if $800 is invested in an account that pays 3% compounded annually, the total amount, a(t), in the account…

if $800 is invested in an account that pays 3% compounded annually, the total amount, a(t), in the account after t years is a(t)=800(1.03)^t. a. find the average rate of change per year of the total amount in the account for the first five years of the investment (from t = 0 to t = 5). b. find the average rate of change per year of the total amount in the account for the second five years of the investment (from t = 5 to t = 10). c. estimate the instantaneous rate of change for t = 5. a. (a(5)/5)-a(0) b. (a(5)-a(0))/5 c. a(5)/a(0) the average rate of change per year of the total amount in the account for the first five years of the investment (from t = 0 to t = 5) is $25.48 per year. (round to the nearest cent as needed.) b. the average rate of change per year of the total amount in the account for the second five years of the investment (from t = 5 to t = 10) is $29.54 per year. (round to the nearest cent as needed.) c. the instantaneous rate of change for t = 5 is about $2 per year. (do not round until the final answer. then round to the nearest cent as needed.)

if $800 is invested in an account that pays 3% compounded annually, the total amount, a(t), in the account after t years is a(t)=800(1.03)^t. a. find the average rate of change per year of the total amount in the account for the first five years of the investment (from t = 0 to t = 5). b. find the average rate of change per year of the total amount in the account for the second five years of the investment (from t = 5 to t = 10). c. estimate the instantaneous rate of change for t = 5. a. (a(5)/5)-a(0) b. (a(5)-a(0))/5 c. a(5)/a(0) the average rate of change per year of the total amount in the account for the first five years of the investment (from t = 0 to t = 5) is $25.48 per year. (round to the nearest cent as needed.) b. the average rate of change per year of the total amount in the account for the second five years of the investment (from t = 5 to t = 10) is $29.54 per year. (round to the nearest cent as needed.) c. the instantaneous rate of change for t = 5 is about $2 per year. (do not round until the final answer. then round to the nearest cent as needed.)

Answer

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. For the function $A(t)=800(1.03)^t$, the average rate of change over the interval $[0,5]$ is $\frac{A(5)-A(0)}{5 - 0}=\frac{A(5)-A(0)}{5}$.

Step2: Calculate $A(0)$ and $A(5)$

When $t = 0$, $A(0)=800(1.03)^0=800$. When $t = 5$, $A(5)=800(1.03)^5=800\times1.159274 = 927.4192$. Then $\frac{A(5)-A(0)}{5}=\frac{927.4192 - 800}{5}=\frac{127.4192}{5}=25.48384\approx25.48$.

Step3: Calculate average rate of change for $[5,10]$

When $t = 10$, $A(10)=800(1.03)^{10}=800\times1.343916379 = 1075.133103$. Then the average rate of change over the interval $[5,10]$ is $\frac{A(10)-A(5)}{10 - 5}=\frac{1075.133103-927.4192}{5}=\frac{147.713903}{5}=29.5427806\approx29.54$.

Step4: Recall instantaneous - rate - of change (derivative)

The derivative of $A(t)=800(1.03)^t$ with respect to $t$ is $A^\prime(t)=800\times(1.03)^t\times\ln(1.03)$. When $t = 5$, $A^\prime(5)=800\times(1.03)^5\times\ln(1.03)=800\times1.159274\times0.0295588 = 26.99\approx27.00$.

Answer:

a. The average rate of change per year for the first five - year period is $$25.48$ per year. b. The average rate of change per year for the second five - year period is $$29.54$ per year. c. The instantaneous rate of change at $t = 5$ is approximately $$27.00$ per year.