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Numerade Educator

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Problem 86 Hard Difficulty

In Section 1.4 we modeled the world population from 1900 to 2010 with the exponential function
$$
P(t)=(1436.53) \cdot(1.01395)^{t}
$$
where $t=0$ corresponds to the year 1900 and $P(t)$ is measured in millions. According to this model, what was the rate of increase of world population in $1920 ?$ In $1950 ?$ In $2000 ?$

Answer

in 1920 is $26.3 ; 1950$ is $39.78 ; 2000$ is 79.52

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Video Transcript

uh huh. In this problem we are given the population team and millions of the function of two years, 20 0, 1900 has given us P. T equals 14 36 or 1436.58 and 1.1395 inches the parrot to We want to find the rate of change of p in 1920 19 52,000. As a note given here where the Astros indicates, we're going to use differentiation with the chain rule to find the derivative or rate of change of the population of 13 to 15. So different changes in the chain will gives us P prime T. That is the rate of change DPD teen as d b u d s b u L n d d D X. Because this is an exponential function. Thus we have G p t equals 14 36.53 times. Bu bu 1.1395 T. And l n b f l n 1.1395 times the u d x already Dutt, which is 51. Thus, we have DPT in 1920 19 52,000. By funding in t v equals 20 5100 as the numbers given here.