The number of hours of daylight D depends upon the latitude and the day t of the year and is...

Question:

The number of hours of daylight D depends upon the latitude and the day t of the year and is given by the equation : {eq}D(t)=12+Asin(\frac{2 \pi}{365}(t-80)) {/eq} where A depends only on the latitude (and not t). For latitude 30 degress, A is about 2.3

1. When is the number of hours of daylight the greatest?

2.When is the number of hours of daylight the least?

3. When is the number of hours of daylight increasing at a rate of 2 minutes per day?

4. when is the number of hours of daylight decreasing at a rate of 2 min per day?

Minimum and Maximum Values of a Function:

When we have a mathematical function that describes the relationship between some physical quantities, then we can employ differentiation to find values where the function has zero slope and thus may have a local minimum or maximum value.

Answer and Explanation:

We are given the function of daylight hours as a function of day of the year. To find minimum and maximum values we need to differentiate it.

The derivative is:

{eq}D'(t)= \displaystyle A\frac{2 \pi}{365}\cos(\frac{2 \pi}{365}(t-80)) {/eq}

To find when daylight hours are maximum or minimum, we set D' equal to zero and solve for t.

{eq}D'(t)= \displaystyle A\frac{2 \pi}{365}\cos(\frac{2 \pi}{365}(t-80))=0 {/eq}

Since {eq}\cos ( \pi/2) = 0 {/eq} and {eq}\cos (3\pi/2)=0 {/eq}

We can solve:

{eq}\displaystyle \frac{2 \pi}{365} (t-80) = \frac{\pi}{2} {/eq}

giving:

{eq}\displaystyle t=80+\frac{365}{4} = 171.25 {/eq}

and

{eq}\displaystyle \frac{2 \pi}{365} (t-80) = \frac{3\pi}{2} {/eq}

giving:

{eq}\displaystyle t=80+\frac{3(365)}{4} = 353.75 {/eq}

1. The number of hours of daylight is the greatest on day: 171 on 21 June,

2. The number of hours of daylight is the least on day: 353 on 20 December.

To find when daylight hours are changing at a rate r, we set D' equal to r and solve for t.

{eq}D'(t)= r {/eq} where r equals {eq}\pm 2/60 {/eq} hours

{eq}D'(t)= \displaystyle A\frac{2 \pi}{365}\cos(\frac{2 \pi}{365}(t-80))= 2/60 {/eq}

Since A = 2.3 we get:

{eq}\displaystyle 0.0395927 \cos(\frac{2 \pi}{365}(t-80))= 2/60 {/eq}

or

{eq}\cos(0.0172(t-80))=0.8419 {/eq}

{eq}0.0172(t-80)=\cos^{-1}(0.8419) = 0.57 {/eq}

giving

{eq}t=113.11 {/eq}

Similarly for a decrease of 2 minutes / day:

{eq}D'(t)= \displaystyle A\frac{2 \pi}{365}\cos(\frac{2 \pi}{365}(t-80))= -2/60 {/eq}

{eq}\displaystyle 0.0395927 \cos(\frac{2 \pi}{365}(t-80))= -1/30 {/eq}

or

{eq}\cos(0.0172(t-80))=-0.8419 {/eq}

{eq}0.0172(t-80)=\cos^{-1}(-0.8419) = 2.5716 {/eq}

giving

{eq}t=229.4 {/eq}

3. The number of hours of daylight is increasing at a rate of 2 min/day on day 113 (24 April)

4. The number of hours of daylight is decreasing at a rate of 2 min/day on day: 229 (18 August)


Learn more about this topic:

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Using Differentiation to Find Maximum and Minimum Values

from Math 104: Calculus

Chapter 9 / Lesson 4
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