Question

A news item is spreading by word of mouth through a population of size 12,000 people. After t days, the number of people (in thousands) who have heard the news is given by the following equation. y = f(t) = 12/1 + 75e^-0.48 t (a) Approximately how many thousand people have heard the news after 6 days? (Give your answer correct to at least three decimal places.) (b) Calculate the derivative f (t). f(t) = (c) At what rate is the news spreading after 8 days? (Give your answer correct to at least four decimal places.) d) Compare the function f(t) to the solutions to the different forms of the logistic growth model from this lesson. Enter the differential equation in terms of y that corresponds to this solution. dy/dt = (e) How many people have heard the news when its rate of spread is 1.35 thousand people per day? (There are two answers. Give your answers correct to the nearest whole person.) people (smaller value) people (larger value) f) At what two times is the news spreading at rate 1.35 thousand people per day? (Give your answers correct to at least three decimal places days (smaller value) days (larger value) (g) What is the fastest rate at which the news spreads? (Give your answer correct to at least three decimal places.) thousand people per day

Answer 1

Answer:

a) 2.303

b) $f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

c) 1.3609

e) $\frac{dy}{dx}=0.04y(12-y)$

f)  4.5 thousand people(smaller)

7.5 thousand people (larger)

e) t=7.93 days (smaller)

t=10.06 days (larger)

g) 1.44

Step-by-step explanation:

The given function is $y=f(t)=\frac{12}{1+75e^{-0.48t}}$

To find how many thousand people have heard the news after 6 days, we substitute to get:

$f(6)=\frac{12}{1+75e^{-0.48*6}}=2.303$  thousands.

b) We rewrite to get: $f(t)=12(1+75e^{-0.48t})^{-1}$

We differentiate using the chain rule to obtain:

$f'(t)=-12(1+75e^{-0.48t})^{-2}\cdot 75e^{-0.48t}\cdot -0.48$

This simplifies to:

$f'(t)=432e^{-0.48t}(1+75e^{-0.48t})^{-2}$

We rewrite as positive index to get:

$f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

c) To find the rate at which the news is spreading after 8 days, we substitute t=8 into $f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$ to get:

$f'(8)=\frac{432e^{-0.48\cdot 8}}{(1+75e^{-0.48\cdot 8})^{2}}=1.3609$

d) The solution to the logistic differential equation:

$\frac{dy}{dt}=ky(M-y)$

is

$y=\frac{M}{1+be^{-Mkt}}$

By comparison, M=12, and $Mk=0.48$

$12k=0.48\\k=0.04$

The differential equation is:

$\frac{dy}{dx}=0.04y(12-y)$

e) To how many people have heard the news when its rate of spread is 1.35 thousand per day, we equate the differential equation to 1.35 and so for t first.

$\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}=1.35$

This gives us: t=10.06 ot t=7.93

We substitute the times into the function to get:

$f(7.93)=\frac{12}{1+75e^{-0.48\cdot7.93}}$=4.5 thousand: smaller value

$f(10.06)=\frac{12}{1+75e^{-0.48\cdot10.06}}=7.5$ thousand: Larger value

f) The two times that the news is spreading at rate 1.35 thousand people per day are:

$t=7.93$ days: Smaller value

$t=10.06$: Larger value

g) To find the fastest rate at which the news spread we take the second derivative and equate it to zero.

$f''(t)=0$

This corresponds to where the horizontal line is tangent to

$f'(t)=\frac{432e^{-0.48t}}{(1+75e^{-0.48t})^{2}}$

From the graph the point of tangency is:

(8.99,1.44)

Therefore the fastest rate at which the news spread is 1.44