Business Calculus Study Guide: Exponential Growth, Decay, Derivatives, and Optimization

Q1. After t hours there are P(t) cells present in a culture, where P(t) = 600e0.08t.

Background

Topic: Exponential Growth Models

This question tests your understanding of exponential growth, initial value, and growth constant in population models.

Key Terms and Formulas

• Exponential growth formula: $P(t) = P_0 e^{kt}$

• $P_0$ = initial population

• $k$ = growth constant

• $t$ = time (hours)

Step-by-Step Guidance

1. Identify the initial population by evaluating $P(0)$.

2. Determine the growth constant $k$ by comparing the formula to $P(t) = P_0 e^{kt}$.

3. To find when the cell count doubles, set $P(t) = 2P_0$ and solve for $t$.

4. To find when $P(t) = 2400$, set up the equation and solve for $t$ using logarithms.

Try solving on your own before revealing the answer!

Final Answer:

Initial population: 600 cells

Growth constant: 0.08

Time to double: 8.66 hours

Time to reach 2400 cells: 18.32 hours

These results are found by plugging values into the exponential growth formula and solving for $t$ using logarithms.

Q2. Determine growth constant k for a population that doubles every 16 days.

Background

Topic: Exponential Growth Rate Calculation

This question tests your ability to calculate the growth constant when given doubling time.

Key Terms and Formulas

• Doubling time formula: $k = \frac{\ln(2)}{T}$, where $T$ is the doubling time.

• $P(t) = P_0 e^{kt}$

Step-by-Step Guidance

1. Set $P(T) = 2P_0$ and substitute into the exponential formula.

2. Take the natural logarithm of both sides to solve for $k$.

3. Plug in $T = 16$ days and calculate $k$.

Try solving on your own before revealing the answer!

Final Answer:

Growth constant $k = 0.0433$

This is calculated using $k = \frac{\ln(2)}{16}$.

Q3. Bacteria culture contains 250 cells and grows at a rate proportional to its size. After one hour, the population increases to 400.

Background

Topic: Exponential Growth, Initial Value, and Rate Calculation

This question tests your ability to set up and solve exponential growth equations given initial and subsequent values.

Key Terms and Formulas

• Exponential growth: $P(t) = P_0 e^{kt}$

• Given $P_0 = 250$, $P(1) = 400$

Step-by-Step Guidance

1. Set up the equation $400 = 250 e^{k \cdot 1}$ and solve for $k$ using logarithms.

2. Write the general expression for $P(t)$ using the calculated $k$.

3. To find the number of cells after 5 hours, substitute $t = 5$ into $P(t)$.

4. To find the growth rate after 5 hours, differentiate $P(t)$ and evaluate at $t = 5$.

Try solving on your own before revealing the answer!

Final Answer:

Growth constant $k = 0.470$

Expression: $P(t) = 250 e^{0.470t}$

Cells after 5 hours: 3455

Growth rate after 5 hours: 1623 cells/hour

Time to reach 2500 cells: 4.99 hours

Q4. Use model data from years 1800 and 1950 to construct exponential model for world population (in millions).

Background

Topic: Exponential Modeling Using Data

This question tests your ability to use two data points to construct an exponential model and make predictions.

Key Terms and Formulas

• Exponential model: $P(t) = P_0 e^{kt}$

• Use two points: $(t_1, P_1)$ and $(t_2, P_2)$

• Solve for $k$ using $k = \frac{\ln(P_2/P_1)}{t_2 - t_1}$

Step-by-Step Guidance

1. Identify $P_0$ and $k$ using the data from 1800 and 1950.

2. Write the exponential model $P(t)$.

3. Predict population for given years by substituting $t$ into $P(t)$.

Try solving on your own before revealing the answer!

Final Answer:

Model: $P(t) = 980 e^{0.0091t}$

Predicted population in 1975: 1953 million

Predicted population in 2005: 3095 million

Model shows population in billions after 100 years, not just a few million more.

Q5. $150,000 deposited into savings account earning interest rate of 4.7% compounded continuously. Find formula for A(t), account balance after t years.

Background

Topic: Continuous Compounding Interest

This question tests your understanding of exponential growth in financial contexts.

