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The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function $$f(t) = 12.25 + 4.75 imes ext{sin} \left( \frac{2\pi t}{365} \right)$$, where $t$ is the number of days after March 21st and $\frac{2\pi t}{365}$ is expressed in radians - Leaving Cert Mathematics - Question 9 - 2015

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Question 9

The-approximate-length-of-the-day-in-Galway,-measured-in-hours-from-sunrise-to-sunset,-may-be-calculated-using-the-function--$$f(t)-=-12.25-+-4.75--imes--ext{sin}-\left(-\frac{2\pi-t}{365}-\right)$$,-where-$t$-is-the-number-of-days-after-March-21st-and-$\frac{2\pi-t}{365}$-is-expressed-in-radians-Leaving Cert Mathematics-Question 9-2015.png

The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function $$f(t) = 12.25 + 4.75 imes ext{sin} \l... show full transcript

Worked Solution & Example Answer:The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function $$f(t) = 12.25 + 4.75 imes ext{sin} \left( \frac{2\pi t}{365} \right)$$, where $t$ is the number of days after March 21st and $\frac{2\pi t}{365}$ is expressed in radians - Leaving Cert Mathematics - Question 9 - 2015

Step 1

Find the length of the day in Galway on June 5th (76 days after March 21st)

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Answer

To find the length of the day on June 5th, we substitute t=76t = 76 into the function:

f(76)=12.25+4.75sin(2π(76)365)f(76) = 12.25 + 4.75 \text{sin} \left( \frac{2\pi (76)}{365} \right)

Calculating the sine term: 2π(76)3651.308497\frac{2\pi (76)}{365} \approx 1.308497

Then, f(76)=12.25+4.75×0.98030416.837f(76) = 12.25 + 4.75 \times 0.980304 \approx 16.837

Converting to hours and minutes gives us approximately 16 hours and 50 minutes.

Step 2

Find a date on which the length of the day in Galway is approximately 15 hours.

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Answer

We need to solve for tt in the equation: f(t)=15f(t) = 15

This gives:

12.25+4.75sin(2πt365)=1512.25 + 4.75 \text{sin} \left( \frac{2\pi t}{365} \right) = 15

Rearranging yields: sin(2πt365)=1512.254.75=0.578947\text{sin} \left( \frac{2\pi t}{365} \right) = \frac{15 - 12.25}{4.75} = 0.578947

Now solving: 2πt3650.6153 or 0.9743\frac{2\pi t}{365} \approx 0.6153 \text{ or } 0.9743

Which gives approximately:

t35.87    April 26t \approx 35.87 \implies \text{April 26}

Step 3

Find $f'(t)$, the derivative of $f(t)$.

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Answer

To find the derivative, we differentiate:

f(t)=0+4.75×2π365cos(2πt365)f'(t) = 0 + 4.75 \times \frac{2\pi}{365} \text{cos} \left( \frac{2\pi t}{365} \right)

This simplifies to:

f(t)=9.5π365cos(2πt365)f'(t) = \frac{9.5\pi}{365} \text{cos} \left( \frac{2\pi t}{365} \right)

Step 4

Hence, or otherwise, find the length of the longest day in Galway.

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Answer

The longest day occurs when cos(2πt365)=1\text{cos} \left( \frac{2\pi t}{365} \right) = 1, which gives:

t=12.25+4.75×1=17 hours.t = 12.25 + 4.75 \times 1 = 17 \text{ hours.}

Step 5

Use integration to find the average length of the day in Galway over the six months from March 21st to September 21st (184 days).

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Answer

To find the average length of the day, we compute:

1baabf(t)dt\frac{1}{b-a} \int_a^b f(t) dt

Where:

a=1, b=184a = 1, \ b = 184

The integral becomes:

1184(12.25+4.75×sin(2πt365))dt\int_1^{184} \left(12.25 + 4.75 \times \text{sin} \left( \frac{2\pi t}{365} \right)\right) dt

This evaluates to approximately 15 hours and 15 minutes.

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