Question

The rate of increase of population of a city is proportional to population present. In 40 years it increased from 30,000 to 40,000. At time $t$ population is $a(b)^{t/40}$. Then $a$ and $b$ are \dots}

• A. 30,000, 2/3
• B. 30,000, 4/3
• C. 40,000, 2/3
• D. 40,000, 3/4

Hint:

In any exponential growth/decay formula $y(t) = a \cdot b^{kt}$, the constant '$a$' represents the initial starting amount (when $t=0$), and '$b$' represents the growth/decay multiplier factor over the specific time interval defined in the exponent.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem describes standard exponential population growth governed by the differential equation $\frac{dP}{dt} \propto P$. We must find the constants $a$ and $b$ that fit the model equation $P(t) = a(b)^{t/40}$ to the given real-world data points.

Step 2: Key Formula or Approach:
Use the two data points provided:
1. At time $t = 0$ (the start), the population is $P(0) = 30,000$.
2. At time $t = 40$ years, the population is $P(40) = 40,000$.
Substitute these directly into the given equation model $P(t) = a(b)^{t/40}$ to solve for the two unknowns.

Step 3: Detailed Explanation:
Apply the first condition ($t=0, P=30000$):
$$30000 = a(b)^{0/40}$$
Since $b^0 = 1$, we get:
$$a = 30000$$
Now the equation is updated to $P(t) = 30000(b)^{t/40}$.
Apply the second condition ($t=40, P=40000$):
$$40000 = 30000(b)^{40/40}$$
$$40000 = 30000(b)^1$$
Divide both sides by 30000 to isolate $b$:
$$b = \frac{40000}{30000} = \frac{4}{3}$$

Step 4: Final Answer:
The values are $a = 30000$ and $b = 4/3$, matching option (b).