Question Statement

If $P (t) = P_{o} e^{k t}$ denotes the growth function of oil and the exponential growth rate of the demand for oil is $10%$ per year, when will the demand be doubled?

Background and Explanation

This problem involves the exponential growth model, where $P_{o}$ is the initial amount, $k$ is the growth constant, and $t$ is time. To find when a value doubles, we determine the time $t$ required for the current population $P (t)$ to reach twice the initial value $P_{o}$ .

Solution

The given growth function is: $P (t) = P_{o} e^{k t} \dots (1)$

Step 1: Find the growth constant $k$

We are told the growth rate is $10%$ per year. This means that after one year ( $t = 1$ ), the demand $P$ will be the original demand plus $10%$ . $P = P_{o} + 0.10 P_{o} = 1.1 P_{o}$

Substitute $t = 1$ and $P = 1.1 P_{o}$ into the growth equation: $1.1 P_{o} = P_{o} e^{k \times 1}$

Divide both sides by $P_{o}$ : $1.1 = e^{k}$

Taking the natural logarithm ( $ln$ ) of both sides: $k = ln 1.1$

Step 2: Set up the equation for doubling demand

We want to find the time $t$ when the demand is doubled, meaning $P (t) = 2 P_{o}$ . Substitute our value for $k$ back into the original equation: $P (t) = P_{o} e^{t l n 1.1}$

Now, set $P (t)$ to $2 P_{o}$ : $2 P_{o} = P_{o} e^{t l n 1.1}$

Step 3: Solve for $t$

Divide both sides by $P_{o}$ : $2 = e^{t l n 1.1}$

Using the property that $e^{l n (x)} = x$ , we can simplify the right side: $2 = (1.1)^{t}$

Take the natural logarithm of both sides to bring down the exponent: $ln 2 = ln (1.1)^{t}$ $ln 2 = t ln (1.1)$

Isolate $t$ : $t = \frac{l n 2}{l n 1.1}$

Using a calculator: $t \approx \frac{0.6931}{0.0953} \approx 7.27$

The demand will be doubled after approximately 7.27 years.

Key Formulas or Methods Used

Exponential Growth Formula: $P (t) = P_{o} e^{k t}$
Logarithmic Identity: $ln (e^{x}) = x$
Power Rule for Logarithms: $ln (a^{b}) = b ln a$
Doubling Condition: Setting $P (t) = 2 P_{o}$

Summary of Steps

Use the $10%$ annual growth rate to establish that at $t = 1$ , $P = 1.1 P_{o}$ .
Solve for the growth constant $k$ by applying natural logarithms.
Substitute $k$ back into the general formula and set the final amount $P (t)$ to $2 P_{o}$ .
Solve the resulting exponential equation for $t$ using logarithms.

Step 1: Find the growth constant

k

We are told the growth rate is

10%

per year. This means that after one year (

t = 1

), the demand

P

will be the original demand plus

10%

P = P_{o} + 0.10 P_{o} = 1.1 P_{o}

Substitute

t = 1

and

P = 1.1 P_{o}

into the growth equation:

1.1 P_{o} = P_{o} e^{k \times 1}

Divide both sides by

P_{o}

1.1 = e^{k}

Taking the natural logarithm (

ln

) of both sides:

k = ln 1.1

Step 3: Solve for

t

Divide both sides by

P_{o}

2 = e^{t l n 1.1}

Using the property that

e^{l n (x)} = x

, we can simplify the right side:

2 = (1.1)^{t}

Take the natural logarithm of both sides to bring down the exponent:

ln 2 = ln (1.1)^{t}

ln 2 = t ln (1.1)

Isolate

t

t = \frac{l n 2}{l n 1.1}

Using a calculator:

t \approx \frac{0.6931}{0.0953} \approx 7.27

The demand will be doubled after approximately 7.27 years.

1.3 Q-7

Question Statement

Background and Explanation

Solution

Step 1: Find the growth constant $k$

Step 2: Set up the equation for doubling demand

Step 3: Solve for $t$

Key Formulas or Methods Used

Summary of Steps

1.3 Q-7

Question Statement

Background and Explanation

Solution

Step 1: Find the growth constant $k$

Step 2: Set up the equation for doubling demand

Step 3: Solve for $t$

Key Formulas or Methods Used

Summary of Steps