Questions: MATH 110 Test #2 Sections 4.1 - 4.5, 5.1 - 5.3 1. There were 210 million licensed drivers in the US at t=0 and 222 million at t=6. Let N represent the number of licensed drivers in the US as a function of time t in years. (a) Model N as an exponential function of year t. (b) How many licensed drivers in the US are predicted at t=15 with the exponential model? 2. Calculate the balance after 2 years if an account initially containing 500 earns (a) nominal rate 3.02% compounded continuously, (b) nominal rate 3.3% compounded quarterly.

Solution Steps

1. (a) To model \( N \) as an exponential function of year \( t \), we can use the general form of an exponential function \( N(t) = N_0 e^{kt} \). We need to find the constants \( N_0 \) and \( k \) using the given data points: \( N(0) = 210 \) million and \( N(6) = 222 \) million.

2. (b) Once we have the exponential model from part (a), we can use it to predict the number of licensed drivers at \( t = 15 \).

3. (a) For continuous compounding, we use the formula \( A = P e^{rt} \), where \( P \) is the principal amount, \( r \) is the nominal interest rate, and \( t \) is the time in years.

4. (b) For quarterly compounding, we use the formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where \( n \) is the number of compounding periods per year.

To model the number of licensed drivers \( N(t) \) as an exponential function, we start with the general form:

\[ N(t) = N_0 e^{kt} \]

Given that \( N(0) = 210 \) million and \( N(6) = 222 \) million, we can find the constant \( k \) using the equation:

\[ k = \frac{\ln\left(\frac{N(6)}{N(0)}\right)}{t_2 - t_1} = \frac{\ln\left(\frac{222}{210}\right)}{6 - 0} \approx 0.00926164185913513 \]

Thus, the exponential model becomes:

\[ N(t) = 210 e^{0.00926164185913513t} \]

Using the model derived in Step 1, we can predict the number of licensed drivers at \( t = 15 \):

\[ N(15) = 210 e^{0.00926164185913513 \cdot 15} \approx 241.2978731847808 \text{ million} \]

Rounding to four significant digits, we have:

\[ N(15) \approx 241.30 \text{ million} \]

For an account with an initial amount \( P = 500 \) dollars earning a nominal rate of \( r = 3.02\% \) compounded continuously, the balance after \( t = 2 \) years is given by:

\[ A = P e^{rt} = 500 e^{0.0302 \cdot 2} \approx 531.1306830611145 \]

Rounding to four significant digits, we find:

\[ A \approx 531.13 \]

For the same initial amount \( P = 500 \) dollars earning a nominal rate of \( r = 3.3\% \) compounded quarterly, the balance after \( t = 2 \) years is calculated as follows:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} = 500 \left(1 + \frac{0.033}{4}\right)^{4 \cdot 2} \approx 533.9687606496701 \]

Rounding to four significant digits, we have:

\[ A \approx 533.97 \]

Final Answer