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 $A(t)$ , 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 $\$ 8000$ ? (c) $\$ 5997.21$ (Round to the nearest cent as needed.) (d) After $\square$ years the balance will reach $\$ 8000$ . (Round to one decimal place as needed.)

Studdy Solution

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 $8000.<br />5. Calculate the growth rate when the balance is$ 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 $t$ : $t = \frac{\ln(2)}{0.045} \approx \frac{0.693}{0.045} \approx 15.4$

STEP 7

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.

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Studdy Solution

STEP 1

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 $8000.<br />5. Calculate the growth rate when the balance is$ 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 $t$ : $t = \frac{\ln(2)}{0.045} \approx \frac{0.693}{0.045} \approx 15.4$

STEP 7

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Studdy Solution

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

STEP 3

STEP 4

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

STEP 5

STEP 6

STEP 7

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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: $\frac{dA}{dt} = 0.045A$