Respuesta :
Answer:
1) x = [ - (12/1000)* e∧ ( - 15/1000)*t + 12/1000 ] /e∧ - (15/1000)*t
2) x = - 0,012 * ( e ∧ 0.18 + 0,012 ) / e∧-0,18
Step-by-step explanation:
1.-Quantity of salt in the tank after t minutes
The rate of change of the quantity of salt in the tank is:
dx(t) /dt = original quantity (0) + input quantity - output quantity (1)
quantity = concentration* rate Then
input quantity = 0.04 Kg/lt * 5 Lt/min + 0.06 Kg/lt * 10Lt/min = 0.2 Kg/min
+ 0.6 Kg/min = 0,8 Kg/lt
output quantity = Output concentration * rate of draining
rate of draining = 15 Lt/min
The input quantity and the output quantity occur at the same rate therefore the volume in the tank is constant 1000Lt.
output quantity = (x/1000 )*15
Plugging these values in equation (1) we get.
dx/dt = 0,8 - ( x/1000)* 15
The last one is a differential first-order equation like
x´ + P(t)*x = q(t)
and the solution is:
x*μ = ∫ q(t)*μ*dt + C
where μ is the integration factor e ∧ ∫p(t)*dt
let´s call b = -15/1000
μ = e ∧ ∫p(t)*dt = e∧∫ b*dt = e∧ b*t = e∧ ( -15/1000)*t
μ = e∧ - (15/1000)*t
Then x*μ = x * e∧ - (15/1000)*t
∫ q(t)*μ*dt = ∫ 0.8 * e∧ - (15/1000)*t*dt = 0.8 * ∫ e∧bt * dt
∫ q(t)*μ*dt = 0.8 * ( 1/b ) e∧bt = - 0,8 *( 15/1000) * e∧ ( - 15/1000)*t
∫ q(t)*μ*dt = - (12/1000)* e∧ ( - 15/1000)*t
x * e∧ - (15/1000)*t = - (12/1000)* e∧ ( - 15/1000)*t + C
Initial condition t = 0 x = 0
0 = - (12 / 1000 )* e⁰ = C
C = 12/1000
x * e∧ - (15/1000)*t = - (12/1000)* e∧ ( - 15/1000)*t + 12/1000
x = [ - (12/1000)* e∧ ( - 15/1000)*t + 12/1000 ] /e∧ - (15/1000)*t
When t = 60 min
x = [ - (12/1000)* e∧ ( - 15/1000)*12 + 12/1000 ] / e∧ - (15/1000) * 12
x = - 0,012 * ( e ∧ 0.18 + 0,012 ) / e∧-0,18
