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Mariana [72]
4 years ago
7

Suppose that news spreads through a city of fixed size of 900000 people at a time rate proportional to the number of people who

have not heard the news.a. Formulate a differential equation and initial condition for y(t) , the number of people who have heard the news t days after it has happened.No one has heard the news at first, so y(0)=0.dy/dt = k ( what ) , where k is the proportionality constantb.) 8 days after a scandal in City Hall was reported, a poll showed that 100000 people have heard the news. Using this information and the differential equation, solve for the number of people who have heard the news after t days.y(t) = what
Mathematics
1 answer:
djverab [1.8K]4 years ago
4 0

Answer:

The relation is ln(\frac{900000}{900000 - y(t)} ) = kt

Step-by-step explanation:

Here we have

\frac{dy(t)}{dt} =k(900000-y(t))

Integrating we get

-ln(900000 - y(t)) = kt + C

With initial condition y(0) = 0 we get

C =  -ln(900000)

Therefore,

ln(\frac{900000}{900000 - y(t)} ) = kt

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The perimeter of a rectangle is 46in and the diagonal is 17 in
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p=2*(L+l)=46

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Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
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Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

4 0
3 years ago
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