A continuous stream of income is modeled by the differential equation with initial condition , where is the balance in dollars after years.
(a) How much money is in the account when it is growing at the rate of per year?
(b) Determine how fast the balance is growing when is in the account.
© Find a formula for .
(d) What is the value of this income stream after 8 years?

Ask by Bates Powers. in the United States

Mar 23,2025

Question

A continuous stream of income is modeled by the differential equation with initial condition , where is the balance in dollars after years.
(a) How much money is in the account when it is growing at the rate of per year?
(b) Determine how fast the balance is growing when is in the account.
© Find a formula for .
(d) What is the value of this income stream after 8 years?

Ask by Bates Powers. in the United States

Mar 23,2025

UpStudy AI · Accepted Answer

To solve the problem, we will follow these steps:

1. **Solve the differential equation** to find the function $$ P(t) $$.
2. **Use the initial condition** to determine the constant of integration.
3. **Answer each part of the question** based on the derived function.

### Step 1: Solve the Differential Equation

The given differential equation is:

$$
P'(t) = 0.06 P(t) + 2400
$$

This is a first-order linear ordinary differential equation. We can solve it using an integrating factor.

The standard form is:

$$
P' - 0.06 P = 2400
$$

The integrating factor $$ \mu(t) $$ is given by:

$$
\mu(t) = e^{\int -0.06 \, dt} = e^{-0.06t}
$$

Multiplying through by the integrating factor:

$$
e^{-0.06t} P' - 0.06 e^{-0.06t} P = 2400 e^{-0.06t}
$$

The left-hand side can be rewritten as the derivative of a product:

$$
\frac{d}{dt}(e^{-0.06t} P) = 2400 e^{-0.06t}
$$

Now, we integrate both sides:

$$
\int \frac{d}{dt}(e^{-0.06t} P) \, dt = \int 2400 e^{-0.06t} \, dt
$$

Let's compute the right-hand side integral.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int 2400e^{-0.06t} dt$$
- step1: Convert the expressions:
  $$\int 2400e^{-\frac{3}{50}t} dt$$
- step2: Use properties of integrals:
  $$2400\times \int e^{-\frac{3}{50}t} dt$$
- step3: Evaluate the integral:
  $$2400\times \frac{1}{-\frac{3}{50}}\times e^{-\frac{3}{50}t}$$
- step4: Simplify:
  $$2400\left(-\frac{50}{3e^{\frac{3}{50}t}}\right)$$
- step5: Multiply the terms:
  $$-2400\times \frac{50}{3e^{\frac{3}{50}t}}$$
- step6: Reduce the fraction:
  $$-800\times \frac{50}{e^{\frac{3}{50}t}}$$
- step7: Multiply the terms:
  $$-\frac{800\times 50}{e^{\frac{3}{50}t}}$$
- step8: Multiply the terms:
  $$-\frac{40000}{e^{\frac{3}{50}t}}$$
- step9: Simplify the expression:
  $$-40000e^{-\frac{3}{50}t}$$
- step10: Add the constant of integral C:
  $$-40000e^{-\frac{3}{50}t} + C, C \in \mathbb{R}$$
The integral of the right-hand side is:

$$
\int 2400 e^{-0.06t} \, dt = -40000 e^{-0.06t} + C
$$

Thus, we have:

$$
e^{-0.06t} P = -40000 e^{-0.06t} + C
$$

Multiplying through by $$ e^{0.06t} $$:

$$
P(t) = -40000 + Ce^{0.06t}
$$

### Step 2: Use the Initial Condition

We know that $$ P(0) = 0 $$:

$$
0 = -40000 + C e^{0}
$$

This simplifies to:

$$
0 = -40000 + C \implies C = 40000
$$

Thus, the function $$ P(t) $$ becomes:

$$
P(t) = -40000 + 40000 e^{0.06t}
$$

### Step 3: Answer Each Part of the Question

#### (a) How much money is in the account when it is growing at the rate of $$ \$10,000 $$ per year?

We need to find $$ P(t) $$ when $$ P'(t) = 10000 $$:

$$
P'(t) = 0.06 P(t) + 2400
$$

Setting $$ P'(t) = 10000 $$:

$$
10000 = 0.06 P(t) + 2400
$$

Solving for $$ P(t) $$:

$$
10000 - 2400 = 0.06 P(t) \implies 7600 = 0.06 P(t) \implies P(t) = \frac{7600}{0.06}
$$

