1.10 Lines and linear applications

When one quantity changes at a constant rate relative to another, a line provides the natural mathematical model.

Slope. The vertical change divided by the horizontal change between two points on a nonvertical line.

The slope through \((x_1,y_1)\) and \((x_2,y_2)\) is

\[m=\frac{y_2-y_1}{x_2-x_1}.\]

Useful line forms are point-slope form \[y-y_1=m(x-x_1)\], slope-intercept form \[y=mx+b\], and general form \[Ax+By+C=0\]. Point-slope form is convenient when a point and slope are known. Slope-intercept form displays the constant rate of change and vertical intercept. In general form, set \(x=0\) to find the vertical intercept and \(y=0\) to find the horizontal intercept.

Vertical lines have undefined slope, horizontal lines have slope zero, parallel lines have equal slopes, and perpendicular nonvertical lines have slopes whose product is \(-1\). Slope has units. If volume is plotted against time, slope is a volume-per-time rate. Its sign indicates whether the response increases or decreases as the input increases.

1.10.1 Intercepts and related lines

The \(y\)-intercept is found by setting \(x=0\), and the \(x\)-intercept is found by setting \(y=0\). For \(y=5x-7\), the intercepts are \((0,-7)\) and \((7/5,0)\).

Worked Example: Finding intercepts

For \(y=5x-7\), setting \(x=0\) gives the \(y\)-intercept \((0,-7)\). Setting \(y=0\) gives \(5x-7=0\), so the \(x\)-intercept is \((7/5,0)\).

Try It. Find both intercepts of \(y=3x+6\).

Check Your Work

The intercepts are \((0,6)\) and \((-2,0)\).

Two distinct nonvertical lines are parallel when their slopes are equal. Two nonvertical lines are perpendicular when their slopes are negative reciprocals. A horizontal and a vertical line are also perpendicular.

A linear model assumes constant change. Its slope must be interpreted with units and within the interval where that assumption is reasonable. Extrapolation far beyond observed values may predict impossible negative populations, volumes, or concentrations, which signals that the linear model should no longer be used.

Worked Example: Modelling tank volume over time

A tank begins with 4.0 m³ and fills at 0.35 m³/min. Its volume after \(t\) minutes is

\[V=0.35t+4.0.\]

The slope is the inflow rate, and the vertical intercept is the initial volume.

Try It. Write a linear model for a tank beginning at 7 m³ and draining at 0.25 m³/min.

Check Your Work

\(V=-0.25t+7\).

Worked Example: Temperature conversion

Water freezes at \((0,32)\) and boils at \((100,212)\) in Celsius-Fahrenheit coordinates. Thus

\[m=\frac{212-32}{100-0}=\frac95,\]

and the conversion equation is

\[F=\frac95C+32.\]

Try It. Use the conversion equation to find the Fahrenheit temperature at \(C=20\).

Check Your Work

\(F=(9/5)(20)+32=68^\circ\text{F}\).

Worked Example: A linear calibration

A sensor gives outputs 2.0 V at 10 mg/L and 5.0 V at 40 mg/L. The slope is \((5-2)/(40-10)=0.1\) V per mg/L. Using the first point gives \(V=0.1C+1\).

Try It. Find the output predicted at 25 mg/L.

Check Your Work

\(V=0.1(25)+1=3.5\) V.

Worked Example: Determining a sensor equation from two readings

A turbidity sensor produces 1.4 V at 0.5 NTU and 3.8 V at 2.5 NTU. The slope is

\[m=\frac{3.8-1.4}{2.5-0.5}=1.2\ \text{V/NTU}.\]

Using \((0.5,1.4)\) gives \(V-1.4=1.2(T-0.5)\), or \(V=1.2T+0.8\).

Try It. A sensor reads 2 V at 10 L/s and 6 V at 30 L/s. Find its linear equation.

Check Your Work

The slope is \((6-2)/(30-10)=0.2\) V per L/s. Using either point gives \(V=0.2q\).

Applied Problem: Interpreting a declining reservoir model. During a controlled drawdown, reservoir volume is modelled by \(V=-1.8t+72\), where \(V\) is in megalitres and \(t\) is in hours. Interpret the slope and intercept, find the volume after 15 h, and determine when the model predicts an empty reservoir.

Check Your Work

The intercept 72 ML is the initial volume. The slope \(-1.8\) ML/h is the net loss rate. At 15 h, \(V=-1.8(15)+72=45\) ML. Setting \(V=0\) gives \(t=40\) h.

To compare two lines, rewrite each equation in slope-intercept form unless it is vertical. Equal slopes identify parallel lines. Slopes with product \(-1\) identify perpendicular lines. A line through a specified point and parallel to a known line uses the known slope; a perpendicular line uses its negative reciprocal.

1.10.2 Practice Problems

1. Find the slope through \((1,2)\) and \((5,10)\).
   Check Your Work

   \(m=2\).
2. Find the line with slope 3 through \((2,1)\).
   Check Your Work

   \(y=3x-5\).
3. Find both intercepts of \(y=4x-8\).
   Check Your Work

   \((0,-8)\) and \((2,0)\).
4. Is \(y=2x+1\) parallel to \(2y=4x-7\)?
   Check Your Work

   Yes. Both slopes are 2.
5. Find a line perpendicular to \(y=(1/2)x+3\) through the origin.
   Check Your Work

   \(y=-2x\).
6. A tank contains 12 m³ initially and fills at 0.6 m³/min. Write its volume model and find the volume after 20 min.
   Check Your Work

   \(V=0.6t+12\); \(V(20)=24\) m³.
7. Flow rises from 18 L/s at 2 min to 30 L/s at 8 min. Find the slope and interpret it.
   Check Your Work

   \(m=(30-18)/(8-2)=2\) L/s per min. Flow increases by 2 L/s each minute.
8. A linear chlorine-sensor model passes through \((0.2,1.1)\) and \((0.8,2.9)\), where the input is mg/L and output is volts. Find the model.
   Check Your Work

   \(m=(2.9-1.1)/(0.8-0.2)=3\). Using the first point gives \(V=3C+0.5\).