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jasenka [17]
3 years ago
9

In a certain region, 6% of a city’s population moves to the surrounding suburbs each year, and 4% of the suburban population mov

es into the city. A 2019 census measures 10,000,000 residents in the city and 800,000 in the suburbs. (a) Set up a difference equation (of the form ~xn+1 = A~xn) that describes this situation, where ~x0 is the initial population in 2019. (b) Estimate the populations in the city and suburbs two years later, in 2021.
Mathematics
1 answer:
Charra [1.4K]3 years ago
3 0

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Following are the differential equation:

\to P(C,S)t+1 = P(C,S)t+(- 0.06Ct +0.04St, 0.06Ct - 0.04St ) \ \ \ \\\\or \ \ \ \ \ \ \ \\\\ \to (0.94Ct + 0.04St, 0.96St + 0.06Ct)

In equation:

\to t = 0, \ as \ 2019\\\\\to Ct = 10,000,000  \\\\\to St = 800,000,000\\

\to P(C, S)1 = (0.94Ct + 0.04St, 0.96St + 0.06Ct) \\\\

                   = (0.94(10000000) + 0.04(800000), 0.96(800000) + 0.06(10000000))\\\\ = (9,400,000 + 32,000), (768,000 +6000000))\\\\ = (9432000, 1368000)

\to P(C, S)2 = (0.94(9432000) + 0.04(1368000), 0.96(1368000) + 0.06(9432000))

                  =(8,866,080 + 54,720), (1,313,280 + 565,920)\\\\=(8,920,800, 1,879,200)

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Harrizon [31]

Answer:

  • <u>0.075 m/min</u>

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

<u>1. Write the equation for the volume of the cone:</u>

      V=\dfrac{1}{3}\pi r^2h

<u />

<u>2. Find the relation between the radius and the height:</u>

  • r = diameter/2 = 5m/2 = 2.5m
  • h = 5.2m
  • h/r =5.2 / 2.5 = 2.08

<u>3. Filling the tank:</u>

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

  • x/y=r/h
  • y = x . h / r

The volume of water inside the cone is:

        V=\dfrac{1}{3}\pi x^2y

        V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3

<u>4. Find the derivative of the volume of water with respect to time:</u>

            \dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}

<u>5. Find x² when the volume of water is 8π m³:</u>

       V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\  11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062m²

<u>6. Solve for dx/dt:</u>

      1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}

      \dfrac{dx}{dt}=0.03596m/min

<u />

<u>7. Find dh/dt:</u>

From y/x = h/r = 2.08:

        y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min

That is the rate at which the water level is rising when there is 8π m³ of water.

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