# Linear Approximations

After studying the differentials, we know that if $\Delta{y}={f{{\left({a}+\Delta{x}\right)}}}-{f{{\left({a}\right)}}}$ and ${d}{y}={f{'}}{\left({x}\right)}\Delta{x}$. This means that $\Delta{x}$ becomes very small, i.e. if we let $\Delta{x}\to{0}$, we can write that ${d}{y}\approx\Delta{y}$.

This can be rewritten as ${f{{\left({a}+\Delta{x}\right)}}}-{f{{\left({a}\right)}}}\approx{f{'}}{\left({x}\right)}\Delta{x}$.

If we now let ${x}={a}+\Delta{x}$, we obtain that ${f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}\approx{f{'}}{\left({x}\right)}{\left({x}-{a}\right)}$.

Definition. The linearization of a function ${y}={f{{\left({x}\right)}}}$ at a point ${a}$ is ${f{{\left({x}\right)}}}\approx{f{{\left({a}\right)}}}+{f{'}}{\left({x}\right)}{\left({x}-{a}\right)}$.

We need this linearization because it might be easy to calculate the value of a function ${f{{\left({a}\right)}}}$ but it might be difficult (or even impossible) to compute the nearby values of ${f{}}$. So, we settle for the easily computed values of the linear function (tangent line).

Example 1. Find the linear approximation of ${y}=\sqrt{{{x}}}$ at ${\left({4},{2}\right)}$. Then approximate $\sqrt{{{3.99}}}$, $\sqrt{{{4.01}}}$, and $\sqrt{{{4.05}}}$.

Since ${f{'}}{\left({x}\right)}=\frac{{1}}{{{2}\sqrt{{{x}}}}}$, we have that ${f{'}}{\left({4}\right)}=\frac{{1}}{{4}}$.

Therefore, linear approximation ${L}{\left({2}\right)}=\frac{{1}}{{4}}{\left({x}-{4}\right)}+{2}$, or $\sqrt{{{x}}}\approx\frac{{1}}{{4}}{x}+{1}$.

$\sqrt{{{3.99}}}\approx\frac{{1}}{{4}}{3.99}+{1}={1.9975}$. The true value is given for comparison: $\sqrt{{{3.99}}}\approx{1.997498435544}$.

$\sqrt{{{4.01}}}\approx\frac{{1}}{{4}}{4.01}+{1}={2.0025}$. The true value is given for comparison: $\sqrt{{{4.01}}}\approx{2.00249843945}$.

$\sqrt{{{4.05}}}\approx\frac{{1}}{{4}}{4.05}+{1}={2.0125}$. The true value is given for comparison: $\sqrt{{{4.05}}}\approx{2.01246117974981}$.

As can be seen, the approximate values are overestimated. That's because the tangent line at ${x}={4}$ lies above ${f{{\left({x}\right)}}}=\sqrt{{{x}}}$.

Let's do another example.

Example 2. Suppose that the temperature of tea is $95^{0}C$. After one minute, its temperature is $85^{0}C$. Find the temperature of the tea in two minutes.

If ${T}{\left({t}\right)}$ represents the temperature of the tea after ${t}$ minutes, ${T}{\left({0}\right)}={95}$, ${T}{\left({1}\right)}={85}$.

In order to make linear approximation with ${a}={1}$, we need to approximate ${T}'{\left({1}\right)}$.

Since ${T}'{\left({1}\right)}=\lim_{{{t}\to{1}}}\frac{{{T}{\left({t}\right)}-{T}{\left({1}\right)}}}{{{t}-{1}}}$, we can't find the exact value of the derivative, but we could estimate ${T}'{\left({1}\right)}$ by taking ${t}={0}$: ${T}'{\left({1}\right)}\approx\frac{{{T}{\left({0}\right)}-{T}{\left({1}\right)}}}{{{0}-{1}}}=\frac{{{95}-{85}}}{{{0}-{1}}}=-{10}$.

Thus, we approximated the instantaneous rate of change by the average rate of change between ${t}={0}$ and ${t}={1}$, which is $-10^{0}C$ per minute. Thus, ${T}{\left({2}\right)}\approx{T}'{\left({1}\right)}{\left({2}-{1}\right)}+{T}{\left({1}\right)}=-{10}{\left({3}-{2}\right)}+{85}={{75}}^{{0}}{C}$.