- Find the total area of the three leaves enclosed by the three-leaf-rose
$r=\sin3\theta$.
- An ellipse is defined by
$x+y+z=1,~x^2+y^2=1$.
At the point $\displaystyle\left(x,y,z\right)=\left(\frac1{\sqrt2},\frac1{\sqrt2},1-\sqrt2\right)$, find the curvature $\kappa=$? - Can you find a series of polynomials
$p_1\left(x\right)+p_2\left(x\right)+p_3\left(x\right)+\cdots$
which is convergent for $\left|x\right|<1$ to the sum $\displaystyle s\left(x\right)=\sum_{n=1}^{\infty}p_n\left(x\right)$, but when $x$ approaches $1$,$\displaystyle\lim_{x\to1}s\left(x\right)\neq\sum_{n=1}^{\infty}p_n\left(1\right)$?
If not, explain why that's impossible. - Cross valley cable is hanged on towers $2$ kilometers apart. Justify that the cable is curved as a catenary: $\displaystyle y=\lambda\cosh\left(\frac{x}{\lambda}\right)$; If the density of the cable is $2$ kg/m and the tension at the middle point $x=0$ is $1.96\times10^5$ newtons, find the length of the cable.
- You may use Lagrange multiplier to find the critical points of the function
$f\left(x,y\right)=-3x^2-2xy+13y^2+16x+32y$
under the constraint$g\left(x,y\right)=x^2+2xy+y^2-x+y=0$.
Indicate which points are local maxima and which are local minima. - 當 $\lambda=0$ 時有 $\left(x,y\right)=\left(3,-1\right)$,此時 $f\left(3,-1\right)=8$;
- 當 $\lambda=6$ 時有 $\left(x,y\right)=\left(0,-1\right)$,此時 $f\left(0,-1\right)=-19$;
- 當 $\lambda=-10$ 時有 $\left(x,y\right)=\left(1,0\right)$,此時有 $f\left(1,0\right)=13$。
訣竅
運用極座標的面積公式求解。解法
運用面積公式可得$\displaystyle A=3\cdot\frac12\int_0^{\frac\pi3}r^2d\theta=\frac32\int_0^{\frac\pi3}\sin^23\theta d\theta=\frac34\int_0^{\frac\pi3}\left(1-\cos6\theta\right)d\theta=\frac34\cdot\frac\pi3=\frac\pi4$.
訣竅
將曲線參數化後依據曲率的公式計算即可。解法
將曲線參數 ${\bf r}\left(t\right)$ 如下$\left\{\begin{aligned} &x\left(t\right)=\cos t,\\&y\left(t\right)=\sin t,\\&z\left(t\right)=1-\cos t-\sin t,\end{aligned}\right.$,$t\in\mathbb{R}$.
那麼曲率公式有$\displaystyle\kappa\left(t\right)=\frac{\left|{\bf r}'\times{\bf r}''\right|}{\left|{\bf r}'\right|^3}=\frac{\left|\left(-\sin t,\cos t,\sin t-\cos t\right)\times\left(-\cos t,-\sin t,\cos t+\sin t\right)\right|}{\left|\left(-\sin t,\cos t,\sin t-\cos t\right)\right|^3}=\frac{\sqrt3}{\sqrt{2-2\cos t\sin t}^3}$.
依坐標取 $\displaystyle t=\frac\pi4$ 代入可得$\displaystyle\kappa\left(\frac\pi4\right)=\sqrt3$.
訣竅
考慮等比級數作為例子。解法
取 $p_k\left(x\right)=\left(-x\right)^k$,那麼當 $\left|x\right|<1$ 時有有$\displaystyle s\left(x\right)=\sum_{n=1}^{\infty}p_n\left(x\right)=\sum_{n=1}^{\infty}\left(-x\right)^n=\frac1{1+x}$.
