Question

A tank contains 100 gal of water and 50 oz of salt.water containing a salt concentration of 1 4 (1 1 2 sin t) oz/gal flows into 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 ta

Answer 1

Step-by-step explanation:

Heres the possible full question and solution:

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?

solution

a)

Consider the tank contains 100gal of water and 50 oz of salt

Assume that the amount of salt in the tank at time t is Q(t).

Then, the rate of change of salt in the tank is given by
`(dQ)/(dt)`.

Here,
`(dQ)/(dt)`=rate of liquid flowing in the tank - rate of liquid flowing out.

Therefore,

`Rate_(in) =2gal/min * (1)/(4) (1+ (1)/(2)sin t)oz/gal\\\\\\ (1)/(2) (1+ (1)/(2)sin t)oz/min\\\\\\Rate_(</p><p>out)=2gal/min *(Q)/(100)oz/gal\\\\(Q)/(50)oz/min`

Therefore,

`(dQ)/(dt)` can be evaluated as shown below:

`(dQ)/(dt)=(1)/(2)(1+(1)/(2)\sin t)-(Q)/(50)\\\\\\(dQ)/(dt)+(1)/(50)Q=(1)/(2)+(1)/(4)\sin t`

The above differential equation is in standard form:

`(dy)/(dt)+Py=G`

Here,
`P=(1)/(50),G=</p><p>(1)/(2)+(1)/(4)\sin t`

The integrating factor is as follows:

`\mu(t)=e^{\int {P}dt}\\\mu(t)=e^{\int {(1)/(50)}dt}\\\mu(t)=e^{(t)/(50)}`

Thus, the integrating factor is
`\mu(t)=e^{(t)/(50)}`

Therefore, the general solution is as follows:

`y\mu(t)=\int {\mu (t)G}dt\\\\Qe^{(t)/(50)}=\int {e^{(t)/(50)}((1)/(2)+(1)/(4)\sin t) dt}\\\\Qe^{(t)/(50)}=(1)/(2)\int {e^{(t)/(50)}dt + (1)/(4)\int {\sin t {e^{(t)/(50)}} dt}\\\\\Qe^{(t)/(50)}=25 {e^{(t)/(50)} + (1)/(4)\int {\sin t {e^{(t)/(50)}} dt}+C...(1)`

Here, C is arbitrary constant of integration.

Solve
`\int {\sin te^{(t)/(50)}} dt}`

Here
`u = e^{(t)/(50)}` and
`v =\sin t`.

Substitute u , v in the below formula:

`\int{u,v}dt=u\int{v}dt-\int(du)/(dt)\int{v}dt\dot dt\\\\\int {e^{(t)/(50)}\sin t}dt=-e^{(t)/(50)}\cos t + (1)/(50)\int{e^{(t)/(50)}\cos t}dt...(2)`

Now, take
`u = e^{(t)/(50)}`,
`v =\sin t`

Therefore,
`\int{e^{(t)/(50)}\cos t} dt=\int {e^{(t)/(50)}\sin t}dt - (1)/(50)\int{e^{(t)/(50)}\sin t}dt...(3)`

Use (3) in equation(2)

`\int {e^{(t)/(50)}\sin t}dt=-e^{(t)/(50)}\cos t + \frac{e^{(t)/(50)}}{50}\sin t - (1)/(2500)\int{e^{(t)/(50)}\sin t}dt\\\\(2501)/(2500)\int{e^{(t)/(50)}\sin t}dt={e^{(t)/(50)}\cos t}+\frac{e^{(t)/(50)}}{50}\sin t\\\\\int{e^{(t)/(50)}\sin t}dt=(2500)/(2501){e^{(t)/(50)}\cos t}+(50)/(2501)e^{(t)/(50)}\sin t...(4)`

Use (4) in equation(l) .

`Qe^{(t)/(50)}=25 e^{(t)/(50)} - (625)/(2501)e^{(t)/(50)}\cos t +(25)/(5002)e^{(t)/(50)}\sin t+C`

Apply the initial conditions
`t =0, Q = 50`.

`50=25-(625)/(2501)+c\\\\c=(63150)/(2501)`

So,
`Qe^{(t)/(50)}=25 e^{(t)/(50)} - (625)/(2501)e^{(t)/(50)}\cos t +(25)/(5002)e^{(t)/(50)}\sin t+(63150)/(2501)`

Therefore, the amount of salt in the tank at any time is as follows:

`Qe^{(t)/(50)}=25 e^{(t)/(50)} - (625)/(2501)e^{(t)/(50)}\cos t +(25)/(5002)e^{(t)/(50)}\sin t+(63150)/(2501)e^{(-t)/(50)}`

b)

sketch the solution curve as shown in attachment as graph 1:

CHECK COMMENT FOR C