Elevation and Road Grade
A road’s elevation tells us how high it is at a particular location. Road grade describes how quickly that elevation changes as we move horizontally along the route. The distinction matters: a road can be at a high elevation and still be level, or be at a low elevation and climb steeply.
Let \(x\) denote horizontal distance measured in the direction of travel, and let \(E(x)\) denote the road’s elevation at that location. The derivative \(E'(x)\), when it exists, measures the instantaneous rate of change of elevation with respect to horizontal distance. In this setting, that rate is the road’s local slope or grade ratio.
For example, a grade ratio of \(0.04\) means that elevation changes by about \(0.04\) units for each horizontal unit near that location. Expressed as a percentage, this is a \(4\%\) grade. A negative value indicates a decrease in elevation as \(x\) increases.
Why the Derivative Represents Local Grade
To estimate the grade across a finite stretch, compare the elevation change with the horizontal distance travelled. Between \(x\) and \(x+h\), the average grade ratio is the change in elevation divided by the change in horizontal distance:
This is the slope of the line joining the two points on the elevation graph. The derivative at \(x\) is the limit of these secant slopes as the second point approaches the first. Therefore the derivative describes the local, instantaneous grade rather than the average grade over a stretch.
provided this limit exists. This also explains why differentiability is required for an instantaneous grade in the usual two-sided sense: the nearby secant slopes must approach one definite value.
Worked Example: A Constant Uphill Grade
Worked Example: A Constant Uphill Grade
Suppose a road’s elevation, in metres, is modeled by \(E(x)=132+0.025x\), where \(x\) is horizontal distance in metres. Find the grade and interpret it.
Use the difference quotient at any location \(x\). First calculate the elevation at \(x+h\):
The change in elevation is therefore
For \(h\ne 0\), the difference quotient is
Its limit as \(h\to 0\) is \(E'(x)=0.025\). Converting the ratio to a percentage gives \(100(0.025)\%=2.5\%\). The road rises \(2.5\) metres per \(100\) horizontal metres.
Check: Substituting the derivative into the grade conversion gives \(100E'(x)\%=100(0.025)\%=2.5\%\), as stated. The result is positive, consistent with the model’s elevation increasing as \(x\) increases.
Positive, Negative, and Zero Grades
The sign of \(E'(x)\) describes the direction of local elevation change relative to the chosen direction of increasing \(x\). The derivative’s sign does not describe whether the elevation itself is positive or negative.
| Derivative value | Grade interpretation |
|---|---|
| \(E'(x)>0\) | Elevation is increasing locally: the road climbs in the direction of increasing \(x\). |
| \(E'(x)<0\) | Elevation is decreasing locally: the road descends in the direction of increasing \(x\). |
| \(E'(x)=0\) | The tangent is horizontal: the road has zero instantaneous grade at that location. |
A zero derivative at one location does not mean the road is level over an entire interval. It says only that the local tangent slope is zero there. The elevation may change on either side of that location.
Worked Example: A Downhill Grade
Suppose \(E(x)=210-0.035x\), with elevation and horizontal distance both measured in metres. Find and interpret the grade.
At \(x+h\), the elevation is
Subtract the elevation at \(x\):
Thus, for \(h\ne 0\),
The difference quotient is constant, so its limit is \(E'(x)=-0.035\). The percentage grade is \(100(-0.035)\%=-3.5\%\). The negative sign means that elevation falls by \(3.5\) metres for each \(100\) horizontal metres travelled in the direction of increasing \(x\).
Check: A \(100\)-metre horizontal change in the model gives \(E(x+100)-E(x)=-0.035(100)=-3.5\) metres. This agrees with the negative \(3.5\%\) grade.
Worked Example: A Zero Grade at One Location
Worked Example: A Zero Grade at One Location
Consider the elevation model \(E(x)=86+0.0002(x-50)^2\), with both variables measured in metres. Find the grade at \(x=50\) and explain what the result does—and does not—say.
