in 2011 a countrys federal receipts (money taken in) totaled $2.12 trillion. in 2013, total federal receipts…

in 2011 a countrys federal receipts (money taken in) totaled $2.12 trillion. in 2013, total federal receipts were $2.71 trillion. assume that the growth of federal receipts, f, can be modeled by an exponential function and use 2011 as the base year (t = 0). a) find the growth rate k to six decimal places, and write the exponential function f(t), for total receipts in trillions of dollars. b) estimate total federal receipts in 2015. c) when will total federal receipts be $12 trillion?

in 2011 a countrys federal receipts (money taken in) totaled $2.12 trillion. in 2013, total federal receipts were $2.71 trillion. assume that the growth of federal receipts, f, can be modeled by an exponential function and use 2011 as the base year (t = 0). a) find the growth rate k to six decimal places, and write the exponential function f(t), for total receipts in trillions of dollars. b) estimate total federal receipts in 2015. c) when will total federal receipts be $12 trillion?

Answer

Explanation:

Step1: Recall exponential - growth formula

The general form of an exponential - growth function is $F(t)=F_0e^{kt}$, where $F_0$ is the initial amount, $k$ is the growth rate, and $t$ is the time. Given that in 2011 ($t = 0$), $F_0=2.12$ (in trillions of dollars). In 2013 ($t = 2$), $F(2)=2.71$. Substitute these values into the formula: $F(2)=F_0e^{k\times2}$.

Step2: Solve for $k$

Substitute $F_0 = 2.12$ and $F(2)=2.71$ into $F(2)=F_0e^{2k}$: [ \begin{align*} 2.71&=2.12e^{2k}\ \frac{2.71}{2.12}&=e^{2k}\ \ln\left(\frac{2.71}{2.12}\right)&=\ln(e^{2k})\ \ln\left(\frac{2.71}{2.12}\right)&=2k\ k&=\frac{1}{2}\ln\left(\frac{2.71}{2.12}\right)\ k&=\frac{1}{2}\ln(1.278302)\ k&\approx0.121229 \end{align*} ] So the exponential function is $F(t)=2.12e^{0.121229t}$.

Step3: Estimate total federal receipts in 2015

In 2015, $t = 4$. Substitute $t = 4$ into $F(t)=2.12e^{0.121229t}$: [ \begin{align*} F(4)&=2.12e^{0.121229\times4}\ &=2.12e^{0.484916}\ &=2.12\times1.623277\ &\approx3.441347 \end{align*} ]

Step4: Find when total federal receipts will be $12$ trillion

Set $F(t)=12$ in $F(t)=2.12e^{0.121229t}$: [ \begin{align*} 12&=2.12e^{0.121229t}\ \frac{12}{2.12}&=e^{0.121229t}\ \ln\left(\frac{12}{2.12}\right)&=\ln(e^{0.121229t})\ \ln\left(\frac{12}{2.12}\right)&=0.121229t\ t&=\frac{\ln\left(\frac{12}{2.12}\right)}{0.121229}\ t&=\frac{\ln(5.660377)}{0.121229}\ t&=\frac{1.734777}{0.121229}\ t&\approx14.31 \end{align*} ] Since $t = 0$ corresponds to 2011, the year will be approximately $2011 + 14.31\approx2025$.

Answer:

a) $k\approx0.121229$, $F(t)=2.12e^{0.121229t}$ b) Approximately $$3.44$ trillion c) Approximately in 2025