2020年7月15日 星期三

國立臺灣大學一百零八學年度研究所碩士班入學考試試題：微積分(A)

  1. (10 pts) Determine the slope of the tangent line to the curve $x^2y^3+y=2$ through the point $\left(1,1\right)$. (Present your derivation.)
    2. 訣竅運用隱函數微分求解即可。
    解法對 $x$ 求導可得

    $\displaystyle2xy^3+3x^2y^2\frac{dy}{dx}+\frac{dy}{dx}=0$

    取 $x=y=1$ 代入可得 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(1,1\right)}=-\frac12$。

  2. (10 pts) When $y=f\left(x\right)$, $f\left(0\right)=1$, and

    $\displaystyle\frac{dy}{dx}=\left(1+y^2\right)\cos\left(x\right),$

    determine $f\left(x\right)$. Present the derivation.
    3. 訣竅運用分離變量法解微分方程即可。
    解法移項可得

    $\displaystyle\frac{dy}{1+y^2}=\cos\left(x\right)dx$

    同取積分可知 $\tan^{-1}y=\sin x+C$。由初始條件可知 $\displaystyle C=\frac\pi4$，因此所求為

    $\displaystyle y\left(x\right)=\tan\left(\sin x+\frac\pi4\right)=\frac{1+\tan\left(\sin x\right)}{1-\tan\left(\sin x\right)}$


  3. (10 pts)
    1. (4 points) Give a sketch of $\Omega$. Here, $\Omega$ is defined by $y>0$, $y^2/16-4\leq x\leq y^2/4-1$, and $1-y^2/4\leq x\leq4-y^2/16$.
    2. (6 points) Evaluate the integral $\displaystyle\int_{\Omega}y^2dA$ over the region $\Omega$.
    4. 訣竅容易注意到邊界皆為拋物線，將此畫出即可，隨後按照邊界範圍寫出適當的積分次序計算即可。
    解法
    1. 將題目給定的邊界繪出如下可得
    2. 由所繪出的圖形可將重積分寫為下列的迭代積分並使用對稱性簡化計算如下

      $\displaystyle\begin{aligned}\int_{\Omega}y^2dA&=2\int_0^3\int_{2\sqrt{x+1}}^{4\sqrt{4-x}}y^2dydx=\frac23\int_0^3\left(64\left(4-x\right)^{3/2}-8\left(x+1\right)^{3/2}\right)dx\\&=\left.-\frac{32}{15}\left(8\left(4-x\right)^{5/2}+\left(x+1\right)^{5/2}\right)\right|_0^3=-\frac{32}{15}\left(8+32-256-1\right)=\frac{6944}{15}\end{aligned}$


  4. (15 pts) Find the area of the region enclosed by this graph, $r=\sin2\theta$. In your answer, it should include the sketch of the region (5 points) and determine the area (10 points). Show your work.
    5. 訣竅運用極座標的概念描點繪圖並使用極座標下的面積公式求解。
    解法首先描點可繪圖如下
    隨後運用極座標面積公式與對稱性可知

    $\displaystyle A=4\cdot\frac12\int_0^{\frac\pi2}\left(\sin2\theta\right)^2d\theta=\int_0^{\frac\pi2}\left(1-\cos4\theta\right)d\theta=\left.\left(\theta-\frac{\sin4\theta}4\right)\right|_0^{\frac\pi2}=\frac\pi2$


  5. (10 pts) Compute the following limit as $x$ goes to $1$.

    $\displaystyle\lim_{x\to1}\sum_{k=1}^{25}\frac{x^k-1}{\ln\left(x\right)}$.