那麼有 $\displaystyle\lim_{x\to1}s\left(x\right)=\frac12$,但 $\displaystyle\sum_{n=1}^{\infty}p_n\left(1\right)=\sum_{n=1}^{\infty}\left(-1\right)^n$ 不存在。訣竅
根據懸鍊線的相關知識進行推導,並由曲線弧長之公式計算之。解法
假定纜線的質料均勻且密度為 $\rho$(單位:牛頓/米),且自纜線上任取相異兩點 $A$ 與 $B$,其中 $A$ 為最低點($x=0$ 處),並記此段曲線長度為 $S$(單位:米)。再者假定 $A$ 處的張力為 $H$(單位:牛頓)而 $B$ 處的張力為 $T$(單位:牛頓),且與水平方向夾角為 $\theta$,那麼由靜力平衡可得\begin{equation}\label{1}\frac{dy}{dx}=\tan\theta=\frac{T\sin\theta}{T\cos\theta}=\frac{\rho S}{H}\end{equation}
此處 $y=y\left(x\right)$ 表曲線之方程式。又由弧長公式有$\displaystyle dS=\sqrt{1+\left(\frac{dy}{dx}\right)^2}$
對 $\left(\ref{1}\right)$ 進行微分有$\displaystyle\frac{d^2y}{dx^2}=\frac{\rho}{H}\frac{dS}{dx}=\frac{\rho}{H}\sqrt{1+\left(\frac{dy}{dx}\right)^2}$
命 $\displaystyle\frac{dy}{dx}=p\left(x\right)$,那麼有$\displaystyle\frac{dp}{dx}=\frac{\rho}{H}\sqrt{1+p^2}$
藉由移項有 $\displaystyle\frac{dp}{\sqrt{1+p^2}}=\frac{\rho}{H}dx$,在 $\left[0,x\right]$ 上同取定積分有$\displaystyle\ln\left(p\left(x\right)+\sqrt{1+p^2\left(x\right)}\right)-\ln\left(p\left(0\right)+\sqrt{1+p^2\left(0\right)}\right)=\frac{\rho}{H}x$
又因在 $A$ 處之斜率為 $0$,故 $p\left(0\right)=0$,從而整理可解得$\displaystyle\frac{dy}{dx}=p\left(x\right)=\frac{\exp\left(\rho x/H\right)-\exp\left(-\rho x/H\right)}{2}$
積分後可得$\displaystyle y\left(x\right)=\frac{H}\rho\frac{\exp\left(\rho x/H\right)+\exp\left(-\rho x/H\right)}2=\frac{H}\rho\cosh\left(\frac{\rho x}{H}\right)$
記 $\displaystyle\lambda=\frac{H}{\rho}$ 即有 $\displaystyle y=\lambda\cosh\left(\frac{x}{\lambda}\right)$。依據題目設定可知 $\rho=2$ 公斤重/米$\,=19.6$ 牛頓/米,$H=1.96\times10^5$ 牛頓,因此 $\lambda=10^4$ 米。使用曲線弧長公式計算有
$\displaystyle s=\int_{-1000}^{1000}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx=\int_{-1000}^{1000}\sqrt{1+\sinh^2\left(\frac{x}{\lambda}\right)^2}dx=\int_{-1000}^{1000}\cosh\left(\frac{x}{\lambda}\right)dx=2\lambda\sinh\left(\frac{1000}{\lambda}\right)=2\cdot10^4\sinh\frac1{10}$
訣竅
運用拉格朗日乘子法即可。解法
考慮拉格朗日乘子函數如下$F\left(x,y,\lambda\right)=f\left(x,y\right)+\lambda g\left(x,y\right)$
那麼解下列聯立方程組$\left\{\begin{aligned} &F_x\left(x,y,\lambda\right)=-6x-2y+16+\lambda\left(2x+2y-1\right)=0,\\&F_y\left(x,y,\lambda\right)=-2x+26y+32+\lambda\left(2x+2y+1\right)=0,\\&F_\lambda\left(x,y,\lambda\right)=x^2+2xy+y^2-x+y=0.\end{aligned}\right.$
我們可以整理第一式與第二式$\left\{\begin{aligned} &\left(2\lambda-6\right)x+\left(2\lambda-2\right)y=\lambda-16,\\&\left(2\lambda-2\right)x+\left(2\lambda+26\right)y=-\lambda-32.\end{aligned}\right.$
若 $\displaystyle\lambda=\frac{10}3$ 時會無解,而當 $\displaystyle\lambda\neq\frac{10}3$ 時恰有一解如下$\displaystyle x=\frac{\left(\lambda+20\right)\left(\lambda-6\right)}{4\left(3\lambda-10\right)},~y=-\frac{\left(\lambda+10\right)\left(\lambda-4\right)}{4\left(3\lambda-10\right)}$
據此代入第三式中可得$\displaystyle\left(x+y\right)^2-x+y=\frac{4\left(\lambda-10\right)^2}{\left(3\lambda-10\right)^2}-\frac{\lambda^2+10\lambda-80}{2\left(3\lambda-10\right)}=0$
此即等價於 $8\left(\lambda-10\right)^2-\left(3\lambda-10\right)\left(\lambda^2+10\lambda-80\right)=0$,這可以因式分解出 $\lambda\left(\lambda+10\right)\left(\lambda-6\right)=0$,故 $\lambda=0$ 或 $\lambda=6$ 或 $\lambda=-10$。(任擇 $4$ 題,每題 $25$ 分)
沒有留言:
張貼留言