According to the US Census, the world population P, in billions, is approximately P = 6.342...

Question:

According to the US Census, the world population P, in billions, is approximately {eq}\displaystyle P = 6.342 e^{0.011 t} {/eq}; where t is in years since January 1, 2004. At what rate was the world's population increasing on January 1, 2005? Give your answer in millions per year.

Exponential Growth and Decay:

Suppose that a time-varying quantity {eq}Q(t) {/eq} obeys some equation of the form {eq}Q(t)=Q_0e^{kt} {/eq}, where {eq}Q_0 {/eq} and {eq}k {/eq} are constants. Then we say that {eq}Q(t) {/eq} is an exponential function of time. The constant {eq}Q_0 {/eq} is called the initial value of {eq}Q(t) {/eq}, and {eq}k {/eq} is called the exponential constant of {eq}Q(t) {/eq}. If {eq}Q(t) {/eq} is an exponential function of {eq}t {/eq}, then the rate of change in {eq}Q(t) {/eq}, {eq}Q'(t) {/eq}, obeys the equation {eq}Q'(t)=kQ(t) {/eq}.

If {eq}k>0 {/eq}, we say that {eq}Q(t) {/eq} is undergoing exponential growth; if {eq}k < 0 {/eq}, we say that {eq}Q(t) {/eq} is undergoing exponential decay.

Answer and Explanation:

The population function {eq}P(t)=6.342e^{0.011t} {/eq} is an exponential function with growth constant {eq}k=0.011 {/eq}. So {eq}P'(t) {/eq} obeys the equation

{eq}\begin{align*} P'(t)&=kP(t)\\ &=0.011P(t)\\ &=0.011(6.342e^{0.011t})\\ &=0.069762e^{0.011t} \, . \end{align*} {/eq}

Since January 1, 2005 is exactly 1 year after January 1, 2004, the desired rate of increase is {eq}P'(1) {/eq}, which can be computed as follows:

{eq}\begin{align*} P'(1)&=0.69762e^{0.011(1)}\\ &=0.69762e^{0.011}\\ &\approx 0.705336 \, . \end{align*} {/eq}

So the population is increasing at a rate of approximately {eq}0.705336 {/eq} billion per year, or {eq}\boxed{705.336\text{ million/year}}\, {/eq}.


Learn more about this topic:

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Exponential Growth vs. Decay

from Math 101: College Algebra

Chapter 10 / Lesson 2
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