Let's calculate $$ P(t) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{7600}{0.06}$$
- step1: Convert the expressions:
  $$\frac{7600}{\frac{3}{50}}$$
- step2: Multiply by the reciprocal:
  $$7600\times \frac{50}{3}$$
- step3: Multiply:
  $$\frac{7600\times 50}{3}$$
- step4: Multiply:
  $$\frac{380000}{3}$$
Calculating $$ P(t) $$:

$$
P(t) = \frac{7600}{0.06} = \frac{380000}{3} \approx 126666.67
$$

So, when the account is growing at the rate of $$ \$10,000 $$ per year, the balance is approximately $$ \$126,666.67 $$.

#### (b) Determine how fast the balance is growing when $$ \$10,000 $$ is in the account.

We need to find $$ P'(t) $$ when $$ P(t) = 10000 $$:

Using the equation:

$$
P'(t) = 0.06 P(t) + 2400
$$

Substituting $$ P(t) = 10000 $$:

$$
P'(t) = 0.06 \times 10000 + 2400
$$

Let's calculate $$ P'(t) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$0.06\times 10000+2400$$
- step1: Multiply the numbers:
  $$600+2400$$
- step2: Add the numbers:
  $$3000$$
Calculating $$ P'(t) $$:

$$
P'(t) = 0.06 \times 10000 + 2400 = 3000
$$

Thus, when there is $$ \$10,000 $$ in the account, the balance is growing at a rate of $$ \$3,000 $$ per year.

#### (c) Find a formula for $$ P(t) $$.

We already derived the formula for $$ P(t) $$:

$$
P(t) = -40000 + 40000 e^{0.06t}
$$

#### (d) What is the value of this income stream after 8 years?

We will calculate $$ P(8) $$:

$$
P(8) = -40000 + 40000 e^{0.06 \times 8}
$$

Let's compute $$ P(8) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$-40000+40000e^{0.06\times 8}$$
- step1: Multiply the numbers:
  $$-40000+40000e^{0.48}$$
- step2: Convert the expressions:
  $$-40000+40000e^{\frac{12}{25}}$$
- step3: Simplify:
  $$-40000+40000\sqrt[25]{e^{12}}$$
Calculating $$ P(8) $$:

$$
P(8) = -40000 + 40000 e^{0.48} \approx -40000 + 40000 \sqrt[25]{e^{12}}
$$

To get a numerical value, we can compute $$ e^{0.48} $$. Let's calculate that.
Calculate the value by following steps:
- step0: Calculate:
  $$e^{0.48}$$
- step1: Convert the expressions:
  $$e^{\frac{12}{25}}$$
- step2: Transform the expression:
  $$\sqrt[25]{e^{12}}$$
The value of $$ e^{0.48} $$ is approximately $$ \sqrt[25]{e^{12}} $$.

Now, substituting this back into the equation for $$ P(8) $$:

$$
P(8) \approx -40000 + 40000 \times 1.616
$$

Calculating this gives us:

$$
P(8) \approx -40000 + 64640 \approx 24640
$$

Thus, the value of this income stream after 8 years is approximately $$ \$24,640 $$.

### Summary of Answers:
(a) $$ \$126,666.67 $$  
(b) $$ \$3,000 $$ per year  
(c) $$ P(t) = -40000 + 40000 e^{0.06t} $$  
(d) $$ \$24,640 $$ after 8 years
(a) The balance is approximately \$126,666.67 when it's growing at \$10,000 per year. (b) The balance is growing at \$3,000 per year when there is \$10,000 in the account. (c) The formula for $$ P(t) $$ is $$ P(t) = -40000 + 40000 e^{0.06t} $$. (d) The value of the income stream after 8 years is approximately \$24,640.

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