Answer:
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
![A(t) = e^{-t/70}e^{C}\\A(t) = Ce^{-t/70} whereC = e^{C}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20e%5E%7B-t%2F70%7De%5E%7BC%7D%5C%5CA%28t%29%20%3D%20Ce%5E%7B-t%2F70%7D%20%20%20%20%20%20%20whereC%20%3D%20e%5E%7BC%7D)
A(0) = 30, So
![A(0) = Ce^{-0/70}\\30 = Ce^{0}\\C = 30](https://tex.z-dn.net/?f=A%280%29%20%3D%20Ce%5E%7B-0%2F70%7D%5C%5C30%20%3D%20Ce%5E%7B0%7D%5C%5CC%20%3D%2030)
![A(t) = 30e^{-t/70}](https://tex.z-dn.net/?f=A%28t%29%20%3D%2030e%5E%7B-t%2F70%7D)
The rate at which the number of grams of salt in the tank is changing at time t is
![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}](https://tex.z-dn.net/?f=dA%28t%29%2Fdt%20%3D%20%5Cfrac%7Bd30e%5E%7B-t%2F70%7D%20%7D%7Bdt%7D%20%5C%5CdA%28t%29%2Fdt%20%3D%20-%5Cfrac%7B30e%5E%7B-t%2F70%7D%20%7D%7B70%7D%5C%5CdA%28t%29%2Fdt%20%3D%20-%5Cfrac%7B3%7D%7B7%7De%5E%7B-t%2F70%7D)
So, the rate at which the number of grams of salt in the tank is changing at time t is
g/min