극한·미분 심화 전체 문제

• Solution 1: We begin by differentiating implicitly. Following the usual convention, we use $y'$ to mean $y'(x)$.

$$\begin{aligned} \arcsin(x+2y)&=x^2+y^2 \qquad\text{Using the chain rule:}\\ \frac{1+2y'}{\sqrt{1-(x+2y)^2}}&=2x+2yy'\\ \frac{1}{\sqrt{1-(x+2y)^2}}+\frac{2y'}{\sqrt{1-(x+2y)^2}}&=2x+2yy'\\ \frac{2y'}{\sqrt{1-(x+2y)^2}}-2yy'&=2x-\frac{1}{\sqrt{1-(x+2y)^2}}\\ y'\left(\frac{2}{\sqrt{1-(x+2y)^2}}-2y\right)&=2x-\frac{1}{\sqrt{1-(x+2y)^2}}\\ y'&=\frac{2x-\frac{1}{\sqrt{1-(x+2y)^2}}}{\frac{2}{\sqrt{1-(x+2y)^2}}-2y} \left(\frac{\sqrt{1-(x+2y)^2}}{\sqrt{1-(x+2y)^2}}\right)\\ y'&=\frac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}} \end{aligned}$$

• Solution 2: We begin by taking the sine of both sides of the equation.

$$\begin{aligned} \arcsin(x+2y)&=x^2+y^2\\ x+2y&=\sin(x^2+y^2) \\ \text{Now, we differentiate implicitly.} \\ 1+2y'&=\cos(x^2+y^2)\cdot(2x+2yy')\\ 1+2y'&=2x\cos(x^2+y^2)+2yy'\cos(x^2+y^2)\\ 2y'-2yy'\cos(x^2+y^2)&=2x\cos(x^2+y^2)-1\\ y'\left(2-2y\cos(x^2+y^2)\right)&=2x\cos(x^2+y^2)-1\\ y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)} \end{aligned}$$

• We used two different methods, and got two answers that look pretty different. However, the answers ought to be equivalent. To see this, we remember that for all values of $x$ and $y$ that we care about (those pairs $(x,y)$ in the domain of our curve), the equality $$\arcsin(x+2y)=x^2+y^2$$ holds. Drawing a triangle:

where the adjacent side (in red) come from the Pythagorean Theorem. Then, $\cos(x^2+y^2)=\sqrt{1-(x+2y)^2}$, so using our second solution:

$$\begin{aligned} y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)}\\ &=\frac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}} \end{aligned}$$

which is exactly the answer from our first solution.

Since $f(0)$ is closer to $g(0)$ than it is to $h(0)$, you would probably want to estimate $f(0) \approx g(0)=1+2\sin (1)$ if you had the means to efficiently figure out what $\sin(1)$ is, and if you were concerned with accuracy. If you had a calculator, you could use this estimation. Also, later in this chapter we will learn methods of approximating $\sin (1)$ that do not require a calculator, but they do require time.

Without a calculator, or without a lot of time, using $f(0)\approx h(0)=0.7$ probably makes the most sense. It isn't as accurate as $f(0) \approx g(0)$, but you get an estimate very quickly, without worrying about figuring out what $\sin(1)$ is.

Remark: when you're approximating something in real life, there probably won't be an obvious “correct" way to do it. There's usually a trade-off between accuracy and ease.

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 $f(0)$.

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.