Lemniscate $(x^2 {\rm{ + }}y^2 )^2 = x^2 – y^2$. At the point $(x,y) = (\frac{{\sqrt 6 }}{4},\frac{{\sqrt 2 }}{4})$ $\frac{{dy}}{{dx}} =?$

$\Rightarrow 2(x^2 + y^2 )(2x + 2yy’) = 2x – 2yy’$
$\Rightarrow (x^2 + y^2 )(2x + 2yy’) = x – yy’$

$\begin{array}{cl} \text{At }(x,y) = (\frac{{\sqrt 6 }}{4},\frac{{\sqrt 2 }}{4}) & \Rightarrow \underbrace {(\frac{6}{{16}} + \frac{2}{{16}})}_{\frac{1}{2}}(\frac{{\sqrt 6 }}{2} + \frac{{\sqrt 2 }}{2}y’) = \frac{{\sqrt 6 }}{4} – \frac{{\sqrt 2 }}{4}y’ \\ & \Rightarrow \frac{{\sqrt 6 }}{4} + \frac{{\sqrt 2 }}{4}y’ = \frac{{\sqrt 6 }}{4} – \frac{{\sqrt 2 }}{4}y’ \\ & \Rightarrow \frac{{\sqrt 2 }}{4}y’ + \frac{{\sqrt 2 }}{4}y’ = 0\\&\Rightarrow \frac{{\sqrt 2 }}{2}y’ = 0 \Rightarrow y’ = 0 \end{array}$
Thus $\frac{{dy}}{{dx}} = 0 ∎$

If $x^3+y^3=3xy$ then $x+y\leq$ max=?

Let $g(x,y) = x^3 + y^3 – 3xy = 0$
By Lagrange’s multipliers.
Let $\nabla f + \lambda \nabla g =\overset{\rightharpoonup}{0}$
$\Rightarrow (1,1) + \lambda (3x^2 – 3y,3y^2 – 3x) =\overset{\rightharpoonup}{0}$
$\Rightarrow \left\{ {\begin{array}{c} {1 + 3\lambda x^2 – 3\lambda y = 0 \cdots (1)} \\ {1 + 3\lambda y^2 – 3\lambda x = 0 \cdots (2)} \end{array}} \right.$, 由此可知 $\lambda \ne 0$
$(1)-(2)$
$\Rightarrow 3\lambda (x^2 – y^2 ) – 3\lambda ( – y + x) = 0$
$\Rightarrow 3\lambda (x – y)(x + y – 1) = 0$
$\Rightarrow \lambda =0$ (不合) 或 $x=y$ 或 $x+y-1=0$

If $x=y$
$\because\ x^3+y^3-3xy=0$
$\therefore\ 2x^3-3x^2=0$
$\Rightarrow x^2 (2x – 3) = 0 \Rightarrow x = 0$ or $x=\frac{3}{2}$
$(x,y)=(0,0)$ or $(\frac{3}{2},\frac{3}{2})$
$\Rightarrow f(x,y) = x + y = 0$ or $3$
If $x+y=1$
$\Rightarrow f(x,y)=x+y=1$
Thus $x+y=\text{max}=3 ∎$

The probability density function of an exponential distribution is $f(x)=3e^{-3x}$ if $x\ge 0$, $f(x)=0$ if $x<0$. Find the expected value (期望值) $E(x)=?$

$\begin{array}{cl} E(x) & =\int_{ – \infty }^\infty {xf(x)} dx \\ & =\int_{ – \infty }^0 {x\underbrace {f(x)}_0} dx + \int_0^\infty {xf(x)} dx \\ & =\int_0^\infty {x(3e^{ – 3x} )} dx\\&=\int_0^\infty {3x \cdot e^{ – 3x} } dx\\&=\left. {(3x)(\frac{{e^{ – 3x} }}{{ – 3}})} \right|_0^\infty – \int_0^\infty {3 \cdot (\frac{{e^{ – 3x} }}{{ – 3}})} dx\\&=\left. {\frac{{ – x}}{{e^{3x} }}} \right|_0^\infty + \int_0^\infty {e^{ – 3x} } dx\\&=(\underbrace {\mathop {\lim }\limits_{x \to \infty } \frac{{ – x}}{{e^{3x} }}}_0 -\underbrace {\frac{{ – 0}}{{e^0 }}}_0) – \left. {\frac{{e^{ – 3x} }}{{ – 3}}}\right|_0^\infty\\&=0 – (0 – \frac{1}{{ – 3}}) = \frac{1}{3} \end{array}$
Thus $E(x) = \frac{{\rm{1}}}{{\rm{3}}} ∎$

