Question

question 9
let (y(t)) represent your bank account balance, in dollars, after (t) years. suppose you start with $50000 in the account. each year the account earns 6% interest, and you deposit $3000 into the account.
this can be modeled with the differential equation:
\\\frac{dy}{dt} = 0.06y + 3000\\
\\y(0) = 50000\\
solve this differential equation for (y(t))
(y(t) =)

question 10
let (y(t)) represent your retirement account balance, in dollars, after (t) years. each year the account earns 9% interest, and you deposit 10% of your annual income. your current annual income is $30000, but it is growing at a continuous rate of 3% per year.
write the differential equation modeling this situation.
\\\frac{dy}{dt} =\\

question 11
a tank contains (50\text{ kg}) of salt and (2000\text{ l}) of water. a solution of a concentration (0.0125\text{ kg}) of salt per liter enters a tank at the rate (5\text{ l/min}). the solution is mixed and drains from the tank at the same rate.
(a) what is the concentration of our solution in the tank initially?
concentration = (kg/l)
(b) find the amount of salt in the tank after 4 hours.
amount = (kg)
(c) find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/l)

Explanation:

🆕 New Concept Discovered: Separable Differential Equations & Mixing Models
Solving first-order differential equations and setting up rate-of-change models.

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Question 9

Step 1: Separate the variables

We start with the given differential equation:
\[ \frac{dy}{dt} = 0.06y + 3000 \]

To solve this, we group all terms involving \( y \) on one side and terms involving \( t \) on the other:
\[ \frac{dy}{0.06y + 3000} = dt \]

We can factor out \( 0.06 \) from the denominator to make integration simpler:
\[ 0.06y + 3000 = 0.06(y + 50000) \]
\[ \frac{dy}{0.06(y + 50000)} = dt \implies \frac{dy}{y + 50000} = 0.06 \, dt \]

Step 2: Integrate both sides

Integrate both sides of the equation:
\[ \int \frac{1}{y + 50000} \, dy = \int 0.06 \, dt \]
\[ \ln|y + 50000| = 0.06t + C \]

Exponentiate both sides to solve for \( y \):
\[ |y + 50000| = e^{0.06t + C} = e^C \cdot e^{0.06t} \]
\[ y + 50000 = A e^{0.06t} \quad (\text{where } A = \pm e^C) \]
\[ y(t) = A e^{0.06t} - 50000 \]

Step 3: Apply the initial condition

We are given the initial condition \( y(0) = 50000 \):
\[ 50000 = A e^{0} - 50000 \]
\[ 50000 = A - 50000 \implies A = 100000 \]

Thus, the solution is:
\[ y(t) = 100000e^{0.06t} - 50000 \]

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Question 10

Step 1: Identify the rate of change components

The rate of change of the retirement account balance, \(\frac{dy}{dt}\), is determined by two factors:
1. **Interest earned**: The account earns \( 9\% \) interest continuously on its current balance \( y \):
\[ \text{Interest Rate} = 0.09y \]
2. **Deposits**: You deposit \( 10\% \) of your annual income. Your current annual income is \(\$30000\), but it grows continuously at a rate of \( 3\% \) per year. This means your income at time \( t \) is modeled by \( 30000e^{0.03t} \).
\[ \text{Deposit Rate} = 0.10 \times 30000e^{0.03t} = 3000e^{0.03t} \]

Step 2: Write the differential equation

Combining both components gives:
\[ \frac{dy}{dt} = 0.09y + 3000e^{0.03t} \]

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Question 11

Step 1: Find the initial concentration

Initially, the tank contains \( 50 \text{ kg} \) of salt in \( 2000 \text{ L} \) of water.
The concentration is:
\[ \text{Concentration} = \frac{\text{Mass of salt}}{\text{Volume of water}} = \frac{50}{2000} = 0.025 \text{ kg/L} \]

Step 2: Set up and solve the differential equation for salt amount

Let \( S(t) \) be the amount of salt (in kg) in the tank at time \( t \) (in minutes).
The rate of change of salt is:
\[ \frac{dS}{dt} = \text{Rate in} - \text{Rate out} \]

* **Rate in**: \( 0.0125 \text{ kg/L} \times 5 \text{ L/min} = 0.0625 \text{ kg/min} \)
* **Rate out**: Since the volume remains constant at \( 2000 \text{ L} \), the concentration leaving is \( \frac{S(t)}{2000} \text{ kg/L} \).
\[ \text{Rate out} = \frac{S(t)}{2000} \times 5 = \frac{5}{2000}S(t) = 0.0025S(t) \]

Thus, the differential equation is:
\[ \frac{dS}{dt} = 0.0625 - 0.0025S \]
\[ \frac{dS}{dt} = -0.0025(S - 25) \]

Separate variables and integrate:
\[ \frac{dS}{S - 25} = -0.0025 \, dt \]
\[ \ln|S - 25| = -0.0025t + C \]
\[ S(t) = 25 + C e^{-0.0025t} \]

Apply the initial condition \( S(0) = 50 \):
\[ 50 = 25 + C \implies C = 25 \]
\[ S(t) = 25 + 25e^{-0.0025t} \]

Step 3: Calculate the salt amount after 4 hours

Note that \( t \) is in minutes. Convert \( 4 \text{ hours} \) to minutes:
\[ t = 4 \times 60 = 240 \text{ minutes} \]

Substitute \( t = 240 \) into the equation:
\[ S(240) = 25 + 25e^{-0.0025 \times 240} \]
\[ S(240) = 25 + 25e^{-0.6} \]
\[ S(240) \approx 25 + 25(0.54881) \approx 25 + 13.72 = 38.72 \text{ kg} \]

Step 4: Find the concentration as time approaches infinity

As \( t \to \infty \), the exponential term \( e^{-0.0025t} \to 0 \).
The amount of salt approaches:
\[ \lim_{t \to \infty} S(t) = 25 \text{ kg} \]

Since the volume remains constant at \( 2000 \text{ L} \), the concentration approaches:
\[ \text{Concentration} = \frac{25}{2000} = 0.0125 \text{ kg/L} \]
*(This matches the concentration of the incoming solution, as expected.)*

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Answer:

Question 9

\( y(t) = 100000e^{0.06t} - 50000 \)

Question 10

\( \frac{dy}{dt} = 0.09y + 3000e^{0.03t} \)

Question 11

(a) \( 0.025 \)
(b) \( 38.72 \)
(c) \( 0.0125 \)