Question

an investment of $10,000 earns interest at an annual rate of 8% compounded continuously. complete parts (a) and (b) below. click the icon to view the derivatives of exponential and logarithmic functions. (a) find the instantaneous rate of change of the amount in the account after 2 years. $ (round to two decimal places as needed).

Explanation:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A(t)=P\cdot e^{rt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. Here, $P = 10000$, $r=0.08$, so $A(t)=10000e^{0.08t}$.

Step2: Differentiate the function

The derivative of $A(t)$ with respect to $t$ using the chain - rule. If $y = e^{u}$ and $u = 0.08t$, then $\frac{dy}{du}=e^{u}$ and $\frac{du}{dt}=0.08$. So, $A^\prime(t)=\frac{d}{dt}(10000e^{0.08t})=10000\times0.08e^{0.08t}=800e^{0.08t}$.

Step3: Evaluate the derivative at $t = 2$

Substitute $t = 2$ into $A^\prime(t)$. $A^\prime(2)=800e^{0.08\times2}=800e^{0.16}$.
Using a calculator, $e^{0.16}\approx1.17351$, so $A^\prime(2)=800\times1.17351 = 938.81$.

Answer:

$938.81$