Why Units Matter in the Mean Value Theorem
A rate is not just a number. It describes how an output changes relative to an input, so its units record that relationship. If a position is measured in kilometers and time in hours, a rate of change has units of kilometers per hour. Changing the time unit changes the numerical value of the rate, even though the physical rate has not changed.
The Mean Value Theorem connects an average rate over an interval to an instantaneous rate at some point in that interval. To use that connection correctly in an application, track the units of the input, output, interval length, secant slope, and derivative together. Rescaling an interval is also useful, but the derivative with respect to the rescaled variable is not generally the same numerical quantity as the derivative with respect to the original input.
Secant Slopes and Derivatives Carry Units
For a function \(f\) on an interval \([a,b]\), the secant slope is
The numerator is an output change, and the denominator is an input change. Thus, if \(f\) is measured in liters and \(x\) in meters, the secant slope has units of liters per meter. The derivative \(f'(x)\), when it exists, has the same units: it describes an instantaneous output change per input change.
The Mean Value Theorem (MVT) states that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), where \(a<b\), then there is at least one \(c\in(a,b)\) such that
Both sides must use consistent units. The theorem does not require the input to be measured in any particular unit, but the endpoints, interval length, function values, and derivative must all be interpreted using the same choices of units.
Worked Example: Converting Time Before Applying the MVT
Worked Example: Converting Time Before Applying the MVT
A position model is \(s(t)=2t^2+1\), where \(s\) is in kilometers and \(t\) is in hours. Find the average velocity from 30 minutes to 90 minutes, then identify a time at which the instantaneous velocity equals that average.
First convert the times to the unit used by the model:
Evaluate the model at the converted endpoints:
The position change is \(5.5-1.5=4\) km, and the elapsed time is \(1.5-0.5=1\) h. Therefore, the average velocity is
The model is a polynomial, so it is continuous on \([0.5,1.5]\) and differentiable on \((0.5,1.5)\). Its derivative is \(s'(t)=4t\), in kilometers per hour. The MVT guarantees some \(c\in(0.5,1.5)\) for which
Check: \(1\) hour is inside the interval, and \(s'(1)=4(1)=4\text{ km/h}\), exactly the average velocity. The corresponding time is \(60\) minutes.
Changing Input Units Changes the Numerical Derivative
Suppose the same input can be measured in two units. If \(x\) is the numerical value in meters and \(X\) is the numerical value in centimeters, then \(X=100x\). A function written using \(x\) must be rewritten using \(X\) before differentiating with respect to \(X\). The output itself has not changed, but its formula in terms of the new numerical input has.
More generally, suppose \(X=kx\) for a positive conversion factor \(k\), and write the same output as \(F(X)=f(X/k)\). By the chain rule,
Thus, the derivative’s numerical value is divided by \(k\) when the input is expressed in a unit that is \(1/k\) as large. For centimeters in place of meters, \(k=100\), so a rate in output per centimeter is one hundredth of the corresponding numerical rate in output per meter.
Worked Example: Converting a Derivative from Per Meter to Per Centimeter
Worked Example: Converting a Derivative from Per Meter to Per Centimeter
Suppose an output \(y\), measured in liters, is modeled by \(y=3x^2\), where \(x\) is the numerical distance in meters. Compare the average rate and the MVT derivative on the interval from \(0.2\) m to \(0.8\) m, then express the result using centimeters.
At the endpoints, the outputs are
The output change is \(1.92-0.12=1.80\) L and the input change is \(0.8-0.2=0.6\) m. The average rate is
The function is a polynomial, so it satisfies the MVT conditions on this interval. Its derivative is \(y'(x)=6x\), with units of liters per meter. Setting it equal to the average rate gives
Now let \(X\) denote the numerical distance in centimeters, so \(x=X/100\). The same output, expressed in terms of \(X\), is
The original interval becomes \(20\leq X\leq 80\). Its endpoint outputs are still \(0.12\) L and \(1.92\) L, so the average rate in these units is
Differentiating the converted formula gives
Set this equal to the average rate:
Check: \(50\) cm lies between \(20\) cm and \(80\) cm, and \(Y'(50)=300/10000=0.03\text{ L/cm}\). This is the same point as \(0.5\) m. The rates also agree after conversion: \(3\text{ L/m}=0.03\text{ L/cm}\), because \(1\text{ m}=100\text{ cm}\).
