QuestionLet $A(t)=3000 e^{0.03 t}$ be the balance in a savings account after $t$ years. Complete parts (a) through ( $f$ ) below. (Kound to the nearest cent as needed.) (d) What differential equation is satisfied by $y=A(t)$ ?
The differential equation that is satisfied by $y=A(t)$ is $A^{\prime}(t)=0.03 A(t)$. (e) Use the results from parts (c) and (d) to determine how fast the balance is growing after 11 years.
The balance is growing at approximately $\$ \square$ per year after 11 years. (Round to the nearest cent as needed.)

Studdy Solution
Evaluate $A'(t)$ at $t = 11$ to determine how fast the balance is growing after 11 years:
$$A'(11) = 90 e^{0.03 \times 11}$$
Calculate the exponent:
$$0.03 \times 11 = 0.33$$
Now, compute $e^{0.33}$ using a calculator:
$$e^{0.33} \approx 1.392$$
Finally, calculate $A'(11)$:
$$A'(11) = 90 \times 1.392$$
$$A'(11) \approx 125.28$$
Therefore, the balance is growing at approximately \$125.28 per year after 11 years.
The balance is growing at approximately $\boxed{125.28}$ per year after 11 years.