The amount of money, $A(t)$, in a savings account that pays $6 \%$ interest, compounded quarterly for $t$ years, when an initial investment of $\$ 8100$ is made, is given by $A(t)=8100(1.015)^{4 t}$.
Find $\frac{A(2)-A(1)}{2-1}$
Interpret the result
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Answer
\(\boxed{527.55}\) is the rate of change of the amount of money in the account with respect to time, between the first and second year. This means that the amount of money in the account increased by about \$527.55 from the first year to the second year.
Step 1 ：Substitute \(t=1\) into the formula \(A(t)=8100(1.015)^{4 t}\) to get \(A(1)\)
Step 2 ：Substitute \(t=2\) into the formula \(A(t)=8100(1.015)^{4 t}\) to get \(A(2)\)
Step 3 ：Subtract \(A(1)\) from \(A(2)\) to get the difference
Step 4 ：Divide the difference by the difference in time (which is 1 year) to get the rate of change
Step 5 ：\(\boxed{527.55}\) is the rate of change of the amount of money in the account with respect to time, between the first and second year. This means that the amount of money in the account increased by about \$527.55 from the first year to the second year.
