선형근사와 미분소

We have no idea what $f(0.93)$ is, but 0.93 is pretty close to 1, and we definitely know $f(1)$. The linear approximation of $f(x)$ about $x=1$ is given by

$$\begin{aligned} f(x) &\approx f(1)+f'(1)(x-1) \\ \text{So, we calculate:} \\ f(1)&=\log(1)=0\\ f'(x)&=\frac{1}{x}\\ f'(1)&=\frac{1}{1}=1 \\ \text{Therefore,} \\ f(x) &\approx 0+1(x-1)=x-1 \\ \text{When $x=0.93$:} \\ f(0.93) &\approx 0.93-1=-0.07 \end{aligned}$$

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.

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

(a)

When $x$ is near 4, the linear approximation of $\Delta y$ is

$$\begin{aligned} \Delta y &\approx s'(4)\Delta x \\ \text{From the problem, $\Delta x= 5-4=1$. Differentiating,} \\ s'(t)&=\sqrt{19.6}\cdot\frac{1}{2\sqrt{x}}=\sqrt{\frac{4.9}{x}},\text{ so }\\ s'(4)&=\sqrt{\frac{4.9}{4}} \\ \text{The linear approximation gives us} \\ \Delta y& \approx \sqrt{\frac{4.9}{4}}(1)\approx1.1 \end{aligned}$$

So moving from 4 metres to 5 metres increases the speed with which you hit the water by about 1.1 metres per second.

(b) Again, we'll use the linear approximation

$$\begin{aligned} \Delta y &\approx s'(a)\Delta x\\ &=\sqrt{\frac{4.9}{a}}\cdot\Delta x \end{aligned}$$

The difference in height between the first two jumps and between the last two jumps is the same, $\Delta x$. The initial height of the first jump is smaller than the initial height of the second jump. So, the value corresponding to $a$ is smaller for the first jump than for the second. Therefore, $\Delta y$ is larger between the first two jumps than it is between the last two jumps. So, the increase in speed from the first jump to the second is larger than the increase in speed from the second jump to the third.

To put that more symbolically, the change in terminal speed between the first two jumps is

$$\begin{aligned} \Delta y &\approx \sqrt{\frac{4.9}{x}}\cdot\Delta x \\ \text{while the change in terminal speed between the next two jumps is} \\ \Delta y &\approx \sqrt{\frac{4.9}{x+\Delta x}}\cdot\Delta x \end{aligned}$$

Since $\sqrt{\dfrac{4.9}{x}}\cdot\Delta x> \sqrt{\dfrac{4.9}{x+\Delta x}}\cdot\Delta x$ (when $x$ and $\Delta x$ are positive), the change in terminal speed is greater between the first two jumps than between the next two jumps.