Question

Pollution begins to enter a lake at time t =0 at a rate​ (in gallons per​ hour) given by the formula f(t)​, where t is the time​ (in hours). At the same​ time, a pollution filter begins to remove the pollution at a rate g (t) as long as the pollution remains in the lake.

Required:
a. How much pollution is in the lake after 12 hours?
b.Use a graphing calculator to find the time when the rate that pollution enters the lake equals the rate the pollution is removed.
c. Find the amount of pollution in the lake at the time found in part b.
d.Use a graphing calculator to find the time when all the pollution has been removed from the lake.

Answer 1

Answer:

a) 71.25 gallons

b) 25 hrs

c) 105 gallons

d) 47.91 hrs

Step-by-step explanation:

Solution:-

- The rate at which the pollution enter the lake is expressed by a function f ( t ) as follows:

                            $f ( t ) = 10*( 1 - e^-^0^.^5^t )$

- Where, the rate at which pollution is removed from the lake via a filter is expressed by a function g ( t ):

                             $g ( t ) = 0.4t$

- We will denote ( x ) as the amount of population in lake at time ( t ).

- We will set up a linear first order ODE using rate of change of pollution ( x ) in the lake at any instance ( t ):

                            $\frac{dx}{dt} = ( flow _ i_n ) - ( flow _ o_u_t )$

- The respective inflow and outflow of pollution from the lake are expressed by the functions f ( t ) and g ( t ). Hence, the ODE becomes:

                           $\frac{dx}{dt} = f ( t ) - g ( t )\\\\\frac{dx}{dt} = 10*( 1 - e^-^0^.^5^t ) - 0.4t$

- Separate the variables and integrate from t = 0 to t = 12 hours.

                         $x = \int\limits^1^2_0 [{10*( 1 - e^-^0^.^5^t ) - 0.4t} ] \, dt \\\\x ( 12 ) = [ 10 * ( t + 2e^-^0^.^5^t ) - 0.2t^2 ] \limits^1^2_0\\\\x ( 12 ) = [ 10 * ( 12 + 2e^-^6 ) - 0.2(12)^2 ] - [ 10* ( 0 + 1 ) - 0.2(0)^2 ]\\\\x ( 12 ) = 71.25 galls$

- We will use the graphing calculator to evaluate the time ( t ) at which the rate of inflow of pollution is equal to the rate at which pollution is removed from the lake. In other words solve the following equation:

                          $f ( t ) = g ( t )\\\\10*( 1 - e^-^0^.^5^t ) = 0.4t$

- Solving for the above equation the following intersection point was observed:

                          t = 25 hours

- We will perform integration of the previously expressed ODE from t = 0 to t = 25 hrs.

                           $x ( 25 ) = [ 10*(t - 2e^0^.^5^t ) - 0.2t^2 ]\limits^2^5 _ 0\\\\x ( 25 ) = [ 10*(25 - 2e^1^2^.^5 ) - 0.2(25)^2 ] - [ 20 ] \\\\x ( 25 ) = 105 galls$

- We will evaluated result of the integration from time t = 0 to time t = t set the amount of pollution ( x ) equal to zero:

                          $x ( t ) = [ 10*( t - 2e^-^0^.^5^t ) - 0.2t^2 ]\limits^t_0\\\\x ( t ) = 10 + 20e^-^0^.^5^t - 0.2t^2 - 20$

- Use the derived result x ( t ) and feed it into the graphing calculator and solve for x ( t ) = 0; hence,

                       10 + 20e^( -0.5*t ) - 0.2t^2 - 20 = 0

- The answer is evaluated as t = 47.91 hr