If $f(x) = (\sin x)^{\cos x}$, $f'(x)=?$

$\begin{array}{cl} f'(x) & =((\sin x)^{\cos x} )’ \\ & =(e^{\ln (\sin x)^{\cos x} } )’\\&=(e^{\cos x\ln (\sin x)} )’\\&=e^{\cos x\ln (\sin x)} \cdot \left[ {( – \sin x)\ln (\sin x) + (\cos x) \cdot \frac{{\cos x}}{{\sin x}}} \right]\\&=(\sin x)^{\cos x} \cdot \left[ {\frac{{\cos ^2 x}}{{\sin x}} – (\sin x)\ln (\sin x)} \right] \end{array}$

Thus $f'(x) = (\sin x)^{\cos x} \cdot \left[ {\frac{{\cos ^2 x}}{{\sin x}} – (\sin x)\ln (\sin x)} \right] ∎$

$f(x) =\frac{1}{{x^2 – x – 2}} =$Taylor series$=\sum {a_n x^n }$, $a_n=?$

◎ 運用幾何級數的性質

$f(x) = \frac{1}{{(x – 2)(x + 1)}} = \frac{a}{{x – 2}} + \frac{b}{{x + 1}} \Rightarrow \left\{ {\begin{array}{l} {a = \frac{1}{3} } \\ {b = – \frac{1}{3}}\end{array}} \right.$
$\begin{array}{cl} \Rightarrow f(x) & =\frac{{\frac{1}{3}}}{{x – 2}} + \frac{{ – \frac{1}{3}}}{{x + 1}}\\&=\frac{1}{3}(\frac{1}{{x – 2}} – \frac{1}{{x + 1}})\\&=\frac{1}{3}(\frac{{ – \frac{1}{2}}}{{1 – \frac{x}{2}}} – \frac{1}{{1 – ( – x)}})\\&=\frac{1}{3}(\underbrace {\sum\limits_{n = 0}^\infty {( – \frac{1}{2}) \cdot (\frac{x}{2})^n } – \sum\limits_{n = 0}^\infty {( – x)^n } }_{\text{as }\left| x \right| < 1})\\&=\frac{1}{3}(\sum\limits_{n = 0}^\infty {\frac{{ – 1}}{{2^{n + 1} }} \cdot x^n } – \sum\limits_{n = 0}^\infty {( – 1)^n x^n } )\\&=\frac{1}{3}\sum\limits_{n = 0}^\infty {(\frac{{ – 1}}{{2^{n + 1} }} – ( – 1)^n } )x^n\\&=\sum\limits_{n = 0}^\infty {\underbrace {\frac{1}{3}(\frac{{ – 1}}{{2^{n + 1} }} – ( – 1)^n {\rm{)}}}_{a_n }} x^n,\ \left| x\right|<1 \end{array}$

Thus $a_n = \frac{1}{3}(\frac{{ – 1}}{{2^{n + 1} }} – ( – 1)^n ) ∎$

A$=(-1,0)$, B$=(0,1)$, C$=(2,2)$. Use the method of least squares to find a line $y=mx+b$ that best fits A, B, C. $m=?\ b=?$

Goal: minimize $d_1^2+d_2^2+d_3^2$
Let $f(m,b) = ( – m + b)^2 + (b – 1)^2 + (2m + b – 2)^2$
$\Rightarrow \nabla f = (f_m ,f_b ) = (0,0)$
$\begin{array}{cl} f_m & =\frac{1}{3}(\frac{1}{{x – 2}} – \frac{1}{{x + 1}}) \\ & =2m – 2b + 8m + 4b – 8 \\&=10m + 2b – 8 = 0 \end{array}$
$\begin{array}{cl} f_b & =2( – m + b) + 2(b – 1) + 2(2m + b – 2) \\ & =2m + 2b + 2b – 2 + 4m + 2b – 4 \\&=2m + 6b – 6 = 0 \end{array}$
$\Rightarrow \left\{ {\begin{array}{c} {5m + b – 4 = 0} \\ {m + 3b – 3 = 0} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{c} {m = \frac{9}{{14}}} \\ {b = \frac{{11}}{{14}}} \end{array}} \right.$