Key Terms and Formulas

• Continuous compounding formula: $A(t) = P e^{rt}$

• $P$ = principal (initial deposit)

• $r$ = interest rate (as decimal)

• $t$ = time (years)

Step-by-Step Guidance

1. Write the formula for $A(t)$ using the given values.

2. To find the balance after 10 years, substitute $t = 10$.

3. To find the growth rate after 10 years, differentiate $A(t)$ and evaluate at $t = 10$.

4. To find how long until the account reaches $500,000$, set $A(t) = 500,000$ and solve for $t$ using logarithms.

Try solving on your own before revealing the answer!

Final Answer:

Formula: $A(t) = 150,000 e^{0.047t}$

Balance after 10 years: $236,900$

Growth rate after 10 years: $11,139$ per year

Time to reach $500,000$: 33.73 years

Q6. $10,000 deposited into savings account earning interest rate of 4.7% compounded continuously. How long until the investment doubles in value?

Background

Topic: Doubling Time for Continuous Compounding

This question tests your ability to solve for time when the investment doubles using continuous compounding.

Key Terms and Formulas

• Continuous compounding: $A(t) = P e^{rt}$

• Doubling time: $A(t) = 2P$

Step-by-Step Guidance

1. Set $A(t) = 2P$ and substitute into the formula.

2. Take the natural logarithm of both sides to solve for $t$.

3. Plug in $r = 0.047$ and calculate $t$.

Try solving on your own before revealing the answer!

Final Answer:

Doubling time: 14.75 years

This is calculated using $t = \frac{\ln(2)}{0.047}$.

Q7. $4000 deposited into savings account earning interest rate of 4.5% compounded continuously. How long until the investment is worth $8000?

Background

Topic: Continuous Compounding and Doubling Value

This question tests your ability to solve for time when the investment doubles using continuous compounding.

Key Terms and Formulas

• Continuous compounding: $A(t) = P e^{rt}$

• Doubling time: $A(t) = 2P$

Step-by-Step Guidance

1. Set $A(t) = 8000$ and $P = 4000$ in the formula.

2. Take the natural logarithm of both sides to solve for $t$.

3. Plug in $r = 0.045$ and calculate $t$.

Try solving on your own before revealing the answer!

Final Answer:

Doubling time: 15.41 years

This is calculated using $t = \frac{\ln(2)}{0.045}$.

Q8. $7500 deposited into savings account with interest compounded continuously. After 7 years, the balance is $9500. Find interest rate.

Background

Topic: Solving for Interest Rate in Continuous Compounding

This question tests your ability to solve for the interest rate given initial and final values and time.

Key Terms and Formulas

• Continuous compounding: $A(t) = P e^{rt}$

• Given $A(7) = 9500$, $P = 7500$

Step-by-Step Guidance

1. Set up the equation $9500 = 7500 e^{7r}$.

2. Take the natural logarithm of both sides to solve for $r$.

3. Calculate $r$ using the values provided.

Try solving on your own before revealing the answer!

Final Answer:

Interest rate: 3.51%

This is calculated using $r = \frac{\ln(9500/7500)}{7}$.

Q9. How much do you need to invest at 6.5% interest compounded continuously so you have $1 million after 40 years?

Background

Topic: Present Value for Continuous Compounding

This question tests your ability to solve for the initial investment needed to reach a future value.

Key Terms and Formulas

• Continuous compounding: $A(t) = P e^{rt}$

• Given $A(40) = 1,000,000$, $r = 0.065$

Step-by-Step Guidance

1. Set up the equation $1,000,000 = P e^{0.065 \cdot 40}$.

2. Take the natural logarithm of both sides to solve for $P$.

3. Calculate $P$ using the values provided.

Try solving on your own before revealing the answer!

Final Answer:

Initial investment needed: $74,149.21

This is calculated using $P = \frac{1,000,000}{e^{0.065 \cdot 40}}$.

Q10. A radioactive isotope has a half-life of 50 years. Suppose we start with a 400g sample.

Background

Topic: Exponential Decay and Half-Life

This question tests your understanding of exponential decay and how to use half-life to model radioactive decay.

Key Terms and Formulas

• Decay formula: $M(t) = M_0 e^{-kt}$

• Half-life formula: $k = \frac{\ln(2)}{T_{1/2}}$

• $M_0$ = initial mass

Step-by-Step Guidance

1. Calculate $k$ using the half-life formula.

2. Write the decay formula for $M(t)$.

3. To find mass after 40 years, substitute $t = 40$ into $M(t)$.

4. To find how much remains after 150 years, substitute $t = 150$.

5. To find how long until only 10g remains, set $M(t) = 10$ and solve for $t$ using logarithms.

Try solving on your own before revealing the answer!

