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Ryan and Kandra are kayaking near Lowe Inlet Marine Provincial Park on Grenville Channel, British Columbia. The current can flow in either direction at up to \(4 \mathrm{km} / \mathrm{h}\) depending on tidal conditions. Ryan and Kandra are capable of kayaking steadily at \(4 \mathrm{km} / \mathrm{h}\) without the current. a) What function relates the time, \(t,\) in hours, it will take them to travel \(4 \mathrm{km}\) along the channel as a function of the speed, \(w\), in kilometres per hour, of the current? What domain is possible for \(w\) in this context? b) Graph the function for an appropriate domain. c) Explain the behaviour of the graph for values at and near its non- permissible value and what the behaviour means in this situation.

Short Answer

Expert verified
The function is \( t(w) = \frac{4}{4 \text{ or } ± w} \) with domain \( -4 < w < 4 \). The graph shows asymptotes at \( w = ±4 \). Kayaking is impossible when the current's speed equals the kayak's speed.

Step by step solution

01

Define the Variables and Understand the Problem

Identify that the total speed while kayaking is affected by the current, and the total speed can be represented as the sum or difference of the kayak's speed and the current speed. Denote the current's speed by the variable \( w \).
02

Establish the Function for Time

Use the relationship between distance, speed, and time: \( \text{time} = \frac{\text{distance}}{\text{speed}} \). Apply this to the problem: the distance is \( 4 \text{ km} \), and the speed would be either \( 4 + w \) km/h when traveling with the current or \( 4 - w \) km/h when traveling against the current.
03

Write Down the Formula

When traveling with the current, the function for time \( t \) is given by \( t = \frac{4 \text{ km}}{4 + w \text{ km/h}} \). When traveling against the current, the function is \( t = \frac{4 \text{ km}}{4 - w \text{ km/h}} \). Therefore, the general function is \( t(w) = \frac{4}{4 \text{ or } + w} \).
04

Determine the Domain

The domain of the function is determined by the permissible values of \( w \) for which the function is defined. Since the speed \( 4 - w \) km/h or \( 4 + w \) km/h cannot be zero (as division by zero is undefined), \( w \) cannot be \( \text{±4} \). The domain of \( w \) in this context is \( -4 < w < 4 \).
05

Graph the Function

Graph the function \( t = \frac{4}{4 \text{ or } + w} \) over the domain \( -4 < w < 4 \). Ensure that the graph correctly represents the function, showing asymptotes at \( w = ±4 \).
06

Analyze the Graph's Behavior

Observe that as \( w \) approaches \( ±4 \), the time \( t \) increases significantly, indicating an asymptote. This indicates that it would take infinitely long to travel 4 km if the current speed matches or opposes the kayak's speed exactly (i.e., 4 km/h). In practical terms, kayaking becomes impossible if the current's speed equals the kayak's speed.

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Key Concepts

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

Speed-Distance-Time Relationship
Understanding the relationship between speed, distance, and time is crucial for solving any motion-related problem. In this kayaking scenario, the distance remains fixed at 4 km, but the total speed changes due to the current. The time taken for travel is derived from the formula: \ \( \text{time} = \frac{\text{distance}}{\text{speed}} \).

This formula means:
  • Time (\( t \)) is directly proportional to distance (\( d \))
  • Time (\( t \)) is inversely proportional to speed (\( v \))

In this problem:
  • When traveling with the current: \( v = 4 + w \)
  • When traveling against the current: \( v = 4 - w \)

Thus, time taken to cover 4 km:
  • With the current: \( t = \frac{4}{4 + w} \)
  • Against the current: \( t = \frac{4}{4 - w} \)
This understanding helps establish a clear link between speed, distance, and time in varying current conditions.
Function Domain and Range
The domain and range of the function describe all possible values of the input (speed of the current) and the output (time taken, respectively). Here, we need to determine the domain of the function \( t(w) \).

