Question

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.
a. Find the amount of salt in the tank at any time.
b. Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.
c. The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

Answer 1

Answer:

Part a)

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b)

Check the attached figure to see the ultimate behavior of the graph.

Part c)

The level = 25, Amplitude = 0.2499

Step-by-step Solution:

Part a)

Given:

$Q(0)=50$

Rate in:

$\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

Rate out:

$\frac{Q}{100}\cdot 2=\frac{Q}{50}$

So, the differential equation would become:

$\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}$

Rewriting the equation:

$\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

As $p(x)$ is the coefficient of $y$, while $q(x)$ is the constant term in the right side of the equation:

$p\left(x\right)=\frac{1}{50}$

$q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

First it is important to determine the function $\mu$ :

$\mu \left(t\right)=e^{\int \:p\left(t\right)dt}$

        $=e^{\int \:\left(\frac{1}{50}\right)dt}$

        $=e^{\frac{t}{50}}$

The general solution then would become:

$y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)$

       $=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c$

       $=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c$

        $=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c$

Evaluate at $t=0$

$50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c$

Solve to c:

$c=25+\frac{1250}{5002}$

$\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}$

$c=25+\frac{625}{2501}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}$

$c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$c=\frac{25\cdot \:2501+625}{2501}$

$c=\frac{63150}{2501}$

$c\approx 25.25$

Therefore, the general solution then would become:

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b) Plot the Solution to see the ultimate behavior of the graph

The graph appears to level off at about the value of $Q=25$.

The graph is attached below.

Part c)

In the graph we note that the level is $Q=25$.

Therefore, the level = 25

The amplitude is the (absolute value of the) coefficient of $cost\:t$ in the general solution (as the coefficient of the sine part is a lot smaller):

Therefore,

                $A=\frac{1250}{5002}\:\approx 2.499$

Keywords: differential equation, word problem

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