선형근사와 미분소

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 $f(0)$.

The linear approximation is a line, passing through $(2,f(2))$, with slope $f'(2)$. That is, the linear approximation to $f(x)$ about $x=2$ is the tangent line to $f(x)$ at $x=2$. It is shown below in red.

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 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.

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 $a=\sqrt{3}$.

Let $f(x)$ be the number of problems finished after $x$ minutes of work. The question tells us that $5\Delta y \approx \Delta x$. So, if $\Delta x = 15$, $\Delta y \approx 3$. That is, in 15 minutes more, you will finish about 3 more problems.

Remark: the math behind this problem is intended to be easy! Looking at symbols like $\Delta y$ and $f\left(x+\Delta x\right)$ can be confusing, but the basic idea is pretty simple.

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.

(a) Since the value and slope of the tangent line match the value and slope of the function at the point of tangency, we know $p(3) = L(3) = -1$ and $p'(3) = L'(3) = -2$. (b) Using the local linearization, $p(2.79) \approx L(2.79) = -2(2.79) + 5 = -0.58$. (c) Since $p''(x) < 0$ for $x < 3$, we know that $p$ is concave down for $x < 3$. This means to the left of the point of tangency at $x = 3$, the tangent line lies above the curve, and thus the estimate in (b) is too large. (d) In the following figure, we see a possible function $p$ that matches the value and slope of the tangent line at the point $(3,-1)$ and that is concave down to the left of $x=3$ and concave up to the right.