Linear approximation and differentials
“All models are wrong but some are useful.”
A known value and a small step
We know that \(\sqrt{4}=2\), but what about \(\sqrt{4.1}\)? The input has increased by just \(0.1\). The derivative tells us how quickly the square root changes near \(4\), so we can use it to estimate the increase in the output.
At \(x=4\), the graph of \(f(x)=\sqrt{x}\) passes through \((4,2)\) with slope \(1/4\). Multiplying the input step by this slope predicts an output increase of \(\frac14(0.1)=0.025\). Adding that increase to the known value gives \(\sqrt{4.1}\approx2.025\).
Why should a straight line work for a curve? Differentiability says that the secant slope approaches the tangent slope as the step shrinks.
Definition
Linearization
If \(f\) is differentiable at \(a\), its linearization at \(a\) is the tangent-line function
\[L(x)=f(a)+f^{\prime}(a)(x-a).\]The linear approximation is \(f(x)\approx L(x)\) for inputs near \(a\). The base input \(a\) stays fixed; the input \(x\) varies.
At the base point the curve and line agree in both value and slope. Away from it they usually differ. The approximation’s usefulness depends on how far we move and how quickly the slopes change; “small” is relative to the problem.
Choose a nearby base input where both the value and derivative are easy to evaluate.
Why differentiability gives this approximation (optional)
For a step \(h\ne0\), subtract the tangent prediction from the actual value and divide by the step:
\[\begin{gathered} \frac{f(a+h)-L(a+h)}h\\ =\frac{f(a+h)-f(a)}h-f^{\prime}(a). \end{gathered}\]The derivative definition gives
\[\lim_{h\to0}\frac{f(a+h)-L(a+h)}h=0.\]This is stronger than saying the error tends to zero: the error becomes small relative to the input step. It does not give a numerical error bound for a particular nonzero step.
Actual change and predicted change
Definition
Differentials
For an input change \(\Delta x\) from \(a\), define
\[\Delta y=f(a+\Delta x)-f(a).\]The differentials are \(dx=\Delta x\) and \(dy=f^{\prime}(a)\,dx\). Thus
\[\begin{aligned} \Delta y&\approx dy,\\ f(a+\Delta x)&\approx f(a)+dy. \end{aligned}\]Here \(\Delta y\) is the actual change along the curve; \(dy\) is the change along the tangent. Neither is the new output value itself.
Example 1
An estimate on the other side
Estimate \(\sqrt[3]{7.94}\). Choose \(f(x)=x^{1/3}\) and \(a=8\), so \(f(8)=2\) and \(f^{\prime}(8)=1/12\). The step is negative:
\[\begin{aligned} dx&=7.94-8=-0.06,\\ dy&=\frac1{12}(-0.06)=-0.005. \end{aligned}\]Therefore \(\sqrt[3]{7.94}\approx1.995\). Subtracting the original input from the new input fixes the sign automatically.
Example 2
A calibration model
A sensor produces \(R(T)\) millivolts at temperature \(T\) in degrees Celsius. At its calibration temperature,
\[\begin{aligned} R(20)&=50\text{ mV},\\ R^{\prime}(20)&=0.8\text{ mV}/\mathrm{°C}. \end{aligned}\]At \(20.3\mathrm{°C}\) the predicted change is \(0.8(0.3)=0.24\text{ mV}\), so the estimated reading is \(50.24\text{ mV}\). The units cancel as they should. These two calibration facts determine the tangent estimate; they do not determine the exact reading away from \(20\).
Relative change and measurement uncertainty
When \(f(a)\ne0\), dividing the predicted change by the original value gives a relative differential \(dy/f(a)\). Multiply its magnitude by \(100\%\) for an estimated percentage change.
Example 3
How a radius error affects volume
A sphere has volume \(V=\frac43\pi r^3\). Differentiating with respect to radius gives
\[dV=4\pi r^2\,dr,\qquad \frac{dV}{V}=3\frac{dr}{r}.\]For a measured radius \(10\text{ cm}\) with possible error of magnitude \(0.02\text{ cm}\), the predicted volume error has magnitude
\[\begin{aligned} \lvert dV\rvert&=4\pi(10)^2(0.02)\\ &=8\pi\text{ cm}^3. \end{aligned}\]The relative estimate is \(3(0.02/10)=0.006\), or \(0.6\%\). The factor \(3\) explains why a small relative radius error is amplified in the volume.
This is a first-order error estimate. The exact change also includes terms involving \((dr)^2\) and \((dr)^3\), which the differential discards.
Warning
A tangent approximation does not by itself certify an error tolerance. In particular, “within \(0.6\%\)” is stronger than “an estimated error of \(0.6\%\).” Later in this module, the mean value theorem will give bounds on actual changes.