Linear Approximations

関連する計算機: 線形近似計算機

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, $$${T}{\left({2}\right)}\approx{T}'{\left({1}\right)}{\left({2}-{1}\right)}+{T}{\left({1}\right)}=-{10}{\left({3}-{2}\right)}+{85}={{75}}^{{0}}{C}$$$.