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In a simple model, the value, \( V \), of a car depends on its age, \( t \), in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 2
Question 9
In a simple model, the value, \( V \), of a car depends on its age, \( t \), in years. The following information is available for car A - its value when new is £20... show full transcript
Worked Solution & Example Answer:In a simple model, the value, \( V \), of a car depends on its age, \( t \), in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 2
Step 1
Use an exponential model to form, for car A, a possible equation linking \( V \) with \( t \).
Answer
To model the value of car A using an exponential function, we can use the general form: [ V = A e^{-kt} ] where ( A ) represents the initial value (when new) and ( k ) is the depreciation constant.
Given that the value when new is ( V(0) = 20000 ) pounds and the value after one year is ( V(1) = 16000 ) pounds, we can set up the equation:
[ 16000 = 20000 e^{-k(1)} ]
To find ( k ), solve for it:
[ e^{-k} = \frac{16000}{20000} = 0.8 ]
Taking the natural logarithm:
[ -k = \ln(0.8) \implies k = -\ln(0.8) ]
Thus, the model for car A can be expressed as: [ V(t) = 20000 e^{-\ln(0.8)t} ]
or more simply: [ V(t) = 20000 \times 0.8^t ]
Step 2
Evaluate the reliability of your model in light of this information.
Answer
Over a 10-year period, the value of car A is monitored and found to be ( V(10) = 2000 ) pounds.
Using our model: [ V(10) = 20000 \times 0.8^{10} \approx 20000 \times 0.1074 \approx 2148 \text{ pounds} ]
Comparing this with the observed value after 10 years, which is 2000 pounds, we find that the model predicts a value of approximately 2148 pounds.
Hence, while the model is not perfectly accurate, it does provide a reasonable estimate, indicating that it is fairly reliable.
Step 3
Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.
Answer
For car B, which depreciates more slowly than car A, we will adjust the depreciation constant ( k ). Assuming that car B has the same initial value as car A, the modified model could be:
[ V_B(t) = 20000 e^{-kt} ]
where ( k ) is now less than the value found for car A (( k < -\ln(0.8) )).
Thus, we can say: [ V_B(t) = 20000 \times 0.9^t ]
if we assume a depreciation rate of 10% per year, as an example.