선형근사와 미분소

$dy = (-\sin(\sin x)\cos x)dx$

$dy = \frac{1}{(x+2)^2}dx$

$dy = ((\ln 3)3^x\ln x + \frac{3^x}{x})dx$

$dy = (\ln x)dx$

$dy = \big(\tan x\big)dx$

$\pm$48in$^2$, or 1/3ft$^2$

Since $10.1$ is pretty close to $10$, we estimate $10.1^3 \approx 10^3=1000$.

Remark: these kinds of approximations are very useful when you are doing computations. It's easy to make a mistake in your work, and having in mind that $10.1^3$ should be about a thousand is a good way to check that whatever answer you have makes sense.

The linear approximation is $L(x)=3x-9$. Since we're approximating at $x=5$, $f(5)=L(5)$, and $f'(5)=L'(5)$. However, there is no guarantee that $f(x)$ and $L(x)$ have the same value when $x \neq 5$. So:

(a) $f(5)=L(5)=6$

(b) $f'(5)=L'(5)=3$

(c) there is not enough information to find $f(0)$.

For any constant $a$, $f(a)=(2a+5)$, and $f'(a)=2$, so our approximation gives us $$f(x) \approx (2a+5)+2(x-a)=2x+5$$

Since $f(x)$ itself is a linear function, the linear approximation is actually just $f(x)$ itself. As a consequence, the linear approximation is perfectly accurate for all values of $x$.

We approximate the function $f(x)=\sqrt{x}$ about $x=4$, since 4 is a perfect square and it is close to 5.

$$\begin{aligned} f(4)&=\sqrt{4}=2\\ f'(x)&=\frac{1}{2\sqrt{x}} \qquad \Rightarrow \qquad f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}\\ f(x) &\approx f(4)+f'(4)(x-4)=2+\frac{1}{4}(x-4)\\ f(5)&\approx 2+\frac{1}{4}(5-4)=\frac{9}{4} \end{aligned}$$

We estimate $\sqrt{5}\approx\dfrac{9}{4}$.

Remark: $\left(\dfrac{9}{4}\right)^2=\dfrac{81}{16}$, which is pretty close to $\dfrac{80}{16}=5$. Our approximation seems pretty good.

We approximate the function $f(x)=\sqrt[5]{x}$. We need to centre the approximation about some value $x=a$ such that we know $f(a)$ and $f'(a)$, and $a$ is not too far from $30$.

$$\begin{aligned} f(x)&=\sqrt[5]{x}=x^{\frac{1}{5}}\\ f'(x)&=\frac{1}{5}x^{-\frac{4}{5}}=\frac{1}{5\sqrt[5]{x}^4} \\ \text{$a$ needs to be a number whose fifth root we know. Since $\sqrt[5]{32}=2$, and 32 is reasonably close to 30, $a=32$ is a great choice.} \\ f(32)&=\sqrt[5]{32}=2\\ f'(32)&=\frac{1}{5\cdot 2^4}=\frac{1}{80} \\ \text{The linear approximation of $f(x)$ about $x=32$ is} \\ f(x) &\approx 2+\frac{1}{80}(x-32) \\ \text{When $x=30$:} \\ f(30) &\approx 2+\frac{1}{80}(30-32)=2-\frac{1}{40}=\frac{79}{40} \end{aligned}$$

We estimate $\sqrt[5]{30}\approx \dfrac{79}{40}$.

Remark: $\dfrac{79}{40} = 1.975$, while $\sqrt[5]{30} \approx 1.97435$. This is a decent estimation.

If $f(x)=x^3$, then $f(10.1)=10.1^3$, which is the value we want to estimate. Let's take the linear approximation of $f(x)$ about $x=10$:

$$\begin{aligned} f(10)&=10^3=1000\\ f'(x)&=3x^2\\ f'(10)&=3(10^2)=300\\ f(a) &\approx f(10)+f'(10)(x-10)\\ &=1000+300(x-10)\\ f(10.1)& \approx 1000+300(10.1-10)=1030 \end{aligned}$$

We estimate $10.1^3 \approx 1030$. If we calculate $10.1^3$ exactly (which is certainly possible to do by hand), we get 1030.301.

