Answer:
Step-by-step explanation:
From the given information:
[tex]R_{in} = ( \dfrac{1}{2} \ lb/gal) (6)\ gal /min \\ \\R_{in} = 3 \ lb/min[/tex]
Given that the solution is pumped at a slower rate of 4gal/min
Then:
[tex]R_{out} = \dfrac{4A}{100+(6-4)t}[/tex]
[tex]R_{out}= \dfrac{2A}{50+t}[/tex]
The differential equation can be expressed as:
[tex]\dfrac{dA}{dt}+ \dfrac{2}{50+t}A = 3 \ \ \ ... (1)[/tex]
Integrating the linear differential equation; we have::
[tex]\int_c \dfrac{2}{50 +t}dt = e^{2In |50+t|[/tex]
[tex]\int_c \dfrac{2}{50 +t}dt = (50+t)^2[/tex]
multiplying above integrating factor fields; we have:
[tex](50 +t)^2 \dfrac{dA}{dt} + 2 (50 + t)A = 3 (50 +t)^2[/tex]
[tex]\dfrac{d}{dt}\bigg [ (50 +t)^2 A \bigg ] = 3 (50 +t)^2[/tex]
[tex](50 + t)^2 A = (50 + t)^3+c[/tex]
A = (50 + t) + c(50 + t)²
Using the given conditions:
A(0) = 20
⇒ 20 = 50 + c (50)⁻²
-30 = c(50) ⁻²
c = -30 × 2500
c = -75000
A = (50+t) - 75000(50 + t)⁻²
The no. of pounds of salt in the tank after 35 minutes is:
A(35) = (50 + 35) - 75000(50 + 35)⁻²
A(35) = 85 - [tex]\dfrac{75000}{7225}[/tex]
A(35) =69.6193 pounds
A(35) [tex]\simeq[/tex] 70 pounds
Thus; the number of pounds of salt in the tank after 35 minutes is 70 pounds.
