the equation a(t)=2000e^{0.06t} gives the balance after t years of an initial investment of 2000 dollars…

Explanation:

Step1: Apply chain - rule for differentiation

The derivative of $y = e^{u}$ with respect to $t$ is $\frac{dy}{dt}=e^{u}\frac{du}{dt}$. Here $A(t)=2000e^{0.06t}$, where $u = 0.06t$. So $\frac{dA}{dt}=2000\times e^{0.06t}\times0.06$.

Step2: Simplify the derivative formula

$\frac{dA}{dt}=120e^{0.06t}$.

Step3: Find $A^{\prime}(6)$

Substitute $t = 6$ into $\frac{dA}{dt}$. So $A^{\prime}(6)=120e^{0.06\times6}=120e^{0.36}$. Using a calculator, $A^{\prime}(6)=120\times1.433329 = 172.00$. The units of $A^{\prime}(t)$ are dollars per year, so $A^{\prime}(6)$ is $172.00$ dollars per year. It represents the rate of change of the balance of the investment at $t = 6$ years.

Step4: Calculate the actual change

$A(7)=2000e^{0.06\times7}=2000e^{0.42}\approx2000\times1.521962 = 3043.92$. $A(6)=2000e^{0.06\times6}=2000e^{0.36}\approx2000\times1.433329 = 2866.66$. The actual change $\Delta A=A(7)-A(6)=3043.92 - 2866.66=177.26$. The approximation is $172$ and the actual change is $177.26$. The difference is $177.26 - 172=5.26$.

Answer:

a. $\frac{dA}{dt}=120e^{0.06t}$ b. $A^{\prime}(6)=172.00$ dollars per year. It represents the rate of change of the investment balance at $t = 6$ years. c. The difference between the actual change ($177.26$) and the approximation ($172$) is $5.26$.