Respuesta :
Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).
Salt flows in at a rate of
[tex]\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}[/tex]
and flows out at a rate of
[tex]\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}[/tex]
so that the net rate of change in the amount of salt in the tank is given by the linear differential equation
[tex]\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))[/tex]
Multiply both sides by the integrating factor, [tex]e^{t/180}[/tex], and rewrite the left side as the derivative of a product.
[tex]e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))[/tex]
[tex]\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))[/tex]
Integrate both sides with respect to t (integrate the right side by parts):
[tex]\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt[/tex]
[tex]\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C[/tex]
Solve for A(t) :
[tex]\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}[/tex]
The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.
[tex]\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}[/tex]
So,
[tex]\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}[/tex]
Recall the angle-sum identity for cosine:
[tex]R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)[/tex]
so that we can condense the trigonometric terms in A(t). Solve for R and θ :
[tex]R \cos(\theta) = -\dfrac{66,096,000}{32,401}[/tex]
[tex]R \sin(\theta) = \dfrac{367,200}{32,401}[/tex]
Recall the Pythagorean identity and definition of tangent,
[tex]\cos^2(x) + \sin^2(x) = 1[/tex]
[tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]
Then
[tex]R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}[/tex]
and
[tex]\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)[/tex]
so we can rewrite A(t) as
[tex]\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}[/tex]
As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of
[tex]24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}[/tex]
and
[tex]24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}[/tex]
which is to say, with amplitude
[tex]2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}[/tex]