    6. 訣竅運用羅必達法則求解即可。
    解法運用羅必達法則可知

    $\displaystyle\lim_{x\to1}\sum_{k=1}^{25}\frac{x^k-1}{\ln\left(x\right)}=\lim_{x\to1}\sum_{k=1}^{25}\frac{kx^{k-1}}{1/x}=\sum_{k=1}^{25}k=\frac{25\cdot\left(25+1\right)}2=13\cdot25=325$


  6. (15 pts) Consider a triangular pyramid having a triangular base with vertices $A$, $B$, and $C$. Denote the opposite sides by $a$, $b$, and $c$ in which $a$ corresponds to vertice $A$, and etc. Let $D$ be the foot of the altitude to the apex (i.e., vertex, tip) of the pyramid. Its height is $h$ which is a fixed height.
    1. (8 pts) Draw lines $DE$, $DF$, and $DG$ from $D$ perpendicular to the sides $a$, $b$, and $c$ of the triangle, with lengths $x$, $y$, and $z$, respectively. Also, draw lines from $D$ to the vertices of the triangle. Show that the area of triangle $ABC$, $T$ is equal to $\left(ax+by+cz\right)/2$. (i.e., $T=\left(ax+by+cz\right)/2$) In your solution, it should includes the plot of curve $C$. Determine which of these triangular pyramids has the minimum surface area.
    2. (7 pts) Compute the lateral surface area of the minimal-area pyramid. The answer should be expressed as a function of $s$, $h$, and $T$. (Note that the lateral surface area does not include the area of the triangle $ABC$.)
    7. 訣竅在第一小題中將三角形作切割可求得底部三角形面積，隨後表達金字塔表面積並透過初等不等式求其極小值；在第二小題中按照題意的指示使用指定的變量表達之。
    解法
    1. 注意到三角形 $ABC$ 可分割為三個三角形 $DAB$、$DBC$ 與 $DCA$，其中各別的面積分別為

      $\displaystyle T=\Delta DAB+\Delta DBC+\Delta DCA=\frac{cz}2+\frac{ax}2+\frac{by}2=\frac{ax+by+cz}2$

      而金字塔總表面積為 $\displaystyle f\left(x,y,z\right)=\frac{ax+by+cz+a\sqrt{x^2+h^2}+b\sqrt{y^2+h^2}+c\sqrt{z^2+h^2}}2$。由於底面是給定的，因此可變動的為 $x,y,z$ 且 $x,y,z$ 滿足 $T=\left(ax+by+cz\right)/2$ 為定值。運用拉格朗日乘子函數設定如下

      $F\left(x,y,z,\lambda\right)=2T+a\sqrt{x^2+h^2}+b\sqrt{y^2+h^2}+c\sqrt{z^2+h^2}+\lambda\left(ax+by+cz-2T\right)$

      據此解下列的聯立方程組

      $\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=\frac{ax}{\sqrt{x^2+h^2}}+a\lambda=0\\&F_y\left(x,y,z,\lambda\right)=\frac{by}{\sqrt{y^2+h^2}}+b\lambda=0\\&F_z\left(x,y,z,\lambda\right)=\frac{cz}{\sqrt{y^2+h^2}}+c\lambda=0\\&F_{\lambda}\left(x,y,z,\lambda\right)=ax+by+cz-2T=0\end{aligned}\right.$

      明顯可知 $x,y,z$ 與 $\lambda$ 皆非零且 $-1<\lambda<0$，故解得 $\displaystyle x=y=z=-\frac{\lambda h}{\sqrt{1-\lambda^2}}$，代入第四式中可得

      $\displaystyle-\frac{\lambda h\left(a+b+c\right)}{\sqrt{1-\lambda^2}}=2T$

      因此 $\displaystyle\lambda=-\frac{2T}{\sqrt{4T^2+h^2\left(a+b+c\right)^2}}$。注意到我們的過程中得到了 $x=y=z$，因此 $D$ 需要選在三角形的內心位置（即三個角平分線的交點。）
    2. 記 $\displaystyle s=\frac{a+b+c}2$，那麼 $\lambda=-\frac{T}{\sqrt{T^2+h^2s^2}}$，如此等號成立條件為 $\displaystyle x=y=z=\frac{T}s$，此時最小的金字塔側面面積為

      $\displaystyle\sqrt{T^2+h^2s^2}$


  7. (15 pts) Let $S$