Math Snap

STEP 1

1. The initial deposit is $4000.
2. The interest rate is $4.5\%$ compounded continuously.
3. The formula for continuous compounding is known.
4. We need to solve for multiple parts: formula, differential equation, future value, time to reach a certain balance, and growth rate at a specific balance.

STEP 2

1. Derive the formula for $A(t)$.
2. Determine the differential equation for $A(t)$.
3. Calculate the balance after 9 years.
4. Determine when the balance will reach $8000.
5. Calculate the growth rate when the balance is $8000.

STEP 3

The formula for the balance with continuous compounding is given by:
$$A(t) = P e^{rt}$$ where $P$ is the principal amount, $r$ is the interest rate, and $t$ is the time in years.
For this problem, $P = 4000$ and $r = 0.045$.
Thus, the formula becomes:
$$A(t) = 4000 e^{0.045t}$$

STEP 4

The differential equation for $A(t)$ is derived from the fact that the rate of change of the balance is proportional to the current balance:
$$\frac{dA}{dt} = rA$$ Substituting the given interest rate:
$$\frac{dA}{dt} = 0.045A$$

STEP 5

To find the balance after 9 years, substitute $t = 9$ into the formula:
$$A(9) = 4000 e^{0.045 \times 9}$$ Calculate:
$$A(9) = 4000 e^{0.405} \approx 4000 \times 1.499 \approx 5997.21$$

STEP 6

To find when the balance reaches $8000, solve the equation:
$$8000 = 4000 e^{0.045t}$$ Divide both sides by 4000:
$$2 = e^{0.045t}$$ Take the natural logarithm of both sides:
$$\ln(2) = 0.045t$$ Solve for $t$:
$$t = \frac{\ln(2)}{0.045} \approx \frac{0.693}{0.045} \approx 15.4$$

The growth rate when the balance is $8000 is given by the derivative:
$$\frac{dA}{dt} = 0.045A$$ Substitute $A = 8000$:
$$\frac{dA}{dt} = 0.045 \times 8000 = 360$$ The solutions are:
(a) $A(t) = 4000 e^{0.045t}$
(b) $\frac{dA}{dt} = 0.045A$
(c) $5997.21
(d) After $15.4$ years the balance will reach $8000.
(e) The balance is growing at $360 per year when it reaches$8000.