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 the iterative formula:

1. First iteration: $T_{1} = \frac{3}{607} (38)^{2} + 162000 = 44.964$
2. Second iteration: $T_{2} = \frac{3}{607} (44.964)^{2} + 162000 = 49.987$
3. Third iteration: $T_{3} = \frac{3}{607} (49.987)^{2} + 162000 = 53.504$

Thus, the values found are:

• $T_{1} = 44.964$
• $T_{2} = 49.987$
• $T_{3} = 53.504$

Using $T_{0} = 38$ is relevant because it represents the starting point or initial estimate for the number of years since 1980, specifically indicating the year 2018 in this context. This serves as a crucial basis for the iteration process which seeks to approximate when oil production will cease.

To find when the country's use of oil equals world production, we set:

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

Translating the year 2029 yields: $t = 2029 - 1980 = 49$

Thus, we need to evaluate both models for $t = 49$:

1. Daily use: $4.5 \times 1.063^{49} = A$
2. World production: Calculate using the original world production model for $t = 49$. Solving this will show that the two values are equal, confirming the year in which they match.