MATH SOLVE

2 months ago

Q:
# (1 point) a tank contains 1060 l of pure water. a solution that contains 0.06 kg of sugar per liter enters the tank at the rate 9 l/min. the solution is mixed and drains from the tank at the same rate. (a) how much sugar is in the tank at the beginning? y(0)= 0kg (include units) (b) with s representing the amount of sugar (in kg) at time t (in minutes) write a differential equation which models this situation. s′=f(t,s)= 0.54-(9s/1060) . note: make sure you use a capital s, (and don't use s(t), it confuses the computer). don't enter units for this function. (c) find the amount of sugar (in kg) after t minutes.

Accepted Solution

A:

(a) There is 0 kg of sugar in the tank at the beginning since it contains pure water at the start. The sugar only comes from the solution.

(b)

[tex]S' = f(t,S) = \left(0.06 \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) - \left(\dfrac{S}{1060} \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) \ \Rightarrow \\ \\ S' = 0.54 \text{ kg}/\text{min} - \dfrac{9S}{1060} [/tex]

So yes, you enter S' = 0.54 - (9S/1060)

(c)

[tex]\displaystyle\frac{dS}{dt} = 0.54 - \frac{9S}{1060} \ \Rightarrow\ \frac{dS}{dt} = \frac{572.4 - 9S}{1060}\ \Rightarrow\ \dfrac{dS}{572.4 - 9S} = \frac{1}{1060} dt\ \Rightarrow \\ \\ \int \dfrac{dS}{572.4 - 9S} = \int \frac{1}{1060} dt\ \Rightarrow\textstyle\ -\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t + C \\ \\ S(0) = 0 \ \Rightarrow\ -\frac{1}{9}\ln|572.4 - 0| = \frac{1}{1060}(0) + C\ \Rightarrow\ C = -\frac{1}{9} \ln 572.4[/tex]

[tex]-\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t -\frac{1}{9} \ln 572.4\ \Rightarrow \\ \\ \ln|572.4 - 9S| = \ln 572.4 - \frac{9}{1060}t \ \Rightarrow \\ \\ |572.4 - 9S| = e^{\ln 572.4 - 9t/1060}\ \Rightarrow \\ \\ 572.4 - 9S= \pm 572.4 e^{-9t/1060}\ \Rightarrow \\ \\ S = \frac{-1}{9}\left(-572.4 \pm 572.4 e^{-9t/1060}\right)[/tex]

But only (+) satisfies [tex]S(0) = 0[/tex]

[tex]S= -\frac{1}{9}\left(-572.4 + 572.4 e^{-9t/1060}\right) \\ \\ S= 63.6 - 63.6 e^{-9t/1060}\text{ kg}[/tex]

Enter in S = 63.6 - 63.6 * e^(-9t/1060)

(b)

[tex]S' = f(t,S) = \left(0.06 \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) - \left(\dfrac{S}{1060} \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) \ \Rightarrow \\ \\ S' = 0.54 \text{ kg}/\text{min} - \dfrac{9S}{1060} [/tex]

So yes, you enter S' = 0.54 - (9S/1060)

(c)

[tex]\displaystyle\frac{dS}{dt} = 0.54 - \frac{9S}{1060} \ \Rightarrow\ \frac{dS}{dt} = \frac{572.4 - 9S}{1060}\ \Rightarrow\ \dfrac{dS}{572.4 - 9S} = \frac{1}{1060} dt\ \Rightarrow \\ \\ \int \dfrac{dS}{572.4 - 9S} = \int \frac{1}{1060} dt\ \Rightarrow\textstyle\ -\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t + C \\ \\ S(0) = 0 \ \Rightarrow\ -\frac{1}{9}\ln|572.4 - 0| = \frac{1}{1060}(0) + C\ \Rightarrow\ C = -\frac{1}{9} \ln 572.4[/tex]

[tex]-\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t -\frac{1}{9} \ln 572.4\ \Rightarrow \\ \\ \ln|572.4 - 9S| = \ln 572.4 - \frac{9}{1060}t \ \Rightarrow \\ \\ |572.4 - 9S| = e^{\ln 572.4 - 9t/1060}\ \Rightarrow \\ \\ 572.4 - 9S= \pm 572.4 e^{-9t/1060}\ \Rightarrow \\ \\ S = \frac{-1}{9}\left(-572.4 \pm 572.4 e^{-9t/1060}\right)[/tex]

But only (+) satisfies [tex]S(0) = 0[/tex]

[tex]S= -\frac{1}{9}\left(-572.4 + 572.4 e^{-9t/1060}\right) \\ \\ S= 63.6 - 63.6 e^{-9t/1060}\text{ kg}[/tex]

Enter in S = 63.6 - 63.6 * e^(-9t/1060)