Half-life, Newton’s Law Of Cooling & Conceptual Logistic Growth (Equilibria & Max Growth)

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1) "The rate of change of a population is proportional to the current population" translates to which differential equation?

DP/dt = k

DP/dt = kP

DP/dt = k(P - M)

DP/dt = k/P

The phrase "proportional to the current population" means the rate is the population P multiplied by a constant k, giving the standard exponential model dP/dt = kP.

Explanation

The phrase "proportional to the current population" means the rate is the population P multiplied by a constant k, giving the standard exponential model dP/dt = kP.

About This Quiz

Half-life, Newtons Law Of Cooling & Conceptual Logistic Growth (Equilibria & Max Growth) - Quiz

Ready to model how things move toward a target value? In this quiz, you’ll explore equations where the rate depends on the difference between a quantity and a surrounding constant, like temperature moving toward room temperature. You’ll practice recognizing Newton’s law of cooling/heating situations, interpreting what the signs and constants... see moremean, and solving practical problems using exponential behavior of differences. Along the way, you’ll strengthen your ability to connect differential equations to everyday processes.
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2) The solution to dQ/dt = -0.08 Q is which of the following?

Q = Q(0) e^(-0.08t)

Q = Q(0) e^(0.08t)

Q = Q(0) / (1 - 0.08t)

Q = 100 e^(-0.08t)

Separate variables: dQ/Q = -0.08 dt. Integrate both sides: ln|Q| = -0.08t + C. Exponentiate: Q = e^C e^(-0.08t). The constant e^C equals the initial amount Q(0), giving Q = Q0 e^(-0.08t).

Explanation

Separate variables: dQ/Q = -0.08 dt. Integrate both sides: ln|Q| = -0.08t + C. Exponentiate: Q = e^C e^(-0.08t). The constant e^C equals the initial amount Q(0), giving Q = Q0 e^(-0.08t).

3) A sample decays to 25% of its original amount in 14 days. What is the half-life?

7 days

10 days

14 days

28 days

25% = 1/4 = (½)², so two half-lives in 14 days → one half-life = 7 days.

Explanation

25% = 1/4 = (½)², so two half-lives in 14 days → one half-life = 7 days.

4) Investment grows continuously at 5% per year. How long to grow from $1000 to $2000?

13.9 years

14 years

20 years

25 years

2000 = 1000 e^(0.05 t) → 2 = e^(0.05 t) → ln 2 = 0.05 t → t = ln 2 / 0.05 ≈ 0.693/0.05 ≈ 13.86 years ≈ 13.9 years.

Explanation

2000 = 1000 e^(0.05 t) → 2 = e^(0.05 t) → ln 2 = 0.05 t → t = ln 2 / 0.05 ≈ 0.693/0.05 ≈ 13.86 years ≈ 13.9 years.

5) A substance decays exponentially. After 10 years, 30% remains. What percentage remains after 20 years?

15%

21%

60%

30% remains after 10 years means multiplication factor of 0.3. After another 10 years (total 20), multiply by 0.3 again → 0.3 * 0.3 = 0.09 = 9%.

Explanation

30% remains after 10 years means multiplication factor of 0.3. After another 10 years (total 20), multiply by 0.3 again → 0.3 * 0.3 = 0.09 = 9%.

6) In exponential decay, what does the negative sign in k indicate?

The quantity is increasing

The quantity is decreasing

The quantity is constant

There is a carrying capacity

When k 0), meaning the quantity is decreasing over time.

Explanation

When k 0), meaning the quantity is decreasing over time.

7) A cold drink warms from 5°C to 15°C in 20 minutes in a 25°C room. Following Newton's law of cooling, how long does it take to reach 20°C?

25 minutes

30 minutes

40 minutes

50 minutes

Newton's law of cooling/heating states that the temperature difference from the ambient temperature decays exponentially. The initial difference from room temperature is 25-5 = 20°C. After 20 minutes, the difference is 25-15 = 10°C, meaning the difference has halved. To reach 20°C, the difference must be 25-20 = 5°C, which is half of 10°C. Since the difference halves every 20 minutes in this scenario, it takes another 20 minutes to reach 5°C difference. Therefore, the total time from the start is 20 + 20 = 40 minutes. This can be verified mathematically using the exponential decay model for the temperature difference: y(t) = 20e^(-kt), where k = ln(2)/20 based on the first 20 minutes. Solving 5 = 20e^(-kt) gives t = 40 minutes.

