4.2 Linear approximations and differentials  (Page 2/12)

Linear approximation of $\text{sin}x$

Find the linear approximation of $f(x)=\text{sin}x$ at $x=\frac{\pi }{3}$ and use it to approximate $\text{sin}(62\text{°}).$

First we note that since $\frac{\pi }{3}$ rad is equivalent to $60\text{°},$ using the linear approximation at $x=\pi \text{/}3$ seems reasonable. The linear approximation is given by

$L(x)=f\left(\frac{\pi }{3}\right)+f\prime \left(\frac{\pi }{3}\right)\left(x-\frac{\pi }{3}\right).$

We see that

$\begin{array}{ccc}\hfill f(x)=\text{sin}x& \Rightarrow \hfill & f\left(\frac{\pi }{3}\right)=\text{sin}\left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}\hfill \\ \hfill f\prime (x)=\text{cos}x& \Rightarrow \hfill & f\prime \left(\frac{\pi }{3}\right)=\text{cos}\left(\frac{\pi }{3}\right)=\frac{1}{2}\hfill \end{array}$

Therefore, the linear approximation of $f$ at $x=\pi \text{/}3$ is given by [link] .

$L(x)=\frac{\sqrt{3}}{2}+\frac{1}{2}\left(x-\frac{\pi }{3}\right)$

To estimate $\text{sin}(62\text{°})$ using $L,$ we must first convert $62\text{°}$ to radians. We have $62\text{°}=\frac{62\pi }{180}$ radians, so the estimate for $\text{sin}(62\text{°})$ is given by

$\text{sin}(62\text{°})=f\left(\frac{62\pi }{180}\right)\approx L\left(\frac{62\pi }{180}\right)=\frac{\sqrt{3}}{2}+\frac{1}{2}\left(\frac{62\pi }{180}-\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}+\frac{1}{2}\left(\frac{2\pi }{180}\right)=\frac{\sqrt{3}}{2}+\frac{\pi }{180}\approx 0.88348.$

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Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for $f(x)={(1+x)}^n$ at $x=0,$ which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form $f(x)={(m+x)}^n$ to estimate roots and powers near a different number $m.$

Approximating roots and powers

Find the linear approximation of $f(x)={(1+x)}^n$ at $x=0.$ Use this approximation to estimate ${(1.01)}^3.$

The linear approximation at $x=0$ is given by

$L(x)=f(0)+f\prime (0)(x-0).$

Because

$\begin{array}{ccc}\hfill f(x)={(1+x)}^n& \Rightarrow \hfill & f(0)=1\hfill \\ \hfill f\prime (x)=n{(1+x)}^{n-1}& \Rightarrow \hfill & f\prime (0)=n,\hfill \end{array}$

the linear approximation is given by [link] (a).

$L(x)=1+n(x-0)=1+nx$

We can approximate ${(1.01)}^3$ by evaluating $L(0.01)$ when $n=3.$ We conclude that

${(1.01)}^3=f(1.01)\approx L(1.01)=1+3(0.01)=1.03.$

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Differentials

We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials . Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values.

When we first looked at derivatives, we used the Leibniz notation $dy\text{/}dx$ to represent the derivative of $y$ with respect to $x.$ Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a meaning to the expressions dy and dx . Suppose $y=f(x)$ is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable $dy$ by

It is important to notice that $dy$ is a function of both $x$ and $dx.$ The expressions dy and dx are called differentials . We can divide both sides of [link] by $dx,$ which yields

This is the familiar expression we have used to denote a derivative. [link] is known as the differential form    of [link] .