Thus $m=\frac{9}{14},\ b=\frac{11}{14} ∎$

$\mathop {\lim }\limits_{x \to 0} (\sin x)^{(\cos x – 1)} =?$

$\begin{array}{cl} \mathop {\lim }\limits_{x \to 0} (\sin x)^{(\cos x – 1)} & =\mathop {\lim }\limits_{x \to 0} e^{\ln (\sin x)^{\cos x – 1} } \\ & =e^{\mathop {\lim }\limits_{x \to 0} \underbrace {(\cos x – 1)}_0 \underbrace {\ln \overbrace {(\sin x)}^0}_{ – \infty }} \end{array}$

$\mathop {\lim }\limits_{x \to 0} (\cos x – 1) \cdot \ln (\sin x)$
$= \mathop {\lim }\limits_{x \to 0} (\cos x – 1) \cdot \ln (\sin x) \cdot \frac{{\cos x + 1}}{{\cos x + 1}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\overbrace {(\cos ^2 – 1)}^{ – \sin ^2 x} \cdot \ln (\sin x)}}{{\cos x + 1}}$
$=\mathop {\lim }\limits_{x \to 0} \underbrace {\frac{{ – 1}}{{\cos x + 1}}}_{ – \frac{1}{2}} \cdot \frac{{\ln (\sin x)}}{{\csc ^2 x}}$
$= – \frac{1}{2} \cdot 0 = 0$

$\begin{array}{cl} ◎\ \mathop {\lim }\limits_{x \to 0} \frac{{\ln (\sin x)}}{{\csc ^2 x}} & \mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\cos x}}{{\sin x}}}}{{2\csc x( – \csc x\cot x)}} \\ & =\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{ – 2\csc x\cot x}} \\ & =\mathop {\lim }\limits_{x \to 0} \frac{{\cos x\sin x\tan x}}{{ – 2}}\\&=0 \end{array}$

Find the area of the surface $z=xy$, $x^2+y^2\leq 1$.

Let $r(x,y)=(x,y,xy)$, $\Omega :x^2+y^2\leq 1$
$\Rightarrow \left\{ {\begin{array}{c} \overset{\rightharpoonup}{r_x}= (1,0,y) \\ \overset{\rightharpoonup}{r_y}= (0,1,x) \end{array}} \right. \Rightarrow \overset{\rightharpoonup}{r_x} \times \overset{\rightharpoonup}{r_y} = ( – y, – x,1)$
$\Rightarrow \parallel \overset{\rightharpoonup}{r_x} \times \overset{\rightharpoonup}{r_y} \parallel = \sqrt {( – y)^2 + ( – x)^2 + 1^2 } = \sqrt {x^2 + y^2 + 1}$

$\begin{array}{cl} \text{Area} & =\iint_{\Omega}\parallel \overset{\rightharpoonup}{r_x} \times \overset{\rightharpoonup}{r_y} \parallel dA \\ & =\iint_{\Omega}\sqrt{x^2+y^2+1}dA \\ &(\because \Omega :\ x^2+y^2\leq 1\therefore \text{let }x=r\cos\theta,\ y=r\sin\theta)\\ & =\int_0^{2\pi } {\int_0^1 {\sqrt {r^2 + 1} } } \cdot rdrd\theta \\ &(\text{Let }u=r^2+1\Rightarrow du=2rdr) \end{array}$
$\begin{array}{cl} \Rightarrow \text{Area} & =\int_0^{2\pi } {\int_1^2 {\sqrt u } } \frac{{du}}{2}d\theta \\ & =\frac{1}{2}\int_0^{2\pi } {\left. {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{u = 1}^{u = 2} } d\theta \\ & =\frac{1}{3}\int_0^{2\pi } {(\sqrt 8 – 1)} d\theta = 2\pi \cdot \frac{{2\sqrt 2 – 1}}{3} \end{array}$
Thus the area of the surface is $2\pi (\frac{{2\sqrt 2 – 1}}{3}). ∎$

$y=f(x)$, $y’ = 2y(10 – y)$, $f(0)=1$, $f(x)=$?

