**Linear Approximation**

Let $y=f(x)$ be a differentiable function. The function $f(x)$ can be approximated by the tangent line to $y=f(x)$ at $a$ if $x$ is near $a$. Such an approximation is called a *linear approximation*.

If $x\approx a$ then $\Delta x=x-a\approx 0$, so we have

\begin{align*}

\frac{\Delta y}{\Delta x}&\approx \frac{dy}{dx}\\

&=f’(a).

\end{align*}

This means that

$$\frac{f(x)-f(a)}{x-a}\approx f’(a),$$

i.e.

\begin{equation}

\label{eq:lineapprox}

f(x)\approx f(a)+f’(a)(x-a).

\end{equation}

The equation \eqref{eq:lineapprox} is called the *linear approximation* or *tangent line approximation* of $f$ at $a$. The linear function

\begin{equation}

L(x):=f(a)+f’(a)(x-a)

\end{equation}

is called the *linearization* of $f$ at $a$. Notice that $L(x)$ is the equation of tangent line to $f$ at $a$.

*Example*. Find the linearlization of $f(x)=\sqrt{x+3}$ at $a=1$ and use it to approximate $\sqrt{3.98}$ and $\sqrt{4.05}$.

*Solution*.

$f’(x)=\frac{1}{2\sqrt{x+3}}$, so

\begin{align*}

L(x)&=f(1)+f’(1)(x-1)\\

&=2+\frac{1}{4}(x-1)\\

&=\frac{x}{4}+\frac{7}{4}.

\end{align*}

When $x\approx 1$, we have the approximation

$$\sqrt{x+3}\approx \frac{x}{4}+\frac{7}{4}.$$

$\sqrt{3.98}$ can be written as $\sqrt{3+0.98}$. Hence,

\begin{align*}

\sqrt{3.98}&\approx \frac{0.98}{4}+\frac{7}{4}\\

&=1.995.

\end{align*}

$\sqrt{4.05}$ can be written as $\sqrt{4+1.05}$. Hence,

\begin{align*}

\sqrt{4.05}&\approx \frac{1.05}{4}+\frac{7}{4}\\

&=2.0125.

\end{align*}

**Differentials**

As seen in the above figure, when $\Delta x\approx 0$, $\Delta x=dx$ and $\Delta y\approx dy$. On the other hand, $\frac{dy}{dx}=f’(x)$. Hence, we obtain

\begin{equation}

\label{eq:differential}

\Delta y\approx f’(x)\Delta x.

\end{equation}

*Example*. The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?

*Solution*. Let $V$ denote the volume of a sphere of radius $r$. Then $V=\frac{4}{3}\pi r^3$. What we are trying to find is $\Delta V$ with $\Delta r=0.05$ cm. As seen in \eqref{eq:differential}, $\Delta V\approx dV$, so we find $dV$ instead because finding $dV$ is easier than findingthe exact error $\Delta V$. Differentiating $V$ with respect to $r$, we obtain

\begin{align*}

dV&=4\pi r^2 dr\\

&=4\pi r^2\Delta r\\

&=4\pi\cdot(21)^2\cdot 0.05\\

&=277.

\end{align*}

So the maximum error in the calculated volume is about 277 $\mbox{cm}^3$.