Explanation

8) Which situation is best modeled by dy/dt = k(y - A)?

Unrestricted population growth

Radioactive decay

Newton's law of cooling

Logistic growth

The rate is proportional to the difference between the object’s temperature y and the ambient temperature A, which is exactly Newton's law of cooling/heating.

Explanation

The rate is proportional to the difference between the object’s temperature y and the ambient temperature A, which is exactly Newton's law of cooling/heating.

9) In logistic growth dy/dt = 0.03 y (1 - y/400), what is the carrying capacity?

0.03

400

1200

1/0.03

The standard logistic form is dy/dt = r y (1 - y/K) where K is carrying capacity. Comparing to 0.03 y (1 - y/400), the denominator is 400, so K = 400.

Explanation

The standard logistic form is dy/dt = r y (1 - y/K) where K is carrying capacity. Comparing to 0.03 y (1 - y/400), the denominator is 400, so K = 400.

10) For the logistic equation dy/dt = k y (L - y), the equilibrium solutions are:

Y = 0 only

Y = L only

Y = 0 and y = L

No equilibrium

Set dy/dt = 0 → k y (L - y) = 0 → y = 0 or y = L. These are the two points where the population does not change.

Explanation

Set dy/dt = 0 → k y (L - y) = 0 → y = 0 or y = L. These are the two points where the population does not change.

11) In logistic growth with carrying capacity L, the growth rate is greatest when the population is:

Near 0

Near L/2

Near L

Above L

The growth rate function k y (L - y) is a quadratic that reaches its maximum at y = L/2, the midpoint between 0 and the carrying capacity.

Explanation

The growth rate function k y (L - y) is a quadratic that reaches its maximum at y = L/2, the midpoint between 0 and the carrying capacity.

12) A logistic model has carrying capacity 10,000. The population will approach 10,000 from below if it starts:

Below 10,000

Above 10,000

Exactly at 10,000

Anywhere

When y 0 (increasing toward 10,000). When y > 10,000, dy/dt

Explanation

When y 0 (increasing toward 10,000). When y > 10,000, dy/dt

13) Which differential equation has a carrying capacity?

Dy/dt = 0.02 y

Dy/dt = -0.02 y

Dy/dt = 0.02 y (50 - y)

Dy/dt = 0.02 (50 - y)

Only the form that includes both y and (C - y) has a limiting value (carrying capacity) as t → ∞. Here C = 50.

Explanation

Only the form that includes both y and (C - y) has a limiting value (carrying capacity) as t → ∞. Here C = 50.

14) In a logistic model, when the population is very small compared to carrying capacity, the growth behaves approximately like:

Linear growth

Exponential growth

No growth

Decay

When y is much smaller than L, (L - y) ≈ L (constant), so dy/dt ≈ (kL) y, which is exponential growth with effective rate kL.

Explanation

When y is much smaller than L, (L - y) ≈ L (constant), so dy/dt ≈ (kL) y, which is exponential growth with effective rate kL.

15) A forest supports a maximum of 1200 deer (logistic model). At what deer population is the growth rate of the herd fastest?

300

600

900

1200

Maximum growth rate in logistic model occurs at half the carrying capacity: 1200 / 2 = 600 deer.

Explanation

Maximum growth rate in logistic model occurs at half the carrying capacity: 1200 / 2 = 600 deer.

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Alva Benedict B. |PhD

College Expert

Alva Benedict B. is an experienced mathematician and math content developer with over 15 years of teaching and tutoring experience across high school, undergraduate, and test prep levels. He specializes in Algebra, Calculus, and Statistics, and holds advanced academic training in Mathematics with extensive expertise in LaTeX-based math content development.

Half-life, Newton’s Law of Cooling & Conceptual Logistic Growth (Equilibria & Max Growth)