Respuesta :
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let c (t ) denote the amount of chlorine (in grams) in the tank at time t .
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(c (t )/(1000 + (10 - 25)t ) g/L) * (25 L/s) = 5c (t ) /(200 - 3t ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.
So the amount of chlorine in the tank changes according to
[tex]\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}[/tex]
which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):
[tex]\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0[/tex]
[tex]\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0[/tex]
[tex]\dfrac{c(t)}{(200-3t)^{5/3}}=C[/tex]
[tex]c(t)=C(200-3t)^{5/3}[/tex]
There are 20 g of chlorine at the start, so c (0) = 20. Use this to solve for C :
[tex]20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}[/tex]
[tex]\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}[/tex]
The amount of chlorine in the tank is a function of time is given as;
[tex]\rm c(t) = \dfrac{1}{200} \sqrt[3]{\dfrac{(200 - 3t)^5}{5}}[/tex]
What is differentiation?
The rate of change of a function with respect to the variable is called differentiation. It can be increasing or decreasing.
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter.
In order to reduce the concentration of chlorine, freshwater is pumped into the tank at a rate of 10 L/s.
The mixture is kept stirred and is pumped out at a rate of 25 L/s.
The pure water is pumped into the tank, so no chlorine is flowing into it, but flowing out at a rate of
[tex]\dfrac{c(t)}{1000+ (10-25)t} * 25 = \dfrac{5c(t)}{200-3t}[/tex]
The amount of liquid in the tank at the start is 1000 L. IF water is pumped in it at a rate of 10 L/s, then after t second, there will be (1000+10t) L of liquid in the tank.
But we are also removing 15 L from the tank per second, so there is a net gain of -15 L of liquid each second. So the volume of the liquid in the tank at time t is (1000 - 15t) L. Then the concentration of the chlorine per unit volume is c(t) divided by this volume.
[tex]\rm \dfrac{dc(t)}{dt} = - \dfrac{5c(t)}{200 - 3t}\\\\\\\dfrac{dc(t)}{dt} + \dfrac{5c(t)}{200 - 3t} = 0[/tex]
Integrate the equation, we have
[tex]\begin{aligned} \dfrac{1}{(200-3t)^{}5/3} * \dfrac{dc(t)}{(200-3t)^{8/3}} + \dfrac{5c(t)}{(200-3t )^{8/3}} &=0\\\\\\\dfrac{d}{dt} \begin{bmatrix} \dfrac{c(t)}{(200 - 3t)^{5/3}} \end{bmatrix} &= 0\\\\\\\dfrac{c(t)}{(200 - 3t)^{5/3}} &= C\\\\\\c(t) &= C * (200 - 3t)^{5/3} \end{aligned}[/tex]
There are 20g of chlorine at the start, so c(0) = 20, then
[tex]20 = C(200)^{5/3} \rightarrow C = \dfrac{1}{200*5^{1/3}}\\\\\\\rightarrow c(t) = \dfrac{1}{200} \sqrt[3]{\dfrac{(200 - 3t)^5}{5}}[/tex]
More about the differentiation link is given below.
https://brainly.com/question/24062595
