MATH SOLVE

6 months ago

Q:
# A tank with a capacity of 100 gallons initially contains 50 gallons of water with 10 pounds of salt in solution. fresh water enters at a rate of 2 gallons per minute and a well-stirred mixture is pumped out at the rate of 1 gallon per minute. compute the amount of salt (in pounds) in the tank at the first moment when the tank is filled.

Accepted Solution

A:

Let [tex]A(t)[/tex] be the amount of salt (in pounds) in the tank at time [tex]t[/tex]. We're given that[tex]A(0)=10\text{ lb}[/tex].

The rate at which the amount of salt in the tank changes is given by the ODE

[tex]A'(t)=\dfrac{2\text{ gal}}{1\text{ min}}\cdot\dfrac{0\text{ lb}}{1\text{ gal}}-\dfrac{1\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}[/tex]

[tex]A'(t)+\dfrac{A(t)}{50+t}=0[/tex]

[tex](50+t)A'(t)+A(t)=0[/tex]

[tex]\bigg((50+t)A(t)\bigg)'=0[/tex]

[tex](50+t)A(t)=C[/tex]

[tex]A(t)=\dfrac C{50+t}[/tex]

Given that [tex]A(0)=10[/tex], we find that

[tex]10=\dfrac C{50+0}\implies C=500[/tex]

so that the amount of salt in the tank is described by

[tex]A(t)=\dfrac{500}{50+t}[/tex]

The tank will be filled when [tex]50+t=100[/tex], or after [tex]t=50[/tex] minutes. At this time, the amount of salt in the tank is

[tex]A(50)=\dfrac{500}{50+50}=5\text{ lb}[/tex]

The rate at which the amount of salt in the tank changes is given by the ODE

[tex]A'(t)=\dfrac{2\text{ gal}}{1\text{ min}}\cdot\dfrac{0\text{ lb}}{1\text{ gal}}-\dfrac{1\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}[/tex]

[tex]A'(t)+\dfrac{A(t)}{50+t}=0[/tex]

[tex](50+t)A'(t)+A(t)=0[/tex]

[tex]\bigg((50+t)A(t)\bigg)'=0[/tex]

[tex](50+t)A(t)=C[/tex]

[tex]A(t)=\dfrac C{50+t}[/tex]

Given that [tex]A(0)=10[/tex], we find that

[tex]10=\dfrac C{50+0}\implies C=500[/tex]

so that the amount of salt in the tank is described by

[tex]A(t)=\dfrac{500}{50+t}[/tex]

The tank will be filled when [tex]50+t=100[/tex], or after [tex]t=50[/tex] minutes. At this time, the amount of salt in the tank is

[tex]A(50)=\dfrac{500}{50+50}=5\text{ lb}[/tex]