Problem 73 - Precalculus: Functions and Graphs

Question

On a clear day with \(D\) hours of daylight, the intensity of sunlight \(I\) (in calories/cm^) may be approximated by $$ I=I_{M} \sin ^{3} \frac{\pi t}{D} \quad \text { for } \quad 0 \leq t \leq D $$ where \(t=0\) corresponds to sunrise and \(I_{\mathrm{M}}\) is the maximum intensity. If \(D=12,\) approximately how many hours after sunrise is \(I=\frac{1}{2} I_{\mathrm{M}} ?\)

Step-by-Step Solution

Answer

Approximately 3 hours after sunrise.

Step 1: Substitute Values into the Equation

First, we need to substitute the given values into the equation. We know that \(D = 12\) and we are looking for \(t\) when \(I = \frac{1}{2}I_M\). Substitute these into the function: \[\frac{1}{2}I_M = I_M \sin^3\left(\frac{\pi t}{12}\right)\].

Step 2: Simplify the Equation

Next, we simplify the equation from Step 1. Divide both sides by \(I_M\): \[ \frac{1}{2} = \sin^3\left(\frac{\pi t}{12}\right) \].

Step 3: Solve for the Sine Function

Take the cube root of both sides to isolate the sine function: \[ \sqrt[3]{\frac{1}{2}} = \sin\left(\frac{\pi t}{12}\right) \].

Step 4: Determine the Angle

Find the angle that satisfies the equation. Use the inverse sine function: \[ \sin^{-1}\left(\sqrt[3]{\frac{1}{2}}\right) = \frac{\pi t}{12} \].

Step 5: Solve for t

Multiply both sides of the equation by 12 to solve for \(t\): \[ t = \frac{12}{\pi}\sin^{-1}\left(\sqrt[3]{\frac{1}{2}}\right) \]. Calculate the value of \(\sin^{-1}\left(\sqrt[3]{0.5}\right)\) to find \(t\).

Step 6: Approximate the Solution

Using a calculator, \(\sin^{-1}\left(\sqrt[3]{0.5}\right) \approx 0.775397497\). Therefore, \[ t \approx \frac{12}{\pi} \times 0.775397497 \approx 2.957 \]. This means the intensity is half of \(I_M\) approximately 3 hours after sunrise.

Key Concepts

Sunlight Intensity

Sunlight intensity can be thought of as the power of sunlight reaching an area, measured in calories per square centimeter in this context. This concept helps us understand how sunlight varies throughout the day.
During a clear day, sunlight intensity changes as the Earth rotates and the angle of the sun changes. This variation can be modeled using trigonometric functions, which provide a way to represent cyclical patterns. The equation used in the original exercise models sunlight intensity as a sinusoidal function of time, illustrating how it gradually increases to a maximum and then decreases as the day progresses.

• Early in the day (sunrise), the intensity starts from zero.
• Midday sees the highest intensity, denoted as maximum intensity, \(I_M\).
• Towards sunset, the intensity decreases back to zero.

Understanding how sunlight intensity changes during the day allows scientists and researchers to better plan for energy usage, such as in solar panels, and understand environmental impacts.

Inverse Sine Function

The inverse sine function, denoted as \( \sin^{-1}(x) \), is used to find the angle whose sine is \(x\). It is also known as arcsine. This function is particularly useful when you have a sine value and need to determine the corresponding angle.
In the context of our problem, once the sine part was isolated, the inverse sine function was used to find the specific angle \( \frac{\pi t}{12} \) that corresponds to a certain value of sine. This is a crucial step in solving trigonometric equations that are based on angles and is a standard approach in solving them.

• The range of the inverse sine function is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• It helps in converting from a sine value back to an angle in radian measure, which is often necessary in problems involving periodic functions.

This function's role is pivotal in problems involving periodic behavior, where we need to trace back from a known function value to the possible angles.

Cube Root

The cube root, notated as \( \sqrt[3]{x} \), is another fundamental mathematical operation, similar to square root, but for finding a number that, when multiplied by itself twice (cubed), gives the original number \(x\).
In the sunlight intensity equation, simplifying the equation involved taking the cube root of both sides. This step addressed the presence of \( \sin^3(x) \), which had resulted from the intensity equation's mathematical form.

• Finding cube roots helps simplify equations where variables are raised to the third power.
• This mathematical operation is important when solving trigonometric functions that include power terms.

Understanding cube roots and practicing how they apply to real-world problems, like in trigonometric equations, strengthens mathematical skills and problem-solving capabilities.

Amplitude of Trigonometric Functions

Amplitude refers to the height of the wave created by a trigonometric function. It represents the maximum value (height) from the function's average value, or midline, to its peak.
In trigonometric functions modeling physical phenomena such as sound waves, light waves, or, in this case, sunlight intensity, amplitude signifies the maximum deviation. It's specifically the peak intensity \( I_M \) when the sinusoidal function hits its maximum.

• Determines the "strength" or "height" of the wave in a trigonometric graph.
• It influences the scale of variation in the function modeling natural phenomena.

Understanding amplitude is crucial for interpreting the scale and scope of real-world cycles and waves, significant in fields like physics, engineering, and environmental studies.