Assuming the population of the earth changes at a rate proportional to the current population...

Question:

Assume that the population of the earth changes at a rate proportional to the current population. Further, it is estimated that at time {eq}t=0 {/eq}, the earth's population was 600 million; at time {eq}t=300 {/eq}, its population was 2.8 billion. Find an expression giving the population of the earth at any time. Assuming that the greatest population the earth can support is 25 billion, when will this limit be reached?

Exponential Growth:

Suppose that a quantity {eq}Q(t) {/eq} grows at a rate which is proportional to its current value. Then we say that {eq}Q(t) {/eq} grows exponentially. In this case, it is always possible to write {eq}Q(t)=Q(0)e^{kt} {/eq} for some positive constant {eq}t {/eq}.

Generally a quantity will grow exponentially when its growth is not restricted in any way. For example, a population will grow exponentially when that population is not experiencing any resource constraints.

Answer and Explanation:

Let {eq}P(t) {/eq} be the population of the earth at time {eq}t {/eq} in billions of people. Then we're given that {eq}P(0)=0.6 {/eq} (since 600 million is 0.6 billion), and {eq}P(300)=2.8 {/eq}. Since the rate of change in {eq}P(t) {/eq} is proportional to {eq}P(t) {/eq} itself, {eq}P(t) {/eq} grows exponentially, and so

{eq}P(t)=P(0)e^{kt}=0.6e^{kt} {/eq}

for some constant {eq}k {/eq}.

We can use the condition {eq}P(300)=2.8 {/eq} to solve for {eq}k {/eq}:

{eq}\begin{align*} 2.8&=P(300)\\ &=0.6e^{300k}\\ \frac{2.8}{0.6}&=e^{300k}|\ \frac{14}{3}&=e^{300k}\\ \ln \frac{14}{3}&=300k&&\text{(taking logarithms of both sides)}\\ \frac{1}{300}\ln\frac{14}{3}&=k \, . \end{align*} {/eq}

So the population at time {eq}t {/eq} in billions of people is {eq}P(t)=0.6\exp\left(\frac{t}{300}\ln\frac{14}{3}\right) {/eq}.

We now want to find the time when the population is 25 billion: that is, when {eq}P(t)=25 {/eq}. This equation can be solved for {eq}t {/eq} as follows:

{eq}\begin{align*} 25&=P(t)\\ &=0.6\exp\left(\frac{t}{300}\ln\frac{14}{3}\right)\\ \frac{25}{0.6}&=\exp\left(\frac{t}{300}\ln\frac{14}{3}\right)\\ \frac{125}{3}&=\exp\left(\frac{t}{300}\ln\frac{14}{3}\right)\\ \ln \frac{125}{3}&=\frac{t}{300}\ln\frac{14}{3}\\ \frac{300\ln(125/3)}{\ln(14/3)}&=t \\ 726 \approx t \, . \end{align*} {/eq}

In summary, the population at time {eq}t {/eq} is {eq}P(t)=0.6\exp\left(\frac{t}{300}\ln\frac{14}{3}\right) {/eq} billion people, and the limiting population will be reached roughly at time {eq}t \approx 726 {/eq}.


Learn more about this topic:

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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
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