Why Unequal Axis Scales Matter
A tangent line may look steep or shallow on a graph, but its visual angle alone does not tell you the numerical value of the derivative. The scale on each axis matters. If one horizontal square represents several hours while one vertical square represents only a few liters, the tangent’s movement in grid squares must be converted into hours and liters before calculating its slope.
The goal is to estimate the contextual derivative: the instantaneous rate of change of the output quantity with respect to the input quantity at a specified point. A graph provides an estimate when the tangent line and its coordinates are read approximately.
Translate Grid Squares into Units
First identify what one grid square represents on each axis. The horizontal scale gives input units per square; the vertical scale gives output units per square. If a tangent moves right by \(r\) horizontal squares and up by \(s\) vertical squares, then its changes in the plotted quantities are obtained by multiplying by their respective scales.
A tangent that moves downward as you move to the right has a negative vertical change, so its rate is negative. The tangent’s sign comes from the direction of its change, not from whether the plotted quantities themselves are positive or negative.
Use a clear pair of points on the tangent line, preferably at grid intersections. The points do not have to be points on the original curve: they are points on the tangent. If the graph is curved, its tangent provides the local slope at the point of interest, while a line through two distant points on the curve would generally give an average rate over an interval instead.
A Derivation from the Graph Scales
Suppose one horizontal square represents \(a\) input units and one vertical square represents \(b\) output units. Let two points on the tangent be separated by \(r\) horizontal squares and \(s\) vertical squares. If the tangent rises, \(s\) is positive; if it falls, \(s\) is negative.
A movement of \(r\) squares corresponds to an input change of \(ra\) input units, because each square represents \(a\) input units.
A movement of \(s\) squares corresponds to an output change of \(sb\) output units, because each square represents \(b\) output units.
By the definition of slope, divide the output change by the input change. Thus the rate is \(\frac{sb}{ra}\), in output units per input unit.
This calculation explains why counting squares alone is not enough. The count \(s/r\) is a slope measured in vertical squares per horizontal square. To obtain the contextual rate, that count must be multiplied by the vertical scale and divided by the horizontal scale. The result is the tangent’s slope in the actual units named by the axes.
This reasoning assumes evenly spaced, linear axes and that the scale per square has been read correctly. For a plotted curve, the result is an estimate of its derivative because the tangent and the points chosen from it are read from a drawing.
Worked Examples
Worked Example: Water Volume Increasing Over Time
A graph shows the volume of water in a tank, in liters, against time, in hours. Each horizontal square represents \(2\) hours, and each vertical square represents \(5\) liters. At the marked time, the tangent rises \(3\) vertical squares while moving right \(2\) horizontal squares. Estimate the rate of change of water volume.
The input change is \(2\) horizontal squares times \(2\) hours per square:
The output change is \(3\) vertical squares times \(5\) liters per square:
Therefore, the estimated derivative is:
Check: the converted rise and run are \(15\) liters and \(4\) hours, and \(3.75(4)=15\) liters. The positive sign agrees with the tangent rising as time increases. At the marked time, the volume is increasing at approximately \(3.75\) liters per hour.
Worked Example: Amount of a Substance Decreasing
A graph shows the amount of a substance remaining, in milligrams, against time, in minutes. Each horizontal square represents \(5\) minutes, and each vertical square represents \(2\) milligrams. The tangent at a particular time falls \(3\) vertical squares as it moves right \(2\) horizontal squares. Estimate the contextual derivative.
Moving right, the horizontal change is positive:
The tangent falls, so its vertical change is negative:
Divide the output change by the input change:
Check: over the chosen tangent segment, a rate of \(-0.6\) milligrams per minute for \(10\) minutes gives \((-0.6)(10)=-6\) milligrams, matching the converted fall. The negative sign indicates that the amount is decreasing at approximately \(0.6\) milligrams per minute at that time.
Worked Example: Production per Worker
A graph shows production, measured in thousands of items, against the number of workers. Each horizontal square represents \(4\) workers, and each vertical square represents \(3\) thousand items. At the specified workforce size, the tangent rises \(2\) vertical squares for every \(5\) horizontal squares. Estimate the rate in items per worker.
The horizontal change is:
The vertical change, still expressed in the graph’s stated units, is:
Thus the estimated rate is:
Since \(1\) thousand items is \(1000\) items, convert the final rate:
Check: \(300\) items per worker across \(20\) workers corresponds to \(300(20)=6000\) items, which is the same as \(6\) thousand items. The tangent rises, so the estimated rate is positive: production is increasing by about \(300\) items per worker at the specified workforce size.
Use Tangent Points, Not the Curve’s Apparent Angle
When estimating from a graph, it helps to select two convenient points on the tangent line and read their grid positions. These points can be away from the point where the tangent touches the curve. Then convert the run and rise separately using the axis scales. If the tangent goes down from left to right, keep the run positive when moving right and give the rise a negative sign.
It is also useful to check that the units make sense. If the vertical axis measures liters and the horizontal axis measures hours, the derivative has units of liters per hour. Reversing the division would produce hours per liter, which describes a different ratio and is not the rate of change of the vertical quantity with respect to the horizontal one.
Common Pitfall: Dividing Square Counts Directly
Suppose the tangent rises \(3\) squares for a run of \(2\) squares. A tempting calculation is \(3/2=1.5\), followed by the claim that the rate is \(1.5\) output units per input unit. That conclusion is generally wrong when the axis scales differ. The number \(1.5\) is only vertical squares per horizontal square.
For example, if a vertical square represents \(5\) liters and a horizontal square represents \(2\) hours, then the actual rise is \(3(5)=15\) liters and the actual run is \(2(2)=4\) hours. The contextual rate is \(15/4=3.75\) liters per hour, not \(1.5\). A quick check is to multiply the proposed rate by the converted run: \(1.5(4)=6\) liters, which does not match the \(15\)-liter rise. The corrected rate gives \(3.75(4)=15\) liters.
Summary
To estimate a contextual derivative from a tangent line when the graph’s axis scales differ:
- Read the input and output units on the two axes.
- Determine how many input units and output units each grid square represents.
- Choose two points on the tangent and find its horizontal and vertical movements in squares.
- Convert the run and rise into input and output units.
- Divide output change by input change, include output-units-per-input-unit, and check the sign.
Keeping the two axis scales separate prevents a visual grid count from being mistaken for a contextual rate.
Check Your Understanding
For each question, use the stated tangent movement and axis scales to estimate the derivative. Include the sign and contextual units.
- The horizontal axis is time in hours, with \(3\) hours per square. The vertical axis is distance in kilometers, with \(4\) kilometers per square. A tangent rises \(2\) squares over a rightward run of \(3\) squares. What is the estimated rate?
- The horizontal axis is time in minutes, with \(2\) minutes per square. The vertical axis is an amount in grams, with \(5\) grams per square. A tangent falls \(4\) squares over a rightward run of \(2\) squares. What is the estimated rate?
- A tangent rises \(4\) vertical squares over \(2\) horizontal squares. Each vertical square represents \(6\) units of output, and each horizontal square represents \(3\) units of input. Find the rate in output units per input unit. Why is \(4/2\) not the contextual rate?
- Explain why a tangent that falls from left to right has a negative rate when the horizontal change is measured moving to the right.
