Short and Sweet Calculus

## 3.12 Linear Approximations

Because $$\lim_{\Delta x\to0}\frac{f(a+\Delta x)-f(a)}{\Delta x}=f'(a)$$, when $$\Delta x=x-a$$ is close to zero, we have $\frac{f(a+\Delta x)-f(a)}{\Delta x}\approx f'(a).$ Multiplying both sides by $$\Delta x$$, we get $f(\underbrace{a+\Delta x}_{x})-f(a)\approx f'(a)\underbrace{(x-a)}_{\Delta x}\tag{\small\ensuremath{\Delta x=x-a}}$ or $\boxed{f(x)\approx f(a)+f'(a)(x-a).}$ This is called the linear approximation of $$f$$ at $$x=a$$. It is also called the tangent line approximation, because the right-hand side $$y=f(a)+f'(a)(x-a)$$ is the equation of the tangent line to $$y=f(x)$$ at $$\left(a,f(a)\right).$$

• The function $$L(x)=f(a)+f'(a)(x-a)$$ is called the linearization of $$f$$ at $$x=a$$.

Example 3.24. Use the linear approximaiton to estimate $$\sin(29^{\circ})$$.

Solution

Because we know the exact value of $$\sin(30^{\circ})=\sin(\pi/6)$$, we can use the linearization of $$f(x)=\sin x$$ at $$x=\pi/6$$ to approximate $$\sin(29^{\circ})$$. The derivative of $$f$$ is $f(x)=\sin x\Rightarrow f'(x)=\cos x$ and so we have $$f(\pi/6)=1/2$$ and $$f'(\pi/6)=\cos(\pi/6)=\sqrt{3}/2$$.

The linearization of $$f$$ at $$x=\pi/6$$ is $L(x)=\underbrace{\frac{1}{2}}_{f(\pi/6)}+\underbrace{\frac{\sqrt{3}}{2}}_{f'(\pi/6)}\underbrace{\Delta x}_{\left(x-\frac{\pi}{6}\right)}.$ Notice that $$\frac{d}{dx}\sin x=\cos x$$ works only when $$x$$ is in radian measure. So $$\Delta x$$ is also in radians. Because $$1^{\circ}=\frac{\pi}{180}\text{ radian,}$$ we have \begin{aligned} \sin(29^{\circ}) & \approx L\left(\frac{\pi}{6}-\frac{\pi}{180}\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(-\frac{\pi}{180}\right)\\ & \approx0.4849.\end{aligned} The ture value of $$\sin(29^{\circ})$$ to 4 digits is 0.4848, respectively.