$\because\ y’=2y(10-y)$
$\therefore\ \int {\frac{1}{{2y(10 – y)}}} dy = \int {dx}$

$\begin{array}{cl} \int {\frac{1}{{2y(10 – y)}}} dy & =\frac{1}{2}\int {(\frac{a}{y} + \frac{b}{{10 – y}}} )dy \\ & =\frac{1}{2}\int {(\frac{{\frac{1}{{10}}}}{y} + \frac{{\frac{1}{{10}}}}{{10 – y}}} )dy \\ &=\frac{1}{{20}}\int {(\frac{1}{y} + \frac{1}{{10 – y}}} )dy\\ & =\frac{1}{{20}}(\ln \left| y \right| – \ln \left| {y – 10} \right|) + C_1 \end{array}$

$\int {dx} = x + C_2$

$\Rightarrow \frac{1}{{20}}\underbrace {(\ln \left| y \right| – \ln \left| {y – 10} \right|)}_{\ln \left| {\frac{y}{{y – 10}}} \right|} + C_1 = x + C_2$
$\Rightarrow \frac{1}{{20}}\ln \left| {\frac{y}{{y – 10}}} \right| = x + C$
$\because\ f(0)=1$
$\therefore\ \underbrace {\frac{1}{{20}}\ln \left| {\frac{1}{{1 – 10}}} \right|}_{\frac{1}{{20}}\ln \frac{1}{9}} = 0 + C$
$\Rightarrow C = \frac{1}{{20}}\ln \frac{1}{9} = \frac{{ – \ln 9}}{{20}}$
$\Rightarrow \frac{1}{{20}}\ln \left| {\frac{y}{{y – 10}}} \right| = x – \frac{{\ln 9}}{{20}}$
$\Rightarrow \ln \left| {\frac{y}{{y – 10}}} \right| = 20x – \ln 9$
$\Rightarrow \left| {\frac{y}{{y – 10}}} \right| = e^{20x – \ln 9} = \frac{{e^{20x} }}{{e^{\ln 9} }} = \frac{1}{9}e^{20x}$
$\Rightarrow \frac{1}{9}e^{20x} = \frac{y}{{y – 10}}$
$\Rightarrow \frac{1}{9}e^{20x} y – \frac{{10}}{9}e^{20x} = y$
$\Rightarrow y(\frac{1}{9}e^{20x} – 1) = \frac{{10}}{9}e^{20x}$
$\Rightarrow y = \frac{{\frac{{10}}{9}e^{20x} }}{{\frac{1}{9}e^{20x} – 1}} = \frac{{10e^{20x} }}{{e^{20x} – 9}}$
Thus $f(x) = \frac{{10e^{20x} }}{{e^{20x} + 9}} ∎$

Polar coordinate $x=r\cos \theta$, $y = r\sin \theta$. Cardioid $r=1-\sin \theta$. At the point $(x,y)=(1,0)$. $\frac{d^2y}{dx^2}=?$

$r=1-\sin\theta=1-\frac{y}{r}$
$\Rightarrow r^2=r-y$
$\Rightarrow x^2 + y^2 = \sqrt {x^2 + y^2 } – y$
$\Rightarrow 2x + 2yy’ = \frac{1}{2}(x^2 + y^2 )^{ – \frac{1}{2}} \cdot (2x + 2yy’) – y’$
At $(x,y)=(1,0)$
$\Rightarrow 2 + 0 = (1 + 0)^{ – \frac{1}{2}} (1 + 0) – y’ = 1 – y’$
$\Rightarrow \left. {y’} \right|_{(1,0)} = – 1$

$\begin{array}{cl} (2x + 2yy’)’ & =2 + 2y’ \cdot y’ + 2y \cdot y^{\prime\prime} \\ & =2 + 2(y’)^2 + 2y \cdot y^{\prime\prime} \cdots (1) \end{array}$
$\begin{array}{cl} ((x^2 + y^2 )^{ – \frac{1}{2}} (x + y{\rm{.}}y’) – y’)’= & \frac{1}{2}(x^2 + y^2 )^{ – \frac{3}{2}} (2x + 2y \cdot y’)(x + y \cdot y’) \\ &+ (x^2 + y^2 )^{ – \frac{1}{2}} (1 + \underbrace {y’ \cdot y’}_{(y’)^2 } + y \cdot y^{\prime\prime}) – y^{\prime\prime} \cdots (2) \end{array}$

At $(x,y)=(1,0)$
$\begin{array}{cl} (1) & =2+2(-1)^2=4 \\ (2) & =\frac{1}{2}(1)^{ – \frac{3}{2}} (2 + 0)(1 + 0) + (1 + ( – 1)^2 + 0) – y^{\prime\prime} \\ & =\frac{1}{2} \cdot 2 + 1 \cdot – y^{\prime\prime} = 1 – y^{\prime\prime} \end{array}$
$\Rightarrow 4 = 1 – y^{\prime\prime} \Rightarrow \left. {y^{\prime\prime}} \right|_{(1,0)} = – 3$
Thus $\left. {\frac{{d^2 y}}{{dx^2 }}} \right|_{(1,0)} = – 3 ∎$