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Sagot :
a. If G(t) = a (1 + r)ᵗ, then G(0) = a corresponds to the starting area of the glacier, which is given to be 142 acres. So a = 142.
The area of the glacier shrinks by 4.4% each year. This means that after 1 year, the area is reduced to
142 - (4.4% of 142) = 142 (1 - 0.044) ≈ 135.75 acres
After another year,
135.75 - (4.4% of 135.75) = 135.75 (1 - 0.044) ≈ 129.78 acres
or equivalently, 142 (1 - 0.044)² acres.
And so on. After t years, the glacier would have an area of
G(t) = 142 (1 - 0.044)ᵗ ⇒ G(t) = 142 × 0.956ᵗ
b. If the year 2007 corresponds to t = 0, then 2012 refers to t = 5. Then the area of the glacier is
G(5) = 142 × 0.956⁵ ⇒ G(5) ≈ 113.39 acres
c. Simply take the difference between G(5) and G(0) :
G(5) - G(0) ≈ 142 - 113.39 ≈ 28.61 acres
d. The average rate of change of G(t) from 2007 to 2012 is given by the difference quotient of G(t) over the interval 0 ≤ t ≤ 5 :
[tex]ARC_{[2007,2012]} = \dfrac{G(5) - G(0)}{5 - 0} \approx \dfrac{113.39 - 142}5[/tex]
so that ARC ≈ -5.72 acres/year. This rate tells us that a little less than 6 acres of glacial ice is lost each year, based on the reduction over the first 5 years.
e. The year 2017 refers to t = 10, so now we compute
[tex]ARC_{[2012,2017]} = \dfrac{G(10) - G(5)}{10 - 5} \approx \dfrac{90.55-113.39}5[/tex]
which comes out to about ARC ≈ -4.57 acres/year. Since this rate is smaller than the ARC between 2007 and 2012, this means that the glacial ice is disappearing at a slower rate in later years.
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