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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 Pure - 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... 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 Pure - Question 11 - 2018 - Paper 1
Step 1
Show that $T$ satisfies the equation
Answer
To show that ( T ) satisfies the equation, we start with the iterative formula provided:
( T = \sqrt{\frac{607T^2 + 162000}{T}} ).
To verify, we can rearrange the equation:
- Multiply both sides by ( T ):
( T^2 = 607T^2 + 162000 ) - Rearranging gives us:
( 0 = 606T^2 + 162000 )
This confirms that the model is valid for predicting the production decline.
Step 2
Use the iterative formula $T_{n+1} = \frac{3}{607}T_n^2 + 162000$, with $T_0 = 38$, to find the values of $T_1, T_2, and T_3$
Answer
Using the initial value of ( T_0 = 38 ):
- Calculate ( T_1 ):
( T_1 = \frac{3}{607}(38^2) + 162000 \approx 44.964 ) - Calculate ( T_2 ):
( T_2 = \frac{3}{607}(T_1^2) + 162000 \approx 49.987 ) - Calculate ( T_3 ):
( T_3 = \frac{3}{607}(T_2^2) + 162000 \approx 53.504 )
Step 3
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.
Answer
To equate both production models, set:
( 10 + 100 \left( \frac{t}{30} \right)^3 - 50 \left( \frac{t}{30} \right)^4 = 4.5 \times 10^{0.63t} ).
By substituting ( t = 49 ):
( 49/30 = 2029-1980 ). This shows that both the use of oil and the world production will equalize in 2029.