The amount of a substance remaining in a solution reduces exponentially over time. An experiment measures the percentage of the substance remaining in the solution. The percentage is measured at the same time each day. The data collected over the first 4 days are given in the table below. Based on the data in the table, estimate which is the first day on which the percentage of the substance in the solution will be less than 0.01%. Day Percentage of substance (%) 1 95 2 42.75 3 19.2375 4 8.6569 A square has sides of length 2 cm. The midpoints of the sides of this square are joined to form another square. This process is continued. The first three squares in the process are shown below. Find the sum of the perimeters of the squares if this process is continued indefinitely. Give your answer in the form a + b√c cm, where a, b, and c ∈ ℕ.

Leaving Cert Mathematics Sequences & Series: The amount of a substance remain

The amount of a substance remaining in a solution reduces exponentially over time - Leaving Cert Mathematics - Question 4 - 2017

Question 4

The amount of a substance remaining in a solution reduces exponentially over time. An experiment measures the percentage of the substance remaining in the solution. ... show full transcript

Worked Solution & Example Answer:The amount of a substance remaining in a solution reduces exponentially over time - Leaving Cert Mathematics - Question 4 - 2017

Step 1

Estimate which is the first day on which the percentage of the substance in the solution will be less than 0.01%.

96%

114 rated

Answer

To solve for the day when the percentage of the substance is less than 0.01%, we first identify that the percentage decreases exponentially, so we can represent this with the formula for a geometric progression:

$T_n = ar^{n-1}$

where:

$T_n$ is the value at day $n$ ,
$a$ is the initial value ( $95$ on day 1),
$r$ is the common ratio calculated as follows:

$r = \frac{T_2}{T_1} = \frac{42.75}{95}$

Calculating the ratio, we find:

$r = \frac{42.75}{95} = 0.450$

Now we need to find $n$ such that:

$T_n = 95(0.450)^{n-1} < 0.01$

By rearranging this inequality, we can solve for $n$ :

$\frac{0.01}{95} > (0.450)^{n-1}$

Taking logarithms on both sides gives us:

$(n-1) \log(0.450) < \log\left(\frac{0.01}{95}\right)$

Now we can find $n$ and the terms of the series:

Using logs and calculations, we eventually estimate that it will be after day 12 when the percentage of the substance will be less than $0.01$ .

Step 2

Find the sum of the perimeters of the squares if this process is continued indefinitely.

99%

104 rated

Answer

The perimeter of the first square is given by:

$P_1 = 4 \times 2 = 8 \text{ cm}$

In the next stage, the side length of the new square is given by half the diagonal of the previous square. For the square with side length 2 cm, the diagonal can be calculated using:

$d = s\sqrt{2} = 2\sqrt{2}$

Hence, the side length of the new square will be:

$s_2 = \frac{2\sqrt{2}}{2} = \sqrt{2}$

So, the perimeter of the second square is:

$P_2 = 4 \times s_2 = 4 \times \sqrt{2}$

Continuing this pattern, we find:

For the third square, the side length is $s_3 = \frac{\sqrt{2}\sqrt{2}}{2} = 1$

Thus, $P_3 = 4\cdot1 = 4$

The total sum of the perimeters as the process continues indefinitely can be expressed as:

$S_{\infty} = 8 + 4\sqrt{2} + 4 + \ldots$

This converges to:

$S_{\infty} = a + b\sqrt{c} = 16 + 8\sqrt{2} ext{ cm}$

The amount of a substance remaining in a solution reduces exponentially over time - Leaving Cert Mathematics - Question 4 - 2017

Worked Solution & Example Answer:The amount of a substance remaining in a solution reduces exponentially over time - Leaving Cert Mathematics - Question 4 - 2017

Estimate which is the first day on which the percentage of the substance in the solution will be less than 0.01%.

Find the sum of the perimeters of the squares if this process is continued indefinitely.

Join the Leaving Cert students using SimpleStudy...