Answer:
(a) After time (t), the amount of salt in the tank is s(t) = 130 − 130e ∧−3t/200 / 3
(b) After 60 minutes, the salt in the tank is s(60) ≈ 25.7153
Step-by-step explanation:
To start with,
Let s(t) = amount of salt in kg of salt at time t.
Then we have:
dt/ds = (rate of salt into tank) − (rate of salt out of tank)
= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) − ( s kg/ 1000 L x 15 L/min)
= 0.25 kg/min + 0.4 kg/min − 15s / 1000 kg/min
So we get the differential equation
dt/ds = 0.65 − 15s / 1000
= 65 / 100 − 15s / 1000
dt/ds = 130 - 3s / 200
We separate s and t to get
1 / 130 - 3s ds = 1 / 200 dt
Then we Integrate,
Thus we have
∫1 / 130 - 3s ds = ∫1 / 200 dt
- 1/3 · ln |130 − 3s| =1 / 200 t + C1
ln |130 − 3s| = - 3 /200 t + C2
|130 − 3s| = e− ³⁺²⁰⁰ ∧ t+C2
|130 − 3s| = C3e
− ∧3t/200
130 − 3s = C4e ∧−3t/200
−3s = −130 + C4e ∧−3t/200
s = 130 - C4e
∧−3t/200 / 3
Since we begin with pure water, we have s(0) = 0. Substituting,
0 = 130 − C4e ∧−3·0/200 / 3
0 = 130 − C4
C4 = 130
So our function is
s(t) = 130 − 130e ∧−3t/200 / 3
After one hour (60 min), we have
s(60) = 130 − 130e ∧−3·60/200 / 3
s(60) ≈ 25.7153
Thus, after one hour there is about 25.72 kg of salt in the tank.
