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yan [13]
1 year ago
15

It is predicted that in t years, the population of a country will be P (t) = 50e ^ (0.02t) million inhabitants. A) What will be

the rate of change of the population in 10 years. B) What will be the relative rate of the population in t years? Is this rate constant?
Mathematics
1 answer:
lakkis [162]1 year ago
5 0
\begin{gathered} \text{Given:} \\ P(t)=50e^{0.02t} \end{gathered}

A) What will be the rate of change of the population in 10 years.

\begin{gathered} \text{Get the derivative of the }P(t)\text{ with respect to }t \\ \frac{dP}{dt}(50e^{0.02t})=50(0.02)e^{0.02t} \\ \frac{dP}{dt}(50e^{0.02t})=e^{0.02t} \end{gathered}

Substitute t = 10, to the derivative of P(t)

\begin{gathered} P^{\prime}(t)=e^{0.02t} \\ P^{\prime}(10)=e^{0.02(10)} \\ P^{\prime}(10)=e^{0.2} \\ P^{\prime}(10)=e^{0.2} \\ P^{\prime}(10)=1.02020134 \\  \\ \text{Round off to two decimal place} \\ P^{\prime}(10)=1.02 \end{gathered}

Therefore, the rate of change of the population in 10 years is 1.02 million.

B) What will be the relative rate of the population in t years? Is this rate constant?​

\begin{gathered} \text{The relative rate of the population in t years is the first derivative of }P(t) \\ P^{\prime}(t)=e^{0.02t} \end{gathered}

The rate is not constant, as it depends on how much time t has passed.

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The sequence is given to be:

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The nth term of an arithmetic sequence is calculated using the formula:

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From the given sequence, the following parameters can be gotten:

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