40–43. Population growth

Starting with an initial value of P(0)=55, the population of a prairie dog community grows at a rate of P′(t)=20−t/5 (prairie dogs/month), for 0≤t≤200, where t is measured in months.

b. Find the population P(t), for 0≤t≤200.

검증된 단계별 안내

Identify the given rate of change of the population, which is the derivative of the population function: $P\'(t) = 20 - \frac{t}{5}$.

Recall that to find the population function $P(t)$, you need to integrate the rate function $P\'(t)$ with respect to $t$: $P(t) = \int P\'(t) \, dt + C$.

Set up the integral: $P(t) = \int \left(20 - \frac{t}{5}\right) dt + C$.

Integrate each term separately: the integral of 20 with respect to $t$ is \$20t$, and the integral of $-\frac{t}{5}$ with respect to $t$ is $-\frac{1}{5} \cdot \frac{t^2}{2} = -\frac{t^2}{10}$.

Combine the results and include the constant of integration $C$: $P(t) = 20t - \frac{t^2}{10} + C$. Use the initial condition $P(0) = 55$ to solve for $C$ by substituting $t=0$ and $P(0)=55$.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Initial Value Problems

An initial value problem involves finding a function that satisfies a given differential equation and meets a specified initial condition. Here, the population function P(t) must satisfy P'(t) = 20 - t/5 and the initial value P(0) = 55, which anchors the solution to a specific starting point.

Antiderivatives and Integration

To find P(t) from its rate of change P'(t), we integrate P'(t) with respect to t. Integration reverses differentiation, allowing us to recover the original function up to a constant, which is then determined using the initial condition.

Applying Initial Conditions to Determine Constants

After integrating the rate function, the solution includes an arbitrary constant. Using the initial condition P(0) = 55, we substitute t = 0 into the integrated function to solve for this constant, ensuring the population model accurately reflects the starting population.

40–43. Population growthStarting with an initial value of P(0)=55, the population of a prairie dog community grows at a rate of P′(t)=20−t/5 (prairie dogs/month), for 0≤t≤200, where t is measured in months.b. Find the population P(t), for 0≤t≤200.