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

STEP 1

1. The principal amount deposited is $4000.
2. The interest rate is 4.5% per annum, compounded continuously.
3. The formula for continuous compounding is $ A(t) = P e^{rt} $, where $ P $ is the principal, $ r $ is the rate, and $ t $ is time in years.

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 $8000.<br />5. Calculate the rate of growth of the balance when it reaches$ 8000.

STEP 3

The formula for continuous compounding is $A(t) = P e^{rt}$ .
Given $P = 4000$ and $r = 0.045$ , substitute these values into the formula:
$A(t) = 4000 e^{0.045t}$

STEP 4

The differential equation for $A(t)$ is derived from the formula for continuous compounding.
The rate of change of the balance $A(t)$ with respect to time $t$ is proportional to the current balance, which gives us:
$A'(t) = 0.045 A(t)$

STEP 5

To find the balance after 9 years, substitute $t = 9$ into the formula:
$A(9) = 4000 e^{0.045 \times 9}$
Calculate the value:
$A(9) = 4000 e^{0.405} \approx 4000 \times 1.499 \approx 5996$
The balance after 9 years is approximately:
$\$5996.00$

STEP 6

To find when the balance will reach $8000, set $ A(t) = 8000 $ and solve for $ t $:
$8000 = 4000 e^{0.045t}$
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 balance will reach $8000 in approximately 15.4 years.

STEP 7

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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Studdy Solution

STEP 1

1. The principal amount deposited is $4000.
2. The interest rate is 4.5% per annum, compounded continuously.
3. The formula for continuous compounding is \( A(t) = P e^{rt} \), where \( P \) is the principal, \( r \) is the rate, and \( t \) is time in years.

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 $8000.<br />5. Calculate the rate of growth of the balance when it reaches$ 8000.

STEP 3

The formula for continuous compounding is $A(t) = P e^{rt}$ .
Given $P = 4000$ and $r = 0.045$ , substitute these values into the formula:
$A(t) = 4000 e^{0.045t}$

STEP 4

The differential equation for $A(t)$ is derived from the formula for continuous compounding.
The rate of change of the balance $A(t)$ with respect to time $t$ is proportional to the current balance, which gives us:
$A'(t) = 0.045 A(t)$

STEP 5

To find the balance after 9 years, substitute $t = 9$ into the formula:
$A(9) = 4000 e^{0.045 \times 9}$
Calculate the value:
$A(9) = 4000 e^{0.405} \approx 4000 \times 1.499 \approx 5996$
The balance after 9 years is approximately:
$\$5996.00$

STEP 6

STEP 7

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Studdy Solution

STEP 1

1. The principal amount deposited is $4000.2. The interest rate is 4.5% per annum, compounded continuously.3. The formula for continuous compounding is \( A(t) = P e^{rt} \), where \( P \) is the principal, \( r \) is the rate, and \( t \) is time in years.

STEP 2

STEP 3

The formula for continuous compounding is A(t)=Pert A(t) = P e^{rt} A(t)=Pert.Given P=4000 P = 4000 P=4000 and r=0.045 r = 0.045 r=0.045, substitute these values into the formula:A(t)=4000e0.045t A(t) = 4000 e^{0.045t} A(t)=4000e0.045t

STEP 4

The differential equation for A(t) A(t) A(t) is derived from the formula for continuous compounding.The rate of change of the balance A(t) A(t) A(t) with respect to time t t t is proportional to the current balance, which gives us:A′(t)=0.045A(t) A'(t) = 0.045 A(t) A′(t)=0.045A(t)

STEP 5

STEP 6

STEP 7

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1. The principal amount deposited is $4000.
2. The interest rate is 4.5% per annum, compounded continuously.
3. The formula for continuous compounding is \( A(t) = P e^{rt} \), where \( P \) is the principal, \( r \) is the rate, and \( t \) is time in years.

The formula for continuous compounding is $A(t) = P e^{rt}$ .
Given $P = 4000$ and $r = 0.045$ , substitute these values into the formula:
$A(t) = 4000 e^{0.045t}$

The differential equation for $A(t)$ is derived from the formula for continuous compounding.
The rate of change of the balance $A(t)$ with respect to time $t$ is proportional to the current balance, which gives us:
$A'(t) = 0.045 A(t)$