Remark: in the previous subsection, we used a constant approximation to estimate $10.1^3 \approx 1000$. That approximation was easy to do in your head, in a matter of seconds. The linear approximation is more accurate, but not much faster than simply calculating $10.1^3$.

There are many possible answers. One is: $$f(x)=\sin x \qquad a=0 \qquad b=\pi$$ We know that $f(\pi)=0$ and $f(0)=0$. Using a constant approximation of $f(x)$ about $x=0$, our estimation is $f(\pi)\approx f(0)=0$, which is exactly the correct value. However, is we make a linear approximation of $f(x)$ about $x=0$, we get $$f(\pi) \approx f(0)+f'(0)(\pi-0)=\sin(0)+\cos(0)\pi=\pi$$ which is not exactly the correct value.

Remark: in reality, we wouldn't estimate $\sin(\pi)$, because we know its value exactly. The purpose of this problem is to demonstrate that fancier approximations are not always more accurate. At the end of this section, we'll talk about how to bound the error of your estimations, to make sure that you are finding something sufficiently accurate.

The linear approximation $L(x)$ of $f(x)$ about $x=a$ is chosen so that $L(a)=f(a)$ and $L'(a)=f'(a)$. So,

$$\begin{aligned} L'(a)&= f'(a)=\dfrac{1}{1+a^2}\\ \frac{1}{4}&=\frac{1}{1+a^2}\\ a&=\pm\sqrt{3} \\ \text{We've narrowed down $a$ to $\sqrt{3}$ or $-\sqrt{3}$. Recall the linear approximation of $f(x)$ about $x=a$ is $f(a)+f'(a)(x-a),$ so its constant term is $f(a)-af'(a)=\arctan(a)-\dfrac{a}{1+a^2}$. We compute this for $a=\sqrt{3}$ and $a=-\sqrt{3}$.} \\ a=\sqrt{3}:~&~\arctan\left(a\right)-\frac{a}{1+a^2} =\arctan\left(\sqrt{3}\right)-\frac{\sqrt{3}}{1+\left(\sqrt{3}\right)^2}=\frac{\pi}{3}-\frac{\sqrt{3}}{4}=\frac{4\pi-\sqrt{27}}{12}\\ a=-\sqrt{3}:~&~\arctan\left(a\right)-\frac{a}{1+a^2} =\arctan\left(-\sqrt{3}\right)-\frac{-\sqrt{3}}{1+\left(-\sqrt{3}\right)^2}=-\frac{\pi}{3}+\frac{\sqrt{3}}{4}=\frac{-4\pi+\sqrt{27}}{12} \\ \text{So, when $a=\sqrt{3}$,} \\ L(x)&=\frac{1}{4}x+\frac{4\pi-\sqrt{27}}{12} \end{aligned}$$

and this does not hold when $a=-\sqrt{3}$. We conclude $a=\sqrt{3}$.

First, let's find the first and second derivatives of $f$ when $x=5$.

$$\begin{aligned} f'(x)&=\frac{1}{1+x^2} & f'(5)&=\frac{1}{26}\\ f''(x)&=\frac{-2x}{\left(1+x^2\right)^2} & f''(5)&=\frac{-10}{26^2} \end{aligned}$$

$$\begin{aligned} \\ \text{The linear approximation for $\Delta y$ when $\Delta x = \dfrac{1}{10}$ is} \\ \Delta y &\approx f'(5)\Delta x = \frac{1}{26}\cdot \frac{1}{10}=\frac{1}{260} \approx 0.003846 \\ \text{The quadratic approximation for $\Delta y$ when $\Delta x = \dfrac{1}{10}$ is} \\ \Delta y &\approx f'(5)\Delta x + \frac{1}{2}f''(5)\left(\Delta x\right)^2 = \frac{1}{26}\cdot\frac{1}{10}+\frac{1}{2}\cdot\frac{-10}{26^2}\cdot\frac{1}{10^2}=\frac{51}{13520}\approx 0.003772 \end{aligned}$$

Remark: $\Delta y = \arctan(5.1)-\arctan(5)\approx 0.003774$.