Respuesta :
Answer:
[tex]y = -\frac{7}{4} e^{4t} + \frac{15}{4}[/tex]
Step-by-step explanation:
Let [tex]y(t)[/tex] be the amount of Radium at any time [tex]t[/tex] .
Since the mass of fish would increase at a rate proportional to the mass with the proportionality constant of 4/yr, we obtain that the time rate of change is proportional to [tex]y(t)[/tex]. Therefore, in that case, the model would be an first order ODE:
[tex]\dfrac{dy}{dt} = 4y[/tex]
However, commercial fishing removes fish mass at a constant rate of 11 million tons per year, which means that the equation is
[tex]\dfrac{dy}{dt} = 4y - 15[/tex]
The initial amount of fish was 2 tons, which gives the initial condition
[tex]y(0) = 2[/tex]
Hence, we obtain the initial value problem
[tex]\dfrac{dy}{dt} = 4y-15, \quad y(0) = 2[/tex]
This is the mathematical model of this process.
Let's solve the ODE. Rearranging terms in the equation gives
[tex]y' = 4y - 15 \iff y' - 4y = - 15[/tex]
This is an linear ODE of the form
[tex]y' + p(t)y = q(t)[/tex]
where [tex]p(t) =-4, q(t) = -15[/tex].
Hence,
[tex]h = \int p \,d t = -4\int dt = -4t +c[/tex]
So, an integrating factor is
[tex]e^h = e^{-4t}[/tex]
and the general solution is given by
[tex]y(t)=e^{-h} \left( c+ \int qe^h \, dt \right)\\ \\\phantom{y(t)} =e^{4t} ( c+ \int (-15) e^{-4t} dt )\\\\\phantom{y(t)} = e^{4t} \left( c + \frac{15}{4}e^{-4t}\right) \\\\\phantom{y(t)} = ce^{4t} + \frac{15}{4}[/tex]
Now, we need to determine [tex]c[/tex]. To do so, we use the initial condition.
Substituting [tex]0[/tex] for [tex]t[/tex] and [tex]2[/tex] for [tex]y[/tex] gives
[tex]2 = y(0) = ce^0 + \frac{15}{4} \implies c = 2 - \frac{15}{4} = -\frac{7}{4}[/tex]
Hence, the particular solution of this problem is:
[tex]y = -\frac{7}{4} e^{4t} + \frac{15}{4}[/tex]
- When will all the fish be gone?
Let [tex]t^*[/tex] be the moment when all fish are gone. Then, since the mass of fish is denoted by [tex]y[/tex],
[tex]y(t^*) = 0[/tex]
We need to solve this equation for [tex]t^*[/tex].
[tex]-\frac{7}{4} e^{4t} + \frac{15}{4} = 0 \implies \frac{7}{4} e^{4t} = \frac{15}{4}\\\phantom{-\frac{7}{4} e^{4t} + \frac{15}{4} = 0} \implies e^{4t} = \frac{15}{7} \\\phantom{-\frac{7}{4} e^{4t} + \frac{15}{4} = 0} \implies 4t =\ln \frac{15}{7}\\\phantom{-\frac{7}{4} e^{4t} + \frac{15}{4} = 0} \implies t =\frac{1}{4}\ln \frac{15}{7} \approx 0.191[/tex]
- If the fishing rate is changed so that the mass of fish remains constant, what should that rate be?
Let the fishing rate be such that the mass of fish remains constant.
Then, we have
[tex]y(t) = 2[/tex]
Therefore,
[tex]-\frac{7}{4} e^{4t} + \frac{r}{4} = 2[/tex]
For [tex]t = 0[/tex],
[tex]2 = -\frac{7}{4} + \frac{r}{4} \implies \frac{15}{4} = \frac{r}{4} \implies r = 15[/tex]
