IDNLearn.com: Your trusted source for accurate and reliable answers. Ask your questions and receive comprehensive, trustworthy responses from our dedicated team of experts.

A small country emits 80,000 kilotons of carbon dioxide per year. In a recent global
agreement, the country agreed to cut its carbon emissions by 2.6% per year for the
next 11 years. In the first year of the agreement, the country will keep its emissions at
80,000 kilotons and the emissions will decrease 2.6% in each successive year. How
many total kilotons of carbon dioxide would the country emit over the course of the
11 year period, to the nearest whole number?


Sagot :

Using an exponential function and integration, it is found that over the entire 11 year period, the country will emit 763,968 kilotons of carbon dioxide.

Exponential function:

A decaying exponential function is modeled by:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which:

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

In this problem:

  • In the first year, the country emits 80,000 kilotons, hence [tex]A(0) = 80000[/tex].
  • Emissions will decrease 2.6% in each successive year, hence [tex]r = 0.026[/tex]

Then, the equation for the amount emitted each year is:

[tex]A(t) = A(0)(1 - r)^t[/tex]

[tex]A(t) = 80000(1 - 0.026)^t[/tex]

[tex]A(t) = 80000(0.974)^t[/tex]

The total amount emitted over the course of the 11 year period is:

[tex]T = \int_{0}^{11} A(t) dt[/tex]

Hence:

[tex]T = \int_{0}^{11} 80000(0.974)^t dt[/tex]

[tex]T = \frac{80000(0.974)^t}{\ln{0.974}}|_{t = 0}^{t = 11}[/tex]

Applying the Fundamental Theorem of Calculus:

[tex]T = \frac{80000(0.974)^{11}}{\ln{0.974}} - \frac{80000(0.974)^0}{\ln{0.974}} = \frac{80000}{\ln{0.974}}((0.974)^{11} - 1) = 763968[/tex]

Over the entire 11 year period, the country will emit 763,968 kilotons of carbon dioxide.

You can learn more about exponential functions at https://brainly.com/question/25537936

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.