Linear approximation and differentials
If you zoom in on a smooth curve, it looks more and more like a straight line. That line is the tangent line. So near a point, we can use the tangent line in place of the curve. Lines are easy to compute with, so this gives quick estimates.
The linearization
The tangent line to $y=f(x)$ at $x=a$ is
$y=f(a)+f'(a)(x-a)$.
(See tangent lines and rates of change.) We call this linear function the linearization of $f$ at $a$, and write
$L(x)=f(a)+f'(a)(x-a)$.
For $x$ near $a$,
$f(x)\approx L(x)$.
This is called the linear approximation, or the tangent line approximation. The symbol $\approx$ means "is approximately equal to."
Example 1: Find the linearization of
$f(x)=\sqrt{x}$ at $a=4$.
Use it to estimate $\sqrt{4.1}$ and $\sqrt{3.9}$.
Solution:
We need $f(4)$ and $f'(4)$:
$f(4)=\sqrt{4}=2$
$f'(x)=\dfrac{1}{2\sqrt{x}}$, so $f'(4)=\dfrac{1}{4}$.
The linearization:
$L(x)=2+\dfrac{1}{4}(x-4)$
Now put each value into $L(x)$:
$\begin{align*}&\sqrt{4.1}\\&\approx 2+\tfrac{1}{4}(0.1)\\&=2.025\end{align*}$
$\begin{align*}&\sqrt{3.9}\\&\approx 2+\tfrac{1}{4}(-0.1)\\&=1.975\end{align*}$
A calculator gives $\sqrt{4.1}\approx 2.02485$ and $\sqrt{3.9}\approx 1.97484$. Both estimates are correct to three decimal places.
The estimate is good only near $a$. For example, $L(9)=3.25$, but $\sqrt{9}=3$. Far from $x=4$, the curve bends away from the line.
Here the curve bends down, below its tangent line. So every estimate from $L(x)$ is a little too big.
Example 2: Find the linearization of
$f(x)=\sin x$ at $a=0$.
Solution:
$f(0)=\sin 0=0$, and $f'(x)=\cos x$, so $f'(0)=\cos 0=1$.
$L(x)=0+1\cdot(x-0)=x$
So for small $x$ (in radians),
$\sin x\approx x$.
Here is how good it is. Set your calculator to radian mode to check these values.
The error is $x-\sin x$.
| $x$ (radians) | $\sin x$ | Error |
|---|---|---|
| $0.5$ | $0.47943$ | $0.02057$ |
| $0.2$ | $0.19867$ | $0.00133$ |
| $0.1$ | $0.09983$ | $0.00017$ |
| $0.05$ | $0.04998$ | $0.00002$ |
This is the small-angle approximation. Physics uses it often, for example for a swinging pendulum.
Example 3: Show that
$(1+x)^k\approx 1+kx$
for $x$ near $0$. Use it to estimate $(1.02)^{10}$.
Solution:
Let $f(x)=(1+x)^k$. Then $f(0)=1$.
By the chain rule, $f'(x)=k(1+x)^{k-1}$, so $f'(0)=k$.
So $L(x)=1+kx$. This gives
$(1+x)^k\approx 1+kx$.
For $(1.02)^{10}$, use $x=0.02$ and $k=10$:
$(1.02)^{10}\approx 1+10(0.02)=1.2$
A calculator gives $(1.02)^{10}\approx 1.21899$. The estimate is off by about $0.02$, because $0.02$ is small but $k=10$ is not.
Differentials
Let $y=f(x)$. Suppose $x$ changes by a small amount. We call this change $dx$ (or $\Delta x$). Then:
The actual change in $y$ is
$\Delta y=f(x+dx)-f(x)$.
The differential of $y$ is
$dy=f'(x)\,dx$.
It is the change along the tangent line.
For small $dx$,
$\Delta y\approx dy$.
This is the linear approximation again, written with changes. The differential $dy$ is usually much easier to compute than $\Delta y$.
The symbols $dy$ and $dx$ now have their own meaning. Their quotient is the derivative:
$\dfrac{dy}{dx}=f'(x)$.
This matches the Leibniz notation for the derivative.
Example 4: Let $y=x^3$. Compare $\Delta y$ and $dy$ when $x$ changes from $2$ to $2.1$.
Solution:
Here $x=2$ and $dx=0.1$.
The actual change:
$\begin{align*}&\Delta y\\&=(2.1)^3-2^3\\&=9.261-8\\&=1.261\end{align*}$
The differential: $dy=3x^2\,dx$, so
$dy=3(2)^2(0.1)=1.2$.
The differential is close to the actual change, and much easier to find.
Estimating errors in measurements
Every measurement has a small error. When we compute a quantity from a measurement, the error carries over. Differentials estimate how big the new error is.
- If a measurement $x$ has an error $dx$, the computed quantity $y=f(x)$ has an error of about $dy=f'(x)\,dx$.
- The relative error is the error divided by the quantity, $\dfrac{dy}{y}$. As a percent, it is called the percentage error.
Example 5: The radius of a ball is measured as $21$ cm, with a possible error of $0.05$ cm. Estimate the possible error in the computed volume. Also find the relative error.
Solution:
The volume of a sphere is
$V=\dfrac{4}{3}\pi r^3$.
So $dV=4\pi r^2\,dr$.
Put in $r=21$ and $dr=0.05$:
$\begin{align*}&dV\\&=4\pi(21)^2(0.05)\\&=88.2\pi\\&\approx 277\ \text{cm}^3\end{align*}$
The relative error:
$\begin{align*}&\dfrac{dV}{V}\\&=\dfrac{4\pi r^2\,dr}{\tfrac{4}{3}\pi r^3}\\&=3\cdot\dfrac{dr}{r}\\&=3\cdot\dfrac{0.05}{21}\\&\approx 0.0071\end{align*}$
So the volume could be off by about $277$ cm$^3$. That is a percentage error of about $0.71\%$.
Notice that the relative error in the volume is $3$ times the relative error in the radius. The exponent $3$ in $r^3$ triples it.
Summary
- The linearization $L(x)$ of $f$ at $a$ is the tangent line at $a$.
- For $x$ near $a$, $f(x)\approx L(x)$. The estimate gets worse farther from $a$.
- The differential is $dy=f'(x)\,dx$. It estimates the actual change $\Delta y$.
- An error $dx$ in a measurement gives an error of about $dy$ in the result.