Question

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At ${40}^{\circ }$ north latitude, the length of a day is approximated by $$D(t)=12-3\cos \left(\frac{2\pi (t+10)}{365}\right)$$ where $D$ is measured in hours and $0\le t\le 365$ is measured in days, with $t=0$ corresponding to January 1 a. Approximately how much daylight is there on March 1 $(t=59)?$ b. Find the rate at which the daylight function changes. c. Find the rate at which the daylight function changes on March $1.$ Convert your answer to units of min/day and explain what this result means. d. Graph the function $y=D^{\prime }(t)$ using a graphing utility. e. At what times of the year is the length of day changing most rapidly? Least rapidly?

Short answer

Short answer: a. On March 1 (t=59), the number of daylight hours is approximately 11.11 hours. b. The rate of change of the daylight function is given by: $$D^{\prime }(t)=3\sin \left(\frac{2\pi (t+10)}{365}\right)\cdot \frac{2\pi }{365}$$ c. The rate of daylight change on March 1 is approximately 2.33 minutes/day, which means that the daylight hours are increasing at a rate of 2.33 minutes per day at this time. d. The graph of the derivative function, D'(t), shows how the rate of daylight change varies throughout the year, with its maximum and minimum values representing the most rapid daylight changes and its zero-crossings representing the least rapid daylight changes. e. The times of most rapid daylight change occur near the solstices (around June 21 and December 21), while the least rapid daylight change occurs near the equinoxes (around March 20 and September 22).

Step-by-step solution

a. Daylight hours on March 1

To find the daylight hours on March 1, we simply substitute $t=59$ into the given function $D(t)$: $$D(59)=12-3\cos \left(\frac{2\pi (59+10)}{365}\right)$$ Now compute the value of this expression.

b. Rate of change of the daylight function

To find the rate at which the daylight function changes, we need to compute the derivative of $D(t)$ with respect to $t$. Using the chain rule, we get: $$D^{\prime }(t)=\frac{d}{dt}[12-3\cos \left(\frac{2\pi (t+10)}{365}\right)]$$ $$D^{\prime }(t)=3\sin \left(\frac{2\pi (t+10)}{365}\right)\cdot \frac{d}{dt}\left[\frac{2\pi (t+10)}{365}\right]$$ $$D^{\prime }(t)=3\sin \left(\frac{2\pi (t+10)}{365}\right)\cdot \frac{2\pi }{365}$$

c. Rate of daylight change on March 1

To find the rate of change of daylight on March 1, we need to evaluate the derivative $D^{\prime }(t)$ at $t=59$: $$D^{\prime }(59)=3\sin \left(\frac{2\pi (59+10)}{365}\right)\cdot \frac{2\pi }{365}$$ Compute the value of this expression and convert its unit to minutes/day: $$\text{rate}=D^{\prime }(59)\cdot 60$$ This result represents the rate at which the daylight hours are increasing or decreasing on March 1. A positive value means daylight hours are increasing, while a negative value indicates they are decreasing.

d. Graphing the derivative function

Using a graphing utility, plot the function $y=D^{\prime }(t)$ over the interval $0\le t\le 365$. This graph will show how the rate of daylight change varies throughout the year.

e. Times of rapid and least rapid daylight change

To find the times of the year when the length of day is changing most rapidly and least rapidly, we can analyze the graph of the derivative function $D^{\prime }(t)$. - The most rapid change corresponds to the maximum and minimum values of the derivative function. - The least rapid change corresponds to the zero-crossings of the derivative function, where the rate of change is zero. Observe the graph of $D^{\prime }(t)$ to identify these points, and calculate the corresponding days.

Key concepts

Rate of Change

The rate of change is a fundamental concept in calculus. It tells us how a quantity changes over time or another variable. In this exercise, we are interested in how the number of daylight hours changes as the days pass. This is expressed as a function that, when differentiated, gives us the rate at which daylight changes.

To determine the rate of change, we need to find the derivative, which will provide us with the speed at which the daylight hours increase or decrease. For instance, if the derivative is positive, it means that the length of day is increasing; if negative, the day length is decreasing. This helps us understand how quickly or slowly seasonal transitions, like the move from winter to spring, are occurring.

This concept of rate of change is applicable in numerous real-world scenarios, helping to make predictions and understand trends, whether it's in weather patterns, stock prices, population growth, or the amount of daylight.

Derivatives

Derivatives are a core part of calculus that allow us to measure how a function changes. They give us the slope of the tangent line to a point on a graph. In our exercise, the derivative of the daylight function, noted as $D^{\prime }(t)$, tells us how the daylight changes at any given day $t$.

To find the derivative of a trigonometric function like our daylight function, we often use rules like the chain rule, which helps in differentiating composite functions. In this specific exercise, finding $D^{\prime }(t)$ involves applying the derivative to the cos function inside the daylength formula. The derivative of cos is $-\sin$, and applying the chain rule helps to find not just the rate but its nature over time.

• The derivative $D^{\prime }(t)=3\sin (\frac{2\pi (t+10)}{365})\cdot \frac{2\pi }{365}$ is derived by manipulating and differentiating the cosine function.
• We can evaluate $D^{\prime }(t)$ for specific days, like March 1, to find exact changes at those times.

Understanding derivatives allows for deeper insight into how the oscillating seasons affect daylight, bringing science to real-life experiences.

Trigonometric Functions

Trigonometric functions play a crucial role in modeling periodic phenomena, such as the variation of daylight hours throughout the year. The function given in the exercise, $D(t)=12-3\cos (\frac{2\pi (t+10)}{365})$, is based on the cosine function, which naturally describes oscillations and cycles.

Trigonometric functions like cosine are essential because they help to easily replicate periodic events. These functions are pivotal in many fields, ranging from engineering to environmental science, wherever there are cyclical events or patterns.

• Cos functions often model natural and rhythmic processes, like daylight length, thanks to their wave-like properties.
• In our exercise, the adjustments made within the cosine function (like the $+10$ shift) help tailor the model to match real-world timing, such as calibrating daylight cycles to fit actual seasonal shifts.

When working through problems like this one, recognizing the pattern of cosine not just in math but in nature furthers our appreciation of how mathematics underlies natural phenomena we observe daily.