급수 심화 전체 문제

$\sqrt{1-{x}^{2}}=1-\frac{{x}^{2}}{2}-\frac{{x}^{4}}{8}-\frac{{x}^{6}}{16}-\frac{5{x}^{8}}{128}+\cdots .$ Thus

${\int }_{-1}^{1}\sqrt{1-{x}^{2}}dx=x-\frac{{x}^{3}}{6}-\frac{{x}^{5}}{40}-\frac{{x}^{7}}{7\cdot 16}-\frac{5{x}^{9}}{9\cdot 128}+\cdots {|}_{-1}^{1}\approx 2-\frac{1}{3}-\frac{1}{20}-\frac{1}{56}-\frac{10}{9\cdot 128}+\text{error}=1.590...$ whereas $\frac{\pi }{2}=1.570...$

(a) The area removed represents a proportion of $\dfrac{\theta}{2\pi}$ of the entire circle, whose area is $\pi(3^2)=9\pi$. So, the area of the sector removed is $$\dfrac{\theta}{2\pi} \cdot9\pi = \dfrac{9}{2}\theta$$

(b) To find $\theta$ from $d$, we cut our triangle (with angle $\theta$ opposite side of length $d$) into two equivalent right triangles, as shown below.

Using the information that the radius of the circle (also the hypotenuse of the right triangle) is 3,

$$\begin{aligned} \sin\left(\frac{\theta}{2}\right)&=\frac{\frac{d}{2}}{3}=\frac{d}{6} \\ \text{Since the question tells us the sector is no more than a quarter of the circle, we know $0 \leq \theta \leq \dfrac{\pi}{2}$, so $0 \leq \dfrac{\theta}{2} \leq \dfrac{\pi}{4}$. This puts $\dfrac{\theta}{2}$ well within the range of arcsine.} \\ \theta&=2\arcsin\left(\frac{d}{6}\right) \end{aligned}$$

(c) First, let's get an expression for the area of the sector in terms of $d$.

$$\begin{aligned} A&=\frac{9}{2}\theta = \frac{9}{2}\left(2\arcsin\left(\frac{d}{6}\right)\right) \\&=9\arcsin\left(\frac{d}{6}\right) \\ \text{Differentiating,} \\ A'(d)&=\frac{9}{\sqrt{1-\left(\frac{d}{6}\right)^2}}\cdot\frac{1}{6}\\ &=\frac{9}{\sqrt{36-d^2}} \\ \text{Let $\Delta A$ is the error in $A$ corresponding to an error of $\Delta d$ in $d$. Since we measured $d$ to be 0.7 instead of $0.68$, in the linear approximation we take $\Delta d = 0.02$ and $d_0=0.68$.} \\ \Delta A &\approx A'(d_0)\cdot\Delta d\\ &=A'\left(0.68\right)\cdot 0.02\\ &=\frac{9}{\sqrt{36-0.68^2}}\cdot0.02\\ &\approx 0.03 \end{aligned}$$

So, the error in $A$ is about 0.03.

Suppose we have a function $V(h)$ that gives the volume of water in the tank as a function of its height.

Let $h_0=0.5$ metres, $\Delta h = -0.05$, and $\Delta V = V(h_0+\Delta h)-V(h_0)=V(0.45)-V(0.5)$. Then, by the linear approximation, $$\Delta V \approx V'(0.5)\cdot \Delta h = -0.05V'(0.5)$$ In order to solve the problem, we will find a function $V(h)$ giving the volume of water in terms of the height, then find $V'(0.5)$, and finally approximate that the change in the volume of the water is $\Delta V \approx -0.05V'(0.5)$.

The water in the tank forms a cone. The volume of a cone of height $h$ and radius $r$ is $$V=\frac{1}{3}\pi r^2 h$$ We need to get rid of the variable $r$. We can do this using similar triangles. The diagram below shows the side view of the tank and the water.

The side view of the tank forms a triangle that is similar to the triangle formed by the side view of the water, so

$$\begin{aligned} \frac{1}{2}&=\frac{2r}{h}\\ r&=\frac{h}{4} \\ \text{Using this, we find our equation for the volume of the water in terms of $h$.} \\ V(h)&=\frac{1}{3}\pi r^2 h = \frac{\pi}{3}\left(\frac{h}{4}\right)^2h=\frac{\pi h^3}{48} \\ \text{Differentiating,} \\ V'(h)&=\frac{\pi h^2}{16}\\ V'(0.5)&=\frac{0.25\pi}{16}=\frac{\pi}{64} \\ \text{Finally, using the approximation $\Delta V \approx -0.05V'(0.5)$,} \\ \Delta V &\approx \frac{-0.05\pi}{64}=-\frac{\pi}{1280}\approx -0.00245 ~\mathrm{m}^3 \end{aligned}$$

We estimate that the volume decreased by about $0.00245$ cubic metres, or about 2450 cubic centimetres.

Let's name $g(x)=x^{1/3}$. Then $g'(x)=\dfrac{1}{3} x^{-2/3}$ and $g''(x)=-\dfrac{2}{9} x^{-5/3}$.

In particular, $g(8)=2$, $g'(8)=\dfrac{1}{12}$, and $g''(x)<0$ for all $x>0$.

The tangent line approximation to $10^{1/3}$ is

$$\begin{aligned} g(10)&\approx g(8)+g'(8)(10-8)\\ &=2+\dfrac{1}{12}(2)=\dfrac{13}{6} \\ \text{Using the error formula:} \\ g(10)&=g(8)+g'(8)(10-8)+\tfrac{1}{2} g''(c)(10-8)^2 \end{aligned}$$

for some $8<c<10$. Since $g''(c)=-\dfrac{2}{9c^{5/3}}$, $g''(c)$ is negative, so $g(10)$ is $\dfrac{13}{6}$ plus some negative quantity. So, the tangent line approximation is too big.

All functions $A$, $B$, and $C$ intersect the function $y=f(x)$ when $x=2$. $B$ is a constant function, so this is the constant approximation. $A$ is the tangent line, so $A$ is the linear approximation. $C$ is a tangent parabola, so $C$ is the quadratic approximation.

(a) To sketch $y=f(x)$, we note the following:

• $f(x)$ is never negative.
• $\lim\limits_{x \to \pm \infty} f(x)=e^0=1$, so the curve has horizontal asymptotes in both directions at $y=1$.
• $\lim\limits_{x \to \pm 0} f(x)=\lim\limits_{x \to \pm 0} \frac{1}{e^{1/x^2}}=\lim\limits_{u \to +\infty}\frac{1}{e^u}=0=f(0)$, so the curve is continuous at $x=0$.
• For $x\neq 0$, $f'(x)=\frac{2}{x^3}e^{-1/x^2}$, so our curve is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$
• For $x\neq 0$, $f''(x)=2x^{-6}(2-3x^2)e^{-1/x^2}$, so our curve is concave up on $(-\sqrt{2/3},\sqrt{2/3})$, and concave down elsewhere.

(b) Since $f^{(n)}(0)=0$ for all whole $n$ (that is, the graph is really quite flat at the origin), and since $f(0)=0$, the Maclaurin series for $f(x)$ is $\displaystyle\sum_{n=0}^\infty \frac{0}{n!}x^n=0$. (c) The Maclaurin series converges for all real values of $x$ (to the constant 0). (d) Since $e^y>0$ for any real $y$, we see $f(x)=0$ only when $x=0$. So, $f(x)$ is only equal to its Maclaurin series at the single point $x=0$.

Remark: the function $f(x)$ is an example of a function whose Maclaurin series converges, but not to $f(x)$! To describe this behaviour, we say $f(x)$ is non-analytic.