Comparing Constant Rates

Comparing Constant Rates exercise

• Convert between different representations of constant rates.

  Hints

  The slope of this line is $\frac{5}{1}$, or $5$. The equation is $y=5x$.

  This line has a slope of $\frac13$. The equation is $y=\frac13x$.

  In the table, as time $t$ increases by $1$, distance $d$ increases by $6$. That means the slope is $\frac61$ or $6$. The equation is $d=6t$. The points in the table are on the graph of the line.

  In the table, as time increases by $7$, distance increases by $1$. That means the slope is $\frac17$ and the equation is $d=\frac17t$. The points in the table are on the graph of the line.

  The slope of this line is $\frac19$. To get a point on the graph of this line, we can plug in $9$ for $x$ and get $1$ for $y$. That gives us the point $(9,1)$. We can also plug in $18$ for $x$ and get $2$ for $y$. That gives us the point $(18,2)$.

  Solution

  The points $(0,0)$ and $(2,1)$ are on one graph. There is a table which contains the same values.

  The points $(0,0)$ and $(1,4)$ are on one graph. There is a table which contains the same values.

  The points $(0,0)$ and $(1,8)$ are on one graph. The slope of this line is $\frac81$. The equation $d=8t$ has the same slope of $8$.

  The points $(0,0)$ and $(4,1)$ are on one graph. The slope of this line is $\frac14$. The equation $d=\frac14 t$ has the same slope of $\frac14$.

  The points $(0,0)$ and $(1,1)$ are on one graph. The slope is $1$. The equation $d=t$, that is, $d=1t$, has the same slope of $1$.

• Compare and contrast the following linear equations.

  Hints

  For the equation $y=3x$, the slope is 3. For the equation $y=\frac13x$, the slope is $\frac13$.

  If the slope of a linear equation is positive, it slopes upwards as we look at the graph from left to right.

  The larger the slope, the steeper the graph.

  All linear equations of the form $y=mx$ have a y-intercept of $0$. That means all such lines go through the point $(0,0)$, the origin.

  All linear equations of the form $y=mx$ are straight lines.

  Solution

  \begin{array}{c|c|c} \text{equation} & \text{slope} & \text{y-intercept}\\ \hline d=8t & 8 & 0\\ d=4t & 4 & 0\\ d=\frac14 t & \frac14 & 0\\ \end{array}

  Here we can see the graphs of these three equations, along with a table organizing some information about them.

  Slope and steepness

  • The slope is the number multiplied by the variable $t$. The larger the slope, the steeper the graph.

  • If the slope is positive, the graph slopes up from left to right.

  • A slope of $\frac41$ is different from a slope of $\frac14$.

  The y-intercept

  • Each graph goes through the origin $(0,0)$, so they all have the same y-intercept of $0$.

  Finding points on the graphs

  \begin{array}{l|l|l|c} \text{equation} & \text{if}& \text{then} &\text{point on graph}\\ \hline d=4t & t=3 &d=4(3)=12 & (3,12) \\ d=\frac14 t & t=12 & d=\frac14(12)=3 & (12,3) \end{array}

• Graph the constant rate equation, $y=2x$.

  Hints

  Given the line $y=5x$, we can find points on the line by substituting values for $x$.

  For example, if $x=3$, then:

  $y=3(5)=15$

  Here is a table showing several points on the line $y=5x$:

  \begin{array}{c|c} x & y\\ \hline 0 & 0\\ 1 & 5\\ 2 & 10\\ 3& 15\\ 10 & 50\\ \end{array}

  The graph of every constant rate equation goes through the origin, that is, the point (0,0).

  Given the line $y=3x$, the slope is $3$, or $\frac31$. That means that from the origin $(0,0)$, we go $1$ unit right and $3$ units up, to arrive at the point $(1,3)$.

  Solution

  This is the graph of $y=2x$.

  If $x=3$, then:

  $y=2(3)=6$

  Here is a table showing several points on the line $y=2x$:

  \begin{array}{c|c} x & y\\ \hline 0 & 0\\ 1 & 2\\ 2 & 4\\ 3& 6\\ 10 & 20\\ \end{array}

• Determine the constant rate equation for each graph.

  Hints

  This is the graph of the line $y=(1/3)x$.

  We can identify two points on the graph as shown, and draw a slope triangle. The slope triangle shows us that for each increase of $3$ in the $x$ direction, there is a corresponding increase of $1$ in the $y$ direction. This means the slope is $\frac13$.

  We type the equation in as $y=(1/3)x$.

  Solution

  How to use a slope triangle to write a constant rate equation:

  1. Connect two points that are on the line and on the grid.
  2. Draw a right triangle as shown to connect the two points.
  3. Count the horizontal and vertical change.
  4. Find the slope by dividing the vertical change by the horizontal change.
  5. Use the slope, $k$, to write the equation in the form $y=kx$.

  Example (see diagram):

  1. We choose the points $(0,0)$ and $(1,7)$.
  2. We draw a slope triangle.
  3. The horizontal change is $7$. The vertical change is $1$.
  4. The slope is $\frac17$.
  5. The equation is $y=\frac17 x$. We type that in as $y=(1/7)x$.

  The slopes and equations for the remaining lines:

  $ \begin{array}{c|l} \text{slope} & \text{line} \\ \hline 3 & y=3x\\ \frac16 & y=(1/6)x\\ 1 & y=x\\ \end{array} $

• Order the equations from the steepest graph to the least steep graph.

  Hints

  The steepness of the graph is determined by the slope. A bigger slope means a steeper graph.

  Given an equation in the form $y=mx$, the slope is $m$.

  The slope of the line $y=10x$ is $10$. The slope of the line $y=\frac19x$ is $\frac19$. The graph of $y=10x$ is steeper than the graph of $y=\frac19$.

  Solution

  • The steepest line is $y=12x$. This is because it has the largest slope, $12$.

  • The least steep line is $y=\frac15 x$. It has the smallest slope, $\frac15$.

  • Remember, the line $y=x$ can be written as $y=1x$. It has a slope of 1.

• Compare the constant rates represented by graphs, tables, and equations.

  Hints

  Given a table like the one shown, we can find the slope by noticing that as $x$ increases by $1$, $y$ increases by $20$. This means the slope is $\frac{20}{1}$, or $20$. The equation is $y=20x$.

  Given a graph as shown, we can connect any two points on the line (and on the grid) in order to show a slope triangle. The slope triangle shows us that for every increase of $1$ in the $x$ direction, there is a corresponding increase of $5$ in the $y$ direction. That means the slope is $\frac51$, or $5$. Therefore, the equation is $y=5x$.

  Given two equations, we can compare their slopes to determine which is steeper. For example, let's compare $y=\frac13 x$ and $y=\frac14 x$. $\frac1/3$ is greater than $\frac14$, so the line $y=\frac13 x$ is steeper.

  Solution

  First, we have the graph of $y=\frac18 x$. The graph goes through the point $(8,1)$. The slope of this line is $\frac18$. This represents a rate of $1$ car every $8$ minutes.

  Second, we have a table. When $x$ increases by $1$, $y$ increases by $11$. The slope of this line is $11$. Cars are passing at a rate of $11$ cars per minute. This line is the steepest of the three, because its slope is the largest. We can represent this constant rate with the equation $y=11x$.

  Third, we have the equation $y=\frac19x$ which has a slope of $\frac19$. This is the smallest slope of the three. In this case, cars are passing at a rate of $1$ car every $9$ minutes.