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 £20000 - its value after one year is £16000 (a) Use an exponential model to form, for car A, a possible equation linking $V$ with $t$. The value of car A is monitored over a 10-year period. Its value after 10 years is £20000 (b) Evaluate the reliability of your model in light of this information. The following information is available for car B - it has the same value, when new, as car A - its value depreciates more slowly than that of car A (c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

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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 8 - 2019 - Paper 1

Question 8

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 £20000 - it... 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 8 - 2019 - Paper 1

Step 1

(a) Use an exponential model to form, for car A, a possible equation linking V with t.

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Answer

To model the value of car A, we can use the exponential decay formula:

$V = A e^{kt}$

where:

$V$ is the value of the car at time $t$ ,
$A$ is the initial value of the car when new,
$k$ is the rate of depreciation,
$t$ is the time in years.

Given that:

$V(0) = 20000$ (initial value), and
$V(1) = 16000$ (value after 1 year),

we can substitute these values into the equation:

From the equation, we know: $20000 e^{k(1)} = 16000$
Dividing both sides by 20000 yields: $e^{k} = rac{16000}{20000} = 0.8$
Taking the natural logarithm of both sides: $k = ext{ln}(0.8)$

Thus, the equation for car A can be expressed as: $V = 20000 e^{ ext{ln}(0.8) t}$

Step 2

(b) Evaluate the reliability of your model in light of this information.

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Answer

To evaluate the reliability of the model, we need to calculate the expected value after 10 years using our derived formula:

Substituting $t = 10$ into our model: $V(10) = 20000 e^{ ext{ln}(0.8) imes 10}$

This simplifies to: $V(10) = 20000 imes 0.8^{10}$

Calculating this yields:

$0.8^{10} \approx 0.1074$ ,
thus, $V(10) \approx 20000 imes 0.1074 \approx 2148$ .

However, the actual value after 10 years is £20000, which indicates that our model greatly underestimates the car's value over long terms. Therefore, the model is not reliable for long-term predictions as it does not account for the situation that the car should not depreciate to such a low value.

Step 3

(c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

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Answer

To adapt the model for car B, which depreciates more slowly but starts with the same initial value, we can adjust the parameter $k$ . If car B's depreciation rate is less severe, we can express $k$ as a smaller negative value (i.e., $k' > k$ ).

Thus, the equation for car B can be written as: $V_B = 20000 e^{k' t},$

where $k'$ is a new depreciation rate that is less negative than the original $k$ . This adjustment will allow the model to reflect the slower depreciation of car B.

In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 1

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 8 - 2019 - Paper 1

(a) Use an exponential model to form, for car A, a possible equation linking V with t.

(b) Evaluate the reliability of your model in light of this information.

(c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

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