A-Level AQA Maths Mechanics Quantities, Units & Modelling: The daily world production of oi

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 the volume of oil in millions of barrels, and $t$ is time in years since 1 January 1980 - AQA - A-Level Maths Mechanics - Question 11 - 2018 - Paper 1

Question 11

$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-the-volume-of-oil-in-millions-of-barrels,-and-$t$-is-time-in-years-since-1-January-1980-AQA-A-Level Maths Mechanics-Question 11-2018-Paper 1.png$

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 the volume o... show full transcript

Worked Solution & Example Answer: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 the volume of oil in millions of barrels, and $t$ is time in years since 1 January 1980 - AQA - A-Level Maths Mechanics - Question 11 - 2018 - Paper 1

Step 1

Show that $T$ satisfies the equation

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Answer

To show that $T$ satisfies the equation, we start with the equation derived from the oil production model at zero output:

$V = 10 + 100 \left( \frac{T}{30} \right)^3 - 50 \left( \frac{T}{30} \right)^4 = 0$

Rearranging gives us:

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

Multiplying through by $\left( \frac{30}{T} \right)^4$ (to eliminate the fractions), we find that:

$T = \frac{3 \sqrt{607 T^2 + 162000}}{T}$

Step 2

Use the iterative formula $T_{n+1} = \frac{3}{607/T_n^2 + 162000/T_n}$, with $T_0 = 38$, to find the values of $T_1$, $T_2$, and $T_3$.

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Answer

Using the iterative formula:

For $T_0 = 38$ : $T_1 = \frac{3}{\frac{607}{38^2} + \frac{162000}{38}} \approx 44.964$
Using $T_1$ to find $T_2$ : $T_2 = \frac{3}{\frac{607}{(44.964)^2} + \frac{162000}{44.964}} \approx 49.987$
Lastly, using $T_2$ to find $T_3$ : $T_3 = \frac{3}{\frac{607}{(49.987)^2} + \frac{162000}{49.987}} \approx 53.504$

Step 3

Explain the relevance of using $T_0 = 38$

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Answer

$T_0 = 38$ corresponds to the year 2018, which is relevant as it provides a current point in time for the iterative calculations, allowing the predictions to be grounded in recent historical data.

Step 4

Use the models to show that the country's use of oil and the world production of oil will be equal during the year 2029.

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Answer

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

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

Substituting $t = 49$ (corresponding to the year 2029) into both equations validates that they yield the same results, showing that the use of oil by the country and the world production will coincide during 2029.

Show that $T$ satisfies the equation

Use the iterative formula $T_{n+1} = \frac{3}{607/T_n^2 + 162000/T_n}$, with $T_0 = 38$, to find the values of $T_1$, $T_2$, and $T_3$.

Explain the relevance of using $T_0 = 38$

Use the models to show that the country's use of oil and the world production of oil will be equal during the year 2029.

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