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Subject Test Math 2

Question 28 of 28

The number of hours of daylight, $d$ , in Hartsville can be modeled by $d = \frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} t)$ , where $t$ is the number of days after March $21$ . The day with the greatest number of hours of daylight has how many more daylight hours than May $1$ ? (March and May have $31$ days each. April and June have $30$ days each.)

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$0.8$ hr

Correct Answer:

Yes

$1.5$ hr

Correct Answer:

No

$2.3$ hr

Correct Answer:

No

$3.0$ hr

Correct Answer:

No

$4.7$ hr

Correct Answer:

No

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The correct answer is A.

To determine how many more daylight hours the day with the greatest number of hours of daylight has than May $1$ , find the maximum number of daylight hours possible for any day and then subtract from that the number of daylight hours for May $1$ .

To find the greatest number of daylight hours possible for any day, notice that the expression $\frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} t)$ is maximized when $\sin (\frac{2 π}{365} t) = 1$ , which corresponds to $\frac{2 π}{365} t = \frac{π}{2}$ , so $t = \frac{365}{4} = 91.25$ . However, for this problem, $t$ must be a whole number, as it represents a count of days after March $21$ . From the shape of the graph of the sine function, either $t = 91$ or $t = 92$ corresponds to the day with the greatest number of hours of daylight, and since $\frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} (91)) > \frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} (92))$ , the expression $\frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} t)$ is maximized when $t = 91$ days after March $21$ . (It is not required to find the day on which the greatest number of hours of daylight occurs, but it is $10 + 30 + 31 + 20$ days after March $21$ ,that is, June $20$ .)

Since May $1$ is $10 + 30 + 1 = 41$ days after March $21$ , the number of hours of daylight for May $1$ is $\frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} (41))$ .

Therefore, the day with the greatest number of hours of daylight has $(\frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} (91))) - (\frac{35}{3} + \frac{7}{3} \sin (\frac{2 π}{365} (41))) \approx 0.8$ more daylight hours than May $1$ .

Question Difficulty:

hard

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Subject Test Math 2

Question 28 of 28

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