For the function \( t(w) = \frac{4}{4 + w} \) or \( t(w) = \frac{4}{4 - w} \), \( w \) cannot be \( \pm 4 \) as these values make the denominator zero,
which leads to an undefined time:
  • For \( w = 4 \), \( 4 + 4 = 0 \)
  • For \( w = -4 \), \( 4 - 4 = 0 \)
Thus, we can define the domain as the set of all \( w \) values that lie between -4 and 4:
\[ -4 < w < 4 \]

The range, on the other hand, represents all possible time values. As the speed decreases towards 0 or increases towards -4 or 4, the time \( t \), gets very large. Thus, the range of the function is:\
  • \( t > 0 \)
This means the time required will always be positive but it can get very large as the current speed approaches the kayak speed.
Asymptotic Behavior
Understanding asymptotic behavior is important to analyze how the function behaves as the input values approach certain critical points. An asymptote is a line that the graph of a function approaches but never touches. In the context of this kayaking problem, the function \( t(w) \) displays vertical asymptotes at\
  • \( w = 4 \)
  • \( w = -4 \)


Analyzing the function \( \ t(w) = \frac{4}{4 + w} \) or \( t(w)= \frac{4}{4 - w} \), the critical points occur when \( w = 4 \) and \( w = -4 \).

As \( w \) approaches 4 or -4:
  • The denominator of the function approaches zero
  • The value of \( t(w) \) increases towards infinity

Graphically, this is shown as the curve getting closer and closer to the vertical line at \( w = 4 \) or \( w = -4 \), but never actually crossing it.
In practical terms for the kayaking scenario, if the current's speed reaches exactly 4 km/h either aiding or opposing the kayak's speed, they would theoretically take an infinite amount of time to travel 4 km, making normal kayaking impossible under these conditions.

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Most popular questions from this chapter

A truck leaves Regina and drives eastbound. Due to road construction, the truck takes \(2 \mathrm{h}\) to travel the first \(80 \mathrm{km} .\) Once it leaves the construction zone, the truck travels at \(100 \mathrm{km} / \mathrm{h}\) for the rest of the trip. a) Let \(v\) represent the average speed, in kilometres per hour, over the entire trip and \(t\) represent the time, in hours, since leaving the construction zone. Write an equation for \(v\) as a function of \(t\) b) Graph the function for an appropriate domain. c) What are the equations of the asymptotes in this situation? Do they have meaning in this situation? Explain. d) How long will the truck have to drive before its average speed is \(80 \mathrm{km} / \mathrm{h} ?\) e) Suppose your job is to develop GPS technology. How could you use these types of calculations to help travellers save fuel?

Predict the location of any asymptotes and points of discontinuity for each function. Then, use technology to check your predictions. a) \(y=\frac{x+2}{x^{2}-4}+\frac{5}{x+2}\) b) \(y=\frac{2 x^{3}-7 x^{2}-15 x}{x^{2}-x-20}\)

A convex lens focuses light rays from an object to create an image, as shown in the diagram. The image distance, \(I\), is related to the object distance, \(b,\) by the function \(I=\frac{f b}{b-f},\) where the focal length, \(f,\) is a constant for the particular lens used based on its specific curvature. When the object is placed closer to the lens than the focal length of the lens, an image is perceived to be behind the lens and is called a virtual image. a) Graph \(I\) as a function of \(b\) for a lens with a focal length of \(4 \mathrm{cm} .\) b) How does the location of the image change as the values of \(b\) change? c) What type of behaviour does the graph exhibit for its non-permissible value? How is this connected to the situation?

Write the equation of a possible rational function with each set of characteristics. a) vertical asymptotes at \(x=\pm 5\) and \(x\) -intercepts of -10 and 4 b) a vertical asymptote at \(x=-4,\) a point of discontinuity at \(\left(-\frac{11}{2}, 9\right)\) and an \(x\) -intercept of 8 c) a point of discontinuity at \(\left(-2, \frac{1}{5}\right)\) a vertical asymptote at \(x=3,\) and an \(x\) -intercept of -1 d) vertical asymptotes at \(x=3\) and \(x=\frac{6}{7},\) and \(x\) -intercepts of \(-\frac{1}{4}\) and 0

\- Mira uses algebra to rewrite the function \(y=\frac{2-3 x}{x-7}\) in an equivalent form that she can graph by hand. \(y=\frac{2-3 x}{x-7}\) \(y=\frac{-3 x+2}{x-7}\) \(y=\frac{-3 x-21+21+2}{x-7}\) \(y=\frac{-3(x-7)+23}{x-7}\) \(y=\frac{-3(x-7)}{x-7}+\frac{23}{x-7}\) \(y=-3+\frac{23}{x-7}\) \(y=\frac{23}{x-7}-3\) a) Identify and correct any errors in Mira's work. b) How might Mira have discovered that she had made an error without using technology? How might she have done so with technology?

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