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Chapter 3: Problem 56

In Exercises \(37-66,\) graph the indicated functions. The voltage \(V\) across a capacitor in a certain electric circuit for a 2-s interval is \(V=2 t\) during the first second and \(V=4-2 t\) during the second second. Here, \(t\) is the time (in \(\mathrm{s}\) ). Plot \(V\) as a function of \(t\)

Short Answer

Expert verified
The graph is a piecewise linear function: increase from (0,0) to (1,2), followed by a decrease from (1,2) to (2,0).

Step by step solution

01

Identify the Time Intervals

The problem specifies that the voltage function has two distinct expressions depending on the time interval. For the first second (when \(0 \leq t < 1\)), the voltage is expressed as \(V = 2t\). For the second second (when \(1 \leq t \leq 2\)), the voltage is expressed as \(V = 4 - 2t\). Let's write down these two pieces clearly.
02

Graph the First Interval

For the interval \(0 \leq t < 1\), the voltage is given by the function \(V = 2t\). This is a linear function with a slope of 2. At \(t = 0\), \(V = 0\), and at \(t = 1\), \(V = 2\). Plot this line segment starting at the origin (0,0) and ending at (1,2).
03

Graph the Second Interval

In the interval \(1 \leq t \leq 2\), the voltage is expressed as \(V = 4 - 2t\). This is also a linear function, but with a negative slope of -2. At \(t = 1\), \(V = 2\), and at \(t = 2\), \(V = 0\). Plot this line segment starting from the point (1,2) and ending at the point (2,0).
04

Combine the Graphs

Since these are piecewise functions defined on consecutive intervals and share a common point, combine the two plotted segments to form the complete graph. Together, the intervals span from \(t = 0\) to \(t = 2\), with a peak at \(t = 1\). The graph starts at (0,0), peaks at (1,2), and returns to (2,0).

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Key Concepts

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

Graphing Linear Functions
Graphing linear functions involves plotting equations of straight lines on a coordinate plane. In our exercise, we have two different linear equations within specified intervals. Each of these equations represents a straight line, but it's essential to understand that each behaves differently within its interval.
  • For the first interval (\(0 \leq t < 1\)), the equation \(V = 2t\) forms a line starting at the origin (0,0) and rising to the point (1,2). The slope here is 2, indicating that for every second, the voltage increases by 2 volts.
  • In the second interval (\(1 \leq t \leq 2\)), the equation \(V = 4 - 2t\) represents a line starting at (1,2) and descending to (2,0). This line has a negative slope of -2, showing that the voltage decreases by 2 volts every second.
By connecting these two linear segments, we form a piecewise continuous graph that accurately represents the relationship between voltage and time over the 2-second interval.
Electric Circuits
Electric circuits involve components like resistors, capacitors, and inductors that collectively work to manage the flow of electric current. In our problem, we specifically deal with a capacitor, which is a device capable of storing and releasing electrical energy.
  • The voltage across the capacitor is described by two separate linear functions, showing how the stored energy changes over time.
  • For the capacitor, the voltage build-up (charging) occurs during the first second, reaching a peak value due to the linear increase described by \(V = 2t\)
  • During the second second, the capacitor discharges, showing a reduction in voltage as indicated by \(V = 4 - 2t\).
This variation in voltage is important for understanding how a capacitor in an electric circuit responds to time-varying signals.
Time Intervals
Time intervals are used to break down a continuous function into segments, allowing us to analyze different behaviors over specific durations. This is particularly helpful in piecewise functions, where each interval can have a different rule.
  • In our example, there are two 1-second intervals, each governed by a distinct linear equation for voltage.
  • These intervals help define when and how the voltage changes, whether it's increasing or decreasing.
  • By identifying these intervals, we can plot the behavior of the function accurately and understand how it evolves through time.
Recognizing these intervals and their corresponding equations is crucial for piecing together the complete graph and understanding the overall function dynamics.

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Most popular questions from this chapter

Graph the indicated functions. The rate \(H\) (in \(\mathrm{W}\) ) at which heat is developed in the filament of an electric light bulb as a function of the electric current \(I\) (in A) is \(H=240 I^{2} .\) Plot \(H\) as a function of \(I\).

Solve the given problems. The stopping distance \(d\) (in \(\mathrm{ft}\) ) of a car going \(v \mathrm{mi} / \mathrm{h}\) is given by \(d=v+0.05 v^{2} .\) Because \(d=f(v),\) find \(f(30), f(2 v),\) and \(f(60),\) using both \(f(v)\) and \(f(2 v)\)

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Use the following table that gives the rate \(R\) of discharge from a tank of water as a function of the height \(H\) of water in the tank. Find the indicated values by linear interpolation. $$\begin{array}{l|c|c|c|c|c|c|c} \text {Height} \text { (ft) } & 0 & 1.0 & 2.0 & 4.0 & 6.0 & 8.0 & 12 \\ \hline \text {Rate }\left(\mathrm{ft}^{3} / \mathrm{s}\right) & 0 & 10 & 15 & 22 & 27 & 31 & 35 \end{array}$$ Find \(R\) for \(H=1.7 \mathrm{ft}.\)

A function and how it is to be shifted is given. Find the shifted function and then display the given function and the shifted function on the same screen of a graphing calculator. y=\sqrt{2 x+1}, \text { up } 1, \text { left } 1
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