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Chapter 7: Problem 58

Length of a Shadow On a day when the sun passes directly overhead at noon, a six-foot-tall man casts a shadow of length $$ S(t)=6\left|\cot \frac{\pi}{12} t\right| $$ where \(S\) is measured in feet and \(t\) is the number of hours since 6 A.M. (a) Find the length of the shadow at \(8 : 00\) A.M. noon, \(2 : 00\) P.M. and \(5 : 45\) P.M. (b) Sketch a graph of the function \(S\) for \(0

Short Answer

Expert verified
At 8:00 AM, the shadow is approximately 10.39 ft; at noon, 0 ft; at 2:00 PM, 3.46 ft; and at 5:45 PM, 51.96 ft. The shadow equals 6 ft at 9:00 AM and 3:00 PM. The shadow becomes extremely long as it approaches 6 PM.

Step by step solution

01

Evaluate the Shadow Length at Specified Times

To find the shadow length at 8:00 A.M., noon, 2:00 P.M., and 5:45 P.M., substitute each time into the function \( S(t) = 6 \left| \cot \frac{\pi}{12} t \right| \). These times correspond to \( t = 2 \), \( t = 6 \), \( t = 8 \), and \( t = 11.75 \), respectively.1. For \( t = 2 \), \[ S(2) = 6 \left| \cot \frac{\pi}{12} \cdot 2 \right| = 6 \left| \cot \frac{\pi}{6} \right| \]. Since \( \cot \frac{\pi}{6} = \sqrt{3} \), \( S(2) = 6\sqrt{3} \approx 10.39 \text{ ft} \).2. For \( t = 6 \), \[ S(6) = 6 \left| \cot \frac{\pi}{12} \cdot 6 \right| = 6 \left| \cot \frac{\pi}{2} \right| \]. The cotangent of \( \pi/2 \) is undefined, meaning the shadow length is 0 feet.3. For \( t = 8 \), \[ S(8) = 6 \left| \cot \frac{\pi}{12} \cdot 8 \right| = 6 \left| \cot \frac{2\pi}{3} \right| \]. Since \( \cot \frac{2\pi}{3} = -\frac{1}{\sqrt{3}} \), \( S(8) = 6 \left( \frac{1}{\sqrt{3}} \right) = 2\sqrt{3} \approx 3.46 \text{ ft} \).4. For \( t = 11.75 \), \[ S(11.75) = 6 \left| \cot \frac{\pi}{12} \cdot 11.75 \right| = 6 \left| \cot \frac{11.75\pi}{12} \right| \]. The cotangent value approximates to casting a longer shadow, \( S(11.75) \approx 51.96 \text{ ft} \).
02

Sketch the Shadow Function Graph

The function \( S(t) \) is periodic and piecewise due to the absolute value of the cotangent component. The graph will oscillate between large positive values and zero (at every multiple of 6 hours, such as noon). There will be symmetry around these zero crossings as well. Graph the function carefully illustrating these features over the domain \( 0 < t < 12 \). The shadow length dips to zero at \( t = 6 \) and grows large near \( t = 12 \).
03

Determine when Shadow Length Equals Man's Height

To find \( t \) when the shadow equals the man's height (6 ft), solve \( 6 = 6 \left| \cot \frac{\pi}{12} t \right| \). This simplifies to \( 1 = \left| \cot \frac{\pi}{12} t \right| \), implying \( \cot \frac{\pi}{12} t = 1 \) or \( \cot \frac{\pi}{12} t = -1 \).- \( \cot \frac{\pi}{4} = 1 \), gives \( \frac{\pi}{12} t = \frac{\pi}{4} \) making \( t = 3 \).- \( \cot \frac{3\pi/4} = -1 \), gives \( \frac{\pi}{12} t = \frac{3 \pi}{4} \) making \( t = 9 \).So the shadow equals 6 feet at 9:00 AM and 3:00 PM.
04

Evaluate Shadow Behavior Near 6 P.M.

As \( t \to 12^, \) evaluate \( S(t) = 6 \left| \cot \frac{\pi}{12} t \right| \). \( \cot \frac{\pi}{12} t \) appears to become very small since \( \cot 0 = \infty \). Therefore, the shadow length becomes extremely large as it approaches 6 PM, indicating elongated shadow close to sunset.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Shadow Length
The concept of shadow length is often explored in trigonometry, particularly when discussing the impact of the sun's position in relation to objects on the ground. It's important to visualize how the length of a shadow changes throughout the day, depending on the sun's angle in the sky.
In the exercise, shadow length is described by the function \( S(t) = 6 \left| \cot \frac{\pi}{12} t \right| \). The cotangent function plays a key role here, dictating how the shadow behaves at different times \( t \), measured in hours since 6 A.M.
  • Morning: As the sun rises, the angle of elevation increases, causing the shadow to shorten until it reaches noon.
  • Noon: At this point, the shadow is the shortest, becoming momentarily zero as the sun passes directly overhead.
  • Afternoon: Gradually, as the sun starts to set, the shadow lengthens again.

This periodic change emphasizes how mathematically captivating and real-life pertinent trigonometric functions can be when modelled with time-dependent variables.
Periodic Functions
Periodic functions repeat their values in regular intervals or periods. In the context of the shadow length function, \( S(t) \), the periodicity arises from the cotangent function, which repeats every half rotation, or \( \pi \) radians, of the angle.
Understanding the periodicity of the cotangent is crucial since this determines how often the shadow length becomes zero or attains maximum values.
The implication of periodicity can be extended to:
  • The concept of cycles: Everyday, the shadow length will follow a predictable pattern of growing and shrinking.
  • Time intervals: Recognizing that the maximum or minimum shadow lengths occur at regular periods helps us compute and intuit significant points, like midday.
Such predictability and repetition form the backbone of periodic functions, and this becomes immensely useful not only in mathematical models but in practical scenarios like measuring time or seasons.
Cotangent
In trigonometry, the cotangent function \( \cot \theta = \frac{1}{\tan \theta} \) is the reciprocal of the tangent. It is defined as the ratio of the adjacent side to the opposite side in a right-angled triangle.
The cotangent function is particularly essential in the given exercise since it influences the shape and behavior of the shadow function \( S(t) \).
Here's why the cotangent is fascinating:
  • It becomes undefined at angles like \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \), hinting at the vertical nature of shadows at these specific times (like noon).
  • It switches signs around these discontinues, offering complementary behavior when shadows lengthen or shorten dramatically.
This reciprocal nature and the introduction of periodic discontinuities allow the cotangent to model phenomena in cycles, like daily shadow changes, with great precision.

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Most popular questions from this chapter

Blood Pressure Each time your heart beats, your blood pressure increases, then decreases as the heart rests between beats. A certain person's blood pressure is modeled by the function $$ p(t)=115+25 \sin (160 \pi t) $$ where \(p(t)\) is the pressure in mmHg at time \(t,\) measured in minutes. (a) Find the amplitude, period, and frequency of \(p .\) (b) Sketch a graph of \(p\) . (c) If a person is exercising, his or her heart beats faster. How does this affect the period and frequency of \(p ?\)

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For an object in simple harmonic motion with amplitude \(a\) and period \(2 \pi / \omega,\) find an equation that models the displacement \(y\) at time \(t\) if (a) \(y=0\) at time \(t=0 : y=\) _________ (b) \(y=a\) at time \(t=0 : y=\) _________

Find the exact value of the trigonometric function at the given real number. (a) \(\cos \left(-\frac{\pi}{4}\right) \quad\) (b) \(\csc \left(-\frac{\pi}{4}\right) \quad\) (c) cot \(\frac{5 \pi}{3}\)
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