Answer:
1.857 minutes
Step-by-step explanation:
From the given information:
Consider the amount of salt present in the tank as x(t) at a given time (t); &
The volume of the solution = V(t)
At t = 0 i.e. (at initial conditions) x(0) = 30; and V(0) = 50
However, the overall increase taken place for 1 gallon per minute is:
V(t) = 50 + t
The amount of salt x(t) at any given point for time (t) is;
[tex]\dfrac{x(t)}{V(t)}= \dfrac{x(t)}{50+t}[/tex]
After 5 gallons of solution exit per minute; the concentration of the salt solution changes at:
[tex]\dfrac{dx(t)}{dt}= -\dfrac{5x(t)}{50+t}[/tex]
Taking the integral of what we have above, we get:
In x(t) = - 5 In(t + 50) + In (C)
In x(t) = In (t+ 50)⁻⁵ + In (C)
In x(t) = In C ( t + 50)⁻⁵
x(t) = C(t + 50)⁻⁵ (General solution)
To estimate the required solution; we apply the initial conditions x(0) = 30;
Thus;
x(0) = C(50)⁻⁵ = 30
⇒ C = 30 × 50⁵
Hence; x(t) = 30 × 50⁵ × (t + 50)⁻⁵
The above expression can be re-written as:
[tex]x(t) = 25 \implies 30 \times \bigg ( \dfrac{50}{t+50} \bigg ) ^5= 25[/tex]
i.e.
[tex]\bigg ( \dfrac{50}{t+50} \bigg ) ^5= \dfrac{25}{30}[/tex]
[tex]\dfrac{50}{t+50}= \bigg ( \dfrac{25}{30}\bigg) ^{\dfrac{1}{5}}[/tex]
[tex]\dfrac{50}{t+50}= 0.964192504[/tex]
[tex]{50}= 0.964192504(t+50)[/tex]
50 = 0.964192504t + 48.2096252
50 - 48.2096252 = 0.964192504t
1.7903748 = 0.964192504t
t = 1.7903748 / 0.964192504
t ≅ 1.857 minutes
We can thereby conclude that the estimated time it will require until there are 25 pounds of salt in the tank is 1.857 minutes.
