A student is asked to memorise a long list of digits, and then write down the list some time later. The proportion, P(t), of the digits recalled correctly after t hours can be modelled by the function: $$P(t) = 0.82 - 0.12 ext{ln}(t + 1)$$ for $0 \, \leq \, t \leq \, 12$ and $t \, \in \, \mathbb{R}$. (c) (i) Find the value of $P'(1)$. (ii) $P''(t)$ is always negative for $0 \leq t \leq 12, \, t \in \mathbb{R}$. What does this tell you about the proportion of digits recalled correctly after t hours, according to this model? (d) Use calculus to show that the graph of $y = P(t)$ has no points of inflection, for $0 \leq t \leq 12, \, t \in \mathbb{R}$.

Leaving Cert Mathematics Differentiation: A student is asked to memorise a

A student is asked to memorise a long list of digits, and then write down the list some time later - Leaving Cert Mathematics - Question 10 - 2022

Question 10

A student is asked to memorise a long list of digits, and then write down the list some time later. The proportion, P(t), of the digits recalled correctly after t ho... show full transcript

Worked Solution & Example Answer:A student is asked to memorise a long list of digits, and then write down the list some time later - Leaving Cert Mathematics - Question 10 - 2022

Step 1

Find the value of $P'(1)$

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Answer

To find $P'(t)$ , we first differentiate the function:

$P'(t) = -0.12 \cdot \frac{1}{t + 1}$

Now, substituting $t = 1$ into $P'(t)$ gives:

$P'(1) = -0.12 \cdot \frac{1}{1 + 1} = -0.12 \cdot \frac{1}{2} = -0.06$

Thus, the value of $P'(1)$ is -0.06.

Step 2

$P''(t)$ is always negative for $0 \leq t \leq 12$

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Answer

The second derivative $P''(t)$ indicates the concavity of the function. Since $P''(t)$ is negative over the interval, this implies that the function $P(t)$ is concave down. Therefore, the proportion of digits recalled correctly decreases as time increases according to this model.

Step 3

Use calculus to show that the graph of $y = P(t)$ has no points of inflection

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Answer

To determine points of inflection, we need to find $P''(t)$ :

Using the first derivative: $P'(t) = -0.12 \cdot \frac{1}{t + 1}$

Differentiating again gives:

$P''(t) = 0.12 \cdot \frac{1}{(t + 1)^2}$

Since $P''(t) > 0$ for all $t \, \in \, [0, 12)$ , there are no points where $P''(t) = 0$ . Therefore, the graph of $y = P(t)$ has no points of inflection in the specified interval.

A student is asked to memorise a long list of digits, and then write down the list some time later - Leaving Cert Mathematics - Question 10 - 2022

Worked Solution & Example Answer:A student is asked to memorise a long list of digits, and then write down the list some time later - Leaving Cert Mathematics - Question 10 - 2022

Find the value of $P'(1)$

$P''(t)$ is always negative for $0 \leq t \leq 12$

Use calculus to show that the graph of $y = P(t)$ has no points of inflection

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