Answer:
[tex]dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex] g/min
Step-by-step explanation:
The mass flow rate dA(t)/dt = mass flowing in - mass flowing out
Since water is pumped in at a rate of 3 L/min, and it contains no salt, the concentration in is thus 0 g/L. the mass flow in is thus 0 g/L × 3 L/min = 0 g/min.
Let A(t) be the mass present at any time, t. The concentration at any time ,t is thus A(t)/volume = A(t)/210. Since water flows out at a rate of 3 L/min, the mass flow out is thus, A(t)/210 g/L × 3 L/min = A(t)/70 g/min.
So, dA(t)/dt = mass flowing in - mass flowing out
dA(t)/dt = 0 g/min - A(t)/70 g/min
dA(t)/dt = - A(t)/70 g/min
Since the tank initially contains 30 g of salt, the initial mass of salt in the tank is 30 g. So A(0) = 30
So, the initial value problem is thus
dA(t)/dt = - A(t)/70 , A(0) = 30
Separating the variables, we have
dA(t)/A(t) = -dt/70
Integrating both sides, we have
∫dA(t)/A(t) = ∫-dt/70
㏑A(t) = -t/70 + C
taking exponents of both sides, we have
A(t) = exp(-t/70 + C)
A(t) = exp(-t/70)expC
[tex]A(t) = e^{-t/70}e^{C}\\A(t) = Ce^{-t/70} whereC = e^{C}[/tex]
A(0) = 30, So
[tex]A(0) = Ce^{-0/70}\\30 = Ce^{0}\\C = 30[/tex]
[tex]A(t) = 30e^{-t/70}[/tex]
The rate at which the number of grams of salt in the tank is changing at time t is
[tex]dA(t)/dt = \frac{d30e^{-t/70} }{dt} \\dA(t)/dt = -\frac{30e^{-t/70} }{70}\\dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex]
So, the rate at which the number of grams of salt in the tank is changing at time t is
[tex]dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex] g/min