Rescaling an Interval to the Unit Interval
An interval \([a,b]\) can be described using a dimensionless variable \(u\) that runs from \(0\) to \(1\). Let \(L=b-a>0\), and define \(x=a+Lu\). Thus \(u=0\) corresponds to \(x=a\), and \(u=1\) corresponds to \(x=b\). Define \(g(u)=f(a+Lu)\).
Rescaling is useful for organizing calculations, but it changes the variable with respect to which the derivative is taken. The chain rule gives \(g'(u)=L f'(a+Lu)\). The factor \(L\) is essential: it converts a rate per unit of \(x\) into a rate per unit of \(u\).
Why the Rescaled MVT Gives the Same Point
Assume \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\). The change of variable \(x=a+(b-a)u\) maps \([0,1]\) onto \([a,b]\). Consequently, \(g(u)=f(a+(b-a)u)\) is continuous on \([0,1]\) and differentiable on \((0,1)\), so the MVT applies to \(g\).
It gives some \(u_*\in(0,1)\) such that
By the chain rule, \(g'(u_*)=(b-a)f'(a+(b-a)u_*)\). Divide by \(b-a>0\), and set \(c=a+(b-a)u_*\). Then \(c\in(a,b)\), and
This derivation explains both why the MVT applies after rescaling and why the interval-length factor cannot be omitted.
Worked Example: Using a Rescaled Interval
Worked Example: Using a Rescaled Interval
Let \(f(x)=x^2\), where \(x\) is measured in meters and \(f(x)\) in square meters. Apply the MVT on \([2,8]\) by rescaling to \(0\leq u\leq1\).
The original endpoint values are \(f(2)=2^2=4\text{ m}^2\) and \(f(8)=8^2=64\text{ m}^2\). The original secant slope is
Now set \(x=2+6u\) m and define \(g(u)=f(2+6u)\). Then
The change in \(g\) from \(u=0\) to \(u=1\) is \(g(1)-g(0)=64-4=60\text{ m}^2\). Since the rescaled interval has length \(1\), its secant slope is \(60\text{ m}^2\) per unit of \(u\). The MVT equation for \(g\) is
Convert back to the original input: \(x_*=2+6(1/2)=5\) m. Check in the original variable: \(f'(x)=2x\), so \(f'(5)=10\text{ m}\), equal to the original secant slope. In the rescaled variable, \(g'(1/2)=12(5)=60\text{ m}^2\) per unit of \(u\). These are consistent because the interval length is \(6\) m, and \(6\text{ m}\cdot10\text{ m}=60\text{ m}^2\).
A Common Unit-Conversion Mistake
A frequent error is to divide by a time measured in minutes and then label the result as a rate per hour. For example, if an object travels \(12\) km in \(60\) minutes, the calculation
is correct, but \(0.2\) is not the numerical rate in kilometers per hour. Since \(60\) min equals \(1\) h, the rate is also
The two rates represent the same motion: \(0.2\text{ km/min}\times60\text{ min/h}=12\text{ km/h}\). The correction is not merely to change the unit label; the numerical value must change with the unit.
Before applying the MVT in a rate problem, check that the function’s input is expressed in the unit used by its formula, compute the interval length in that same unit, and label the derivative using output per input. If the interval is rescaled, include the interval-length factor when relating the new derivative to the original one.
Check Your Understanding
For each question, keep track of the units as well as the numerical calculations.
- A function \(p(t)\) measures position in meters and time \(t\) in seconds. What units does its secant slope have, and what units does \(p'(t)\) have?
- A model uses hours, but the requested interval is from 15 minutes to 75 minutes. What endpoints in hours should be used?
- If \(X=100x\) converts meters to centimeters, how does the derivative with respect to \(X\) relate to the derivative with respect to \(x\)?
- When \(x=a+(b-a)u\), what factor relates \(g'(u)\), for \(g(u)=f(x)\), to \(f'(x)\)?
- Why is a rate of \(0.4\) km per minute not the same numerical value as the corresponding rate in kilometers per hour?
