Math / Calculus

QuestionFour thousand dollars is deposited into a savings account at $4.5 \%$ interest compounded continuously. (a) What is the formula for $A(t)$ , the balance after $t$ years? (b) What differential equation is satisfied by $\mathrm{A}(\mathrm{t})$ , the balance after t years? (c) How much money will be in the account after 9 years? (d) When will the balance reach $\$ 8000$ ? (e) How fast is the balance growing when it reaches $\$ 8000$ ? (b) $A^{\prime}(t)=0.045 A$ (c) $\$ \square$ (Round to the nearest cent as needed.)

Studdy Solution
The rate of growth of the balance when it reaches $8000 is given by the derivative:
$A'(t) = 0.045 A(t)$
Substitute $A(t) = 8000$ :
$A'(t) = 0.045 \times 8000 = 360$
The balance is growing at a rate of $360 per year when it reaches$ 8000.
The solutions are: (a) $A(t) = 4000 e^{0.045t}$ (b) $A'(t) = 0.045 A(t)$ (c) $\$5996.00$ (d) Approximately 15.4 years (e) $\$360$ per year

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