# Population y grows according to the equation { \frac{dy}{dt} = ky }, where k is a constant and...

## Question:

Population y grows according to the equation {eq}\frac{dy}{dt} = ky {/eq}, where k is a constant and t is measured in years. If the population doubles every 10 years, then the value of k is.....

a. 0.069

b. 0.200

c. 0.301

d. 3.322

e. 5.000

## Population Growth

Population growth is typically modeled using an exponential function: {eq}P(t) = P_0 e^{kt} {/eq} where {eq}P_0 {/eq} is the initial population and {eq}k {/eq} is the growth constant. This equation is derived from the assumption that the rate of growth is proportional to the current population. Other growth expressions can be derived based on alternative assumptions.

## Answer and Explanation:

We solve for the function {eq}y {/eq} as follows:

$$\begin{align*} \frac{dy}{dt} &= ky \\ \frac{dy}{y} &= k \ dt &\color{blue}{\text{Separable DE}}\\ \int \frac{dy}{y} &= k \int\ dt \\ \ln y &= kt + C\\ e^{\ln y} &= e^{kt+C} \\ y &= e^{kt}e^C &\color{blue}{\text{Recall: }x^{a+b} = (x^a)(x^b)}\\ y &= Ce^{kt} &\color{blue}{\text{Note: }e^C \text{ is still a constant}} \end{align*} $$

Suppose {eq}y_0 {/eq} is the initial population. It follows that:

$$\begin{align*} y_0 &= Ce^{0} \\ y_0 &= C \end{align*} $$

The equation thus becomes {eq}y = y_0e^{kt} {/eq}.

It was noted that after 10 years, the population doubles. Hence, {eq}y = 2y_0 {/eq} at {eq}t = 10 {/eq}. From this condition, we can solve for the value of {eq}k {/eq}

$$\begin{align*} 2y_0 &= y_0e^{10k} \\ 2 &= e^{10k} \\ \ln 2 &= \ln \left (e^{10k} \right ) \\ \ln 2 &= 10k &\color{blue}{\text{Recall: }\ln e^x = x}\\ \frac{\ln 2}{10} &= k \\ 0.069 &\approx k \end{align*} $$

Therefore, {eq}\boxed{k \approx 0.069} {/eq}

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10