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Chapter 1: Problem 3

Altitude: A helicopter takes off from the roof of a building that is 200 feet above the ground. The altitude of the helicopter increases by 150 feet each minute. a. Use a formula to express the altitude of a helicopter as a function of time. Be sure to explain the meaning of the letters you choose and the units. b. Express using functional notation the altitude of the helicopter 90 seconds after takeoff, and then calculate that value. c. Make a graph of altitude versus time covering the first 3 minutes of the flight. Explain how the description of the function is reflected in the shape of the graph.

Short Answer

Expert verified
a. \( h(t) = 200 + 150t \); b. 425 feet; c. A line from (0,200) to (3,650) shows linear growth.

Step by step solution

01

Understanding the Problem

The helicopter starts from an altitude of 200 feet and gains 150 feet per minute. We need to describe the altitude as a function, calculate the altitude at a specific time, and create a graph of this function.
02

Create the Function Formula

Let \( t \) represent time in minutes and \( h(t) \) represent the altitude in feet. The helicopter starts at 200 feet and gains altitude at a rate of 150 feet per minute. Thus, the function for altitude is:\[ h(t) = 200 + 150t \]Here, 200 is the initial altitude, and 150 is the rate of ascent.
03

Calculate Altitude at Specific Time

Given time is 90 seconds, which is equivalent to 1.5 minutes. Substitute 1.5 for \( t \) in the function:\[ h(1.5) = 200 + 150(1.5) \]Calculate:\[ h(1.5) = 200 + 225 = 425 \]Thus, the altitude at 90 seconds is 425 feet.
04

Graph the Function

The function \( h(t) = 200 + 150t \) is a linear equation with a y-intercept of 200 and a slope of 150, meaning that the graph is a straight line starting at 200 on the y-axis. For the first 3 minutes, calculate altitude for key points:- At \( t=0 \), \( h(0) = 200 \).- At \( t=1 \), \( h(1) = 350 \).- At \( t=2 \), \( h(2) = 500 \).- At \( t=3 \), \( h(3) = 650 \).Plotting these points shows a straight line sloping upwards, illustrating constant ascent.

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Key Concepts

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

Rate of Change
The rate of change is essentially how one quantity, such as altitude, changes related to another, such as time. In the context of this exercise, it's crucial because it tells us how quickly the helicopter climbs into the sky. For the helicopter:
  • The starting altitude is 200 feet.
  • The altitude increases by 150 feet for every passing minute.
This gives us a rate of change of 150 feet per minute. If you imagine the helicopter鈥檚 journey skyward, this rate is what describes the steady rise.
However, it鈥檚 more than just numbers; it tells a story about motion and speed. Knowing how fast the altitude changes helps in predicting future paths or stops.
In equations, the rate of change often shows up as the coefficient of the variable representing time. In this example, \[h(t) = 200 + 150t\] the 150 next to the \(t\) indicates the helicopter's consistent rate of ascent with time.
Understanding rate of change is foundational in describing how systems behave over time, not just in heights but financial trends, speeds, and more.
Linear Function
Linear functions are a type of function that generates a straight line when graphed. They follow a pattern of consistent change, like stepping up a hill at a constant pace.
The general formula for these functions is \( y = mx + b \), where:
  • \(m\) represents the slope or rate of change of the function.
  • \(b\) stands for the y-intercept, representing the starting point on a graph.
In our helicopter scenario, this becomes \( h(t) = 200 + 150t \), a perfect example of a linear function! \(150\) is how steep or fast the line climbs, and \(200\) is where the helicopter starts its journey in terms of altitude.
Linear functions are prized for their predictability. Once you鈥檝e got your formula, you can determine where you鈥檒l land just by plugging in a value for \( t \) (how much time has passed).
What鈥檚 handy is the way these functions can represent many real-world events, not just helicopter altitudes but also things like income predictions, expenses, or much more!
Graphing Functions
Graphing functions is like capturing the behavior of an equation over time on paper. In this activity, we plot the helicopter's altitude against time.
  • Time, \( t \), is often on the x-axis.
  • Altitude, \( h(t)\), is vertical on the y-axis.
By using the formula \(h(t) = 200 + 150t\), we place points on the graph for every time check. For example:
  • At \( t = 0 \), altitude is 200 feet.
  • At \( t = 1 \), altitude rises to 350 feet.
  • Keep going this way.
This steadily increasing line beautifully showcases our linear function at work.
Consistently spaced points mean that the helicopter's ascent is consistent鈥攋ust like the line shows.
Graphing isn鈥檛 just for math class; we use these snapshots to interpret data in fields like economics, science, and engineering. They become visual summaries of the story behind the numbers.

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Most popular questions from this chapter

Flushing chlorine: City water, which is slightly chlorinated, is being used to flush a tank of heavily chlorinated water. The concentration \(C=C(t)\) of chlorine in the tank \(t\) hours after flushing begins is given by \(C=0.1+2.78 e^{-0.37 t}\) milligrams per gallon a. What is the initial concentration of chlorine in the tank? b. Express the concentration of chlorine in the tank after 3 hours using functional notation, and then calculate its value.

The \(3 x+1\) problem: Here is a mathematical function \(f(n)\) that applies only to whole numbers \(n\). If a number is even, divide it by 2 . If it is odd, triple it and add 1. For example, 16 is even, so we divide by 2: \(f(16)=\frac{16}{2}=8\). On the other hand, 15 is odd, so we triple it and add \(1: f(15)=3 \times 15+1=46\). a. Apply the function \(f\) repeatedly beginning with \(n=1\). That is, calculate \(f(1), f\) (the answer from the first part), \(f\) (the answer from the second part), and so on. What pattern do you see? b. Apply the function \(f\) repeatedly beginning with \(n=5\). How many steps does it take to get to 1 ? c. Apply the function \(f\) repeatedly beginning with \(n=7\). How many steps does it take to get to \(1 ?\) d. Try several other numbers of your own choosing. Does the process always take you back to 1 ? (Note: We can't be sure what your answer will be here. Every number that anyone has tried so far leads eventually back to 1 , and it is conjectured that this happens no matter what number you start with. This is known to mathematicians as the \(3 x+1\) conjecture, and it is, as of the writing of this book, an unsolved problem. If you can find a starting number that does not lead back to 1 , or if you can somehow show that the path always leads back to 1 , you will have solved a problem that has eluded mathematicians for a number of years. Good hunting!)

Head and aquifers: This is a continuation of Exercise 21. In underground water supplies such as aquifers, the water normally permeates some other medium such as sand or gravel. The head for such water is determined by first drilling a well down to the water source. When the well reaches the aquifer, pressure causes the water to rise in the well. The head is the height to which the water rises. In this setting, we get the pressure using Pressure \(=\) Density \(\times 9.8 \times\) Head \(.\) Here density is in kilograms per cubic meter, head is in meters, and pressure is in newtons per square meter. (One newton is about a quarter of a pound.) A sandy layer of soil has been contaminated with a dangerous fluid at a density of 1050 kilograms per cubic meter. Below the sand there is a rock layer that contains water at a density of 990 kilograms per cubic meter. This aquifer feeds a city water supply. Test wells show that the head in the sand is \(4.3\) meters, whereas the head in the rock is \(4.4\) meters. A liquid will flow from higher pressure to lower pressure. Is there a danger that the city water supply will be polluted by the material in the sand layer?

Tubeworm: An article in Nature reports on a study of the growth rate and life span of a marine tubeworm. \({ }^{12}\) These tubeworms live near hydrocarbon seeps on the ocean floor and grow very slowly indeed. Collecting data for creatures at a depth of 550 meters is extremely difficult. But for tubeworms living on the Louisiana continental slope, scientists developed a model for the time \(T\) (measured in years) required for a tubeworm to reach a length of \(L\) meters. From this model the scientists concluded that this tubeworm is the longest-lived noncolonial marine invertebrate known. The model is $$ T=14 e^{1.4 L}-20 \text {. } $$ A tubeworm can grow to a length of 2 meters. How old is such a creature? (Round your answer to the nearest year.)

Sound pressure and decibels: Sound exerts a pressure \(P\) on the human ear. \({ }^{6}\) This pressure increases as the loudness of the sound increases. If the loudness \(D\) is measured in decibels and the pressure \(P\) in dynes \({ }^{7}\) per square centimeter, then the relationship is given by $$ P=0.0002 \times 1.122^{D} . $$ b. A decibel level of 120 causes pain to the ear and can result in damage. What is the corresponding pressure level on the ear?
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