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

Explanation:

Step1: Differentiate $A(t)$

We know that if $y = ae^{bx}$, then $y^\prime=abe^{bx}$. Here $A(t)=2000e^{0.06t}$, so $\frac{dA}{dt}=2000\times0.06e^{0.06t}=120e^{0.06t}$.

Step2: Evaluate $\frac{dA}{dt}$ at $t = 6$

Substitute $t = 6$ into $\frac{dA}{dt}=120e^{0.06t}$. We get $A^\prime(6)=120e^{0.06\times6}=120e^{0.36}\approx172$. The unit of $A^\prime(6)$ is dollars per year, which means that at $t = 6$ years, the balance of the investment is increasing at a rate of approximately 172 dollars per year.

Step3: Find the actual change

First, find $A(7)$ and $A(6)$. $A(7)=2000e^{0.06\times7}=2000e^{0.42}$ and $A(6)=2000e^{0.06\times6}=2000e^{0.36}$. Then the actual change $\Delta A=A(7)-A(6)=2000e^{0.42}-2000e^{0.36}=2000(e^{0.42}-e^{0.36})\approx2000(1.5219 - 1.4333)=2000\times0.0886 = 177.2$.

Step4: Calculate the percentage - error

The approximation is 172 and the actual change is 177.2. The percentage - error $=\left|\frac{172 - 177.2}{177.2}\right|\times100=\left|\frac{- 5.2}{177.2}\right|\times100\approx2.93%$.

Answer:

a. $\frac{dA}{dt}=120e^{0.06t}$ b. $A^\prime(6)\approx172$ dollars per year. It means that at $t = 6$ years, the balance of the investment is increasing at a rate of approximately 172 dollars per year. c. The percentage - error between the approximation of 172 and the actual change of approximately 177.2 is about $2.93%$.