At \(x=50\), the elevation is
At a nearby location \(50+h\),
Therefore, for \(h\ne 0\), the difference quotient at \(50\) is
As \(h\to 0\), \(0.0002h\to 0\), so \(E'(50)=0\). The grade at \(x=50\) is \(100(0)\%=0\%\). The road has a horizontal tangent there.
This does not mean the model is constant. For example, \(E(40)=86+0.0002(-10)^2=86.02\) metres, while \(E(50)=86\) metres. Also, \(E(60)=86+0.0002(10)^2=86.02\) metres. Elevation changes away from \(50\), even though the instantaneous grade at \(50\) is zero.
Check: The difference quotient is \(0.0002h\), whose limit at \(h=0\) is \(0\). That verifies the zero derivative and the zero-percent grade at the specified location.
Instantaneous Grade and Average Grade
For a whole road segment from \(x=a\) to \(x=b\), where \(a\ne b\), the average grade ratio is
This is the slope of the secant line across the segment. The instantaneous grade \(E'(x)\), by contrast, concerns the local change at a single location. These values need not be the same. A segment may contain both uphill and downhill portions whose net elevation change is small, even though the local grades are not small everywhere.
When the model is linear, the grade is constant, so the average grade over any segment equals the instantaneous grade at every point. For a curved elevation graph, the grade can vary with location. A grade calculation should therefore identify whether it refers to a particular point or an entire segment.
Units and Percentage Conversion
If \(E\) is measured in metres and \(x\) in metres, then \(E'(x)\) has units of metres per metre. The units cancel, leaving a dimensionless ratio. In road practice, that ratio is commonly reported as a percentage by multiplying by \(100\). Thus a ratio of \(0.06\) corresponds to a \(6\%\) grade, not a \(0.06\%\) grade.
If elevation and horizontal distance are recorded in different units, convert them to compatible units before interpreting the ratio. For instance, a rate in metres per kilometre is not yet the metres-per-metre grade ratio: \(1\) kilometre is \(1000\) metres. Unit conversion is part of interpreting the numerical derivative correctly.
Worked Example: Converting Units Before Reporting Grade
Suppose the measured elevation rate is \(24\) metres per kilometre. Convert it to a percentage grade.
First convert the horizontal distance to metres. Since \(1\) kilometre equals \(1000\) metres, the rate becomes
The dimensionless grade ratio is \(0.024\). As a percentage, it is
If the original rate is positive, this describes a \(2.4\%\) climb in the direction of increasing distance; if it is negative, it describes a \(2.4\%\) descent.
Check: A \(2.4\%\) grade means a rise of \(2.4\) metres per \(100\) horizontal metres. Over \(1000\) metres, that corresponds to \(24\) metres, matching the stated rate.
A Common Percentage-Grade Mistake
A frequent error is to report the decimal derivative itself as the percentage. For example, if \(E'(x)=0.04\), writing “the road has a \(0.04\%\) grade” is incorrect. The number \(0.04\) is the grade ratio; multiplying it by \(100\) gives the percentage grade:
A related interpretation error is to infer grade from the elevation value alone. A road at \(E(x)=500\) metres is not necessarily climbing: grade depends on \(E'(x)\), not on \(E(x)\). For example, a constant elevation model has a positive elevation value but derivative zero.
Check Your Understanding
Use the definitions and interpretations in this tutorial to answer the following.
- If \(E'(x)=0.018\) and elevation and horizontal distance use the same units, what is the percentage grade, and what does its sign mean?
- A road has derivative \(-0.027\) at a location. How many metres does its elevation change per \(100\) horizontal metres in the direction of increasing \(x\)?
- Why is the average grade over a segment generally not the same quantity as the instantaneous grade at one point?
- A model has \(E'(a)=0\). What local geometric conclusion follows, and why does this not establish that the road is level on an interval around \(a\)?
- A rate is reported as \(15\) metres per kilometre. What is the corresponding percentage grade ratio, assuming the rate is positive?