Final Answer:

Decay constant $k = 0.0139$

Mass after 40 years: 144g

Mass after 150 years: 50g

Time to reach 10g: 263.5 years

Q11. Radioactive isotope Carbon-14 has half-life of 5730 years. Plant fiber found contains 75% of carbon-14 levels found in living plants. Find age of this plant fiber.

Background

Topic: Radioactive Decay and Carbon Dating

This question tests your ability to use exponential decay to estimate the age of a sample based on remaining isotope.

Key Terms and Formulas

• Decay formula: $M(t) = M_0 e^{-kt}$

• Half-life formula: $k = \frac{\ln(2)}{T_{1/2}}$

• Given $M(t)/M_0 = 0.75$

Step-by-Step Guidance

1. Calculate $k$ using the half-life formula.

2. Set up the equation $0.75 = e^{-kt}$ and solve for $t$ using logarithms.

3. Plug in the values and calculate $t$.

Try solving on your own before revealing the answer!

Final Answer:

Age of plant fiber: 1371 years

This is calculated using $t = \frac{\ln(0.75)}{-k}$.

Q12. Iodine-131 has half-life of 8 days. A food product has 5 times the allowable limit for human consumption. How long does the product need to be stored before it reaches allowable limit?

Background

Topic: Radioactive Decay and Safety Limits

This question tests your ability to use exponential decay to determine safe storage times for radioactive substances.

Key Terms and Formulas

• Decay formula: $M(t) = M_0 e^{-kt}$

• Half-life formula: $k = \frac{\ln(2)}{T_{1/2}}$

• Given $M(t)/M_0 = 1/5$

Step-by-Step Guidance

1. Calculate $k$ using the half-life formula.

2. Set up the equation $1/5 = e^{-kt}$ and solve for $t$ using logarithms.

3. Plug in the values and calculate $t$.

Try solving on your own before revealing the answer!

Final Answer:

Storage time: 13.58 days

This is calculated using $t = \frac{\ln(1/5)}{-k}$.

Q13. A person is injected with 150 milligrams of penicillin at time t = 0. Let F(t) be the amount (in mg) of penicillin present in person's bloodstream t hours after injection.

Background

Topic: Exponential Decay in Biological Systems

This question tests your ability to model drug decay and calculate biological half-life.

Key Terms and Formulas

• Decay formula: $F(t) = F_0 e^{-kt}$

• Biological half-life: $k = \frac{\ln(2)}{T_{1/2}}$

• Given $F_0 = 150$ mg

Step-by-Step Guidance

1. Set up the decay formula for penicillin.

2. To find how much remains after 3 hours, substitute $t = 3$.

3. To find biological half-life, set $F(t) = 75$ and solve for $t$ using logarithms.

Try solving on your own before revealing the answer!

Final Answer:

Amount after 3 hours: 43.05 mg

Biological half-life: 1.77 hours

Q14. Calculate the derivative of the following functions.

Background

Topic: Differentiation Rules (Product, Quotient, Chain)

This question tests your ability to apply differentiation rules to various functions.

Key Terms and Formulas

• Product Rule: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

• Quotient Rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

• Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Step-by-Step Guidance

1. Identify which rule applies to each function.

2. Apply the rule step-by-step, writing out derivatives of each part.

3. Combine terms as needed, but stop before simplifying to the final answer.

Try solving on your own before revealing the answer!

Final Answer:

Derivatives calculated using appropriate rules for each function.

See worksheet for detailed steps and results.

Q15. Find the absolute max and min values of the following function on the closed interval [-3,1]: f(x) = x^2 + x + 18

Background

Topic: Optimization on Closed Intervals

This question tests your ability to find critical points and evaluate endpoints to determine absolute extrema.

Key Terms and Formulas

• Critical points: where $f'(x) = 0$ or $f'(x)$ DNE

• Evaluate $f(x)$ at critical points and endpoints

Step-by-Step Guidance

1. Find $f'(x)$ and solve for critical points.

2. Evaluate $f(x)$ at critical points and endpoints.

3. Compare values to determine absolute max and min.

Try solving on your own before revealing the answer!

Final Answer:

Absolute max: 27 at $x = 1$

Absolute min: 18 at $x = -3$