# A lake holds a population of 600,000 trout. The natural growth rate of the trout population is...

## Question:

A lake holds a population of 600,000 trout. The natural growth rate of the trout population is 5%. Fishing removes 35,000 trout each year from the lake. Set up a differential equation modeling the fish population as a function of time. Solve it to find out if and when the fish population will be completely depleted.

## Integration:

A mathematical quantity that represents the summation of the infinitesimal small function or data to form as complete data or function is known as integration. It widely used in the engineering field to analyse complex systems.

Given Data:

• The number of population lake holds is: {eq}N = 600000\;{\rm{trouts}} {/eq}
• The natural growth rate of trout population is: {eq}G = 5\% = 0.05 {/eq}
• The number of trout remove per year is: {eq}T = 35000 {/eq}

There is natural growth rate of trout and per year trout remove from the lake

The expression for differential equation for fish population as function of time is

{eq}\dfrac{{dN}}{{dt}} = GN - T {/eq}

Substitute and solve the above expression

{eq}\begin{align*} \dfrac{{dN}}{{dt}} &= 0.05N - 35000\\ \dfrac{{dN}}{{dt}} &= \dfrac{{N - 700000}}{{20}}\\ \dfrac{{dN}}{{N - 700000}} &= \dfrac{{dt}}{{20}} \end{align*} {/eq}

Integrate the above expression

{eq}\begin{align*} \int {\dfrac{{dN}}{{N - 700000}}} &= \int {\dfrac{{dt}}{{20}}} \\ \ln \left[ {N - 700000} \right] &= \dfrac{t}{{20}} + \ln K\\ N - 700000 &= K{e^{\dfrac{t}{{20}}}}\\ N &= K{e^{\dfrac{t}{{20}}}} + 700000 \cdots\cdots\rm{(I)} \end{align*} {/eq}

Thus the differential equation for fish population as function of time is {eq}N = K{e^{\dfrac{t}{{20}}}} + 700000 {/eq}

Initially lake has {eq}N = 600000 {/eq} at {eq}t = 0 {/eq}

Substitute and solve the above expression

{eq}\begin{align*} 600000 &= K{e^{\dfrac{0}{{20}}}} + 700000\\ 600000 &= K{e^0} + 700000\\ - 100000 &= K \end{align*} {/eq}

If population is completely depleted then number of population will be zero

{eq}N = 0 {/eq}

Substitute and solve the expression (I)

{eq}\begin{align*} 0 &= \left( { - 100000} \right){e^{\dfrac{t}{{20}}}} + 700000\\ - 700000 &= \left( { - 100000} \right){e^{\dfrac{t}{{20}}}}\\ 7 &= {e^{\dfrac{t}{{20}}}}\\ \ln 7 &= \ln {e^{\dfrac{t}{{20}}}}\\ \ln 7 &= \dfrac{t}{{20}}\ln e\\ t &= 38.91\;{\rm{years}} \end{align*} {/eq}

Thus the fish population will be depleted at {eq}38.91\;{\rm{years}} {/eq} 