,500 people have heard the news?
(GRAPHS CANNOT COPY)
(d) Approximately when will the news spread at the rate of 2500 people per day?
(e) Use equations (3) and (4) to determine the differential equation satisfied by $f(t)$
(f) At what rate will the news spread when half the potential audience has heard the news?

Key Concepts

Solving Exponential Equations with Logarithms

When an exponential model is equated to a particular value, solving for time typically involves isolating the exponential term and applying logarithmic operations. This method is essential for determining the time at which the process reaches a prescribed level, a common requirement in problems involving growth, decay, or spread phenomena.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In the context of exponential models, they provide a compact representation of the dynamic behavior of the system. For instance, the exponential saturation model naturally leads to a first?order linear differential equation that describes how the rate of change is proportional to the difference between the current state and the maximum achievable state.

Exponential Saturation Model

This concept involves functions of the form f(t) = L(1 - e^(-kt)), where L is the limiting value (or carrying capacity) and k is a positive rate constant. It models processes that approach a maximum value over time, such as the spread of information, population growth in a confined environment, or the accumulation of a substance in a system.

Instantaneous Rate of Change

In any continuously changing process, the instantaneous rate of change refers to the derivative of the function with respect to time. It indicates how fast the quantity is changing at any given moment. This concept is central to calculus and is used to determine speeds, growth rates, and other dynamics in various applications including modeling spread or diffusion.

Transcript

00:01 For this problem, if we consider a function, we have to consider a function that represents the number of people who heard the news.

00:13 So f of t, f of t is equal to a 50 ,000 times 1 minus e to the negative 0 .3c.

00:45 So start off, let's t represent the number of days.

00:50 So we know the total audience is $50 ,000.

00:55 So the objective is to determine the number of people who hear the news after 10 days.

01:04 So a function, f of t will be the number of people who heard the news after t8, which is this function fft.

01:19 So now we have to determine the total number of people who heard the news after 10 days.

01:25 So that becomes f of 10.

01:28 So plugging sent into original functions, we have 47 ,510 people.

01:49 That's the amount of people that heard the news after 10 days.

01:54 So now we could graph this function as all.

02:06 So for part b, the objective is to determine the rate at which the news spread initially.

02:16 So the rate of spreading the news will be given by df over dt to df over dt.

02:37 We have 50 ,000 times a negative e to the negative 0 .3t times negative 0 .3.

03:07 Right.

03:08 So now if you have, if you have, if you take the derivative, of dfdt, so f prime of t, f prime of t, if you take the derivative, it gives you 15 ,000 times e to the negative 0 .3 t.

03:41 So at the initial stage, when t is equal to zero, we could put f of zero into our derivative function.

03:57 F of zero gives you 15 ,000.

03:59 So the rate at which the news initially spread reaches about 15 ,000 people per day...

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