a particular countrys exports of goods are increasing exponentially. the value of the exports, ( t ) years…
a particular countrys exports of goods are increasing exponentially. the value of the exports, ( t ) years after 2008, can be approximated by ( v(t)=1.6e^{0.041t} ) where ( t = 0 ) corresponds to 2008 and ( v ) is in billions of dollars.\na) estimate the value of the countrys exports in 2008 and 2022.\nb) what is the doubling - time for the value of the countrys exports?\na) the value of the countrys exports in 2008 is $ $ billion. (simplify your answer. round to the nearest tenth as needed. do not include the $ symbol in your answer.)
Answer
Explanation:
Step1: Find value in 2008
Since (t = 0) corresponds to 2008 and (V(t)=1.6e^{0.041t}), substitute (t = 0) into the formula. When (t = 0), (V(0)=1.6e^{0.041\times0}=1.6e^{0}). Since (e^{0}=1), then (V(0)=1.6).
Step2: Find value in 2022
The number of years from 2008 to 2022 is (t=2022 - 2008=14). Substitute (t = 14) into (V(t)=1.6e^{0.041t}), we get (V(14)=1.6e^{0.041\times14}=1.6e^{0.574}). Using a calculator, (e^{0.574}\approx1.775), so (V(14)=1.6\times1.775 = 2.84).
Step3: Find the doubling - time
We want to find (t) when (V(t)=2\times1.6 = 3.2). Set (V(t)=3.2) in the equation (V(t)=1.6e^{0.041t}), we have (3.2=1.6e^{0.041t}). Divide both sides by 1.6: (\frac{3.2}{1.6}=e^{0.041t}), so (2 = e^{0.041t}). Take the natural logarithm of both sides: (\ln(2)=\ln(e^{0.041t})). Since (\ln(e^{x})=x), we get (\ln(2)=0.041t). Solve for (t): (t=\frac{\ln(2)}{0.041}). Using a calculator, (\ln(2)\approx0.693), so (t=\frac{0.693}{0.041}\approx16.9).
Answer:
a) 1.6 b) 2.8 c) 16.9