The daily world production of oil can be modelled using $$V = 10 + 100\left(\frac{t}{30}\right)^3 - 50\left(\frac{t}{30}\right)^4$$ where $V$ is volume of oil in millions of barrels, and $t$ is time in years since 1 January 1980 - AQA - A-Level Maths Pure - Question 11 - 2018 - Paper 1

Using $T_{n+1} = \frac{3}{607T_n^2 + 162000}$ with $T_0 = 38$:

1. Calculate $T_1$: $T_1 = \frac{3}{607 \cdot 38^2 + 162000} = 44.964$ (to three decimal places)

2. Calculate $T_2$: $T_2 = \frac{3}{607 \cdot T_1^2 + 162000} = 49.987$ (to three decimal places)

3. Calculate $T_3$: $T_3 = \frac{3}{607 \cdot T_2^2 + 162000} = 53.504$ (to three decimal places)

Using $T_0 = 38$ is relevant as it represents the current year 2018. This gives a starting point in the model, reflecting the initial state of oil production and allowing subsequent values to be calculated iteratively. Hence, deriving accurate projections for future years becomes more reliable.

To find when world production equals the country's oil use:

1. Use the equation from the world production:

   $V = 10 + 100 \left(\frac{t}{30}\right)^3 - 50 \left(\frac{t}{30}\right)^4$

2. Set $V = 4.5 \times 1.063^t$ for the country's usage:

   $4.5 \times 1.063^t = 10 + 100 \left(\frac{t}{30}\right)^3 - 50 \left(\frac{t}{30}\right)^4$

3. Solving these equations will show that both are equal at $t = 49.50$ which correlates to the year 2029:

   $1980 + 49 = 2029$. Thus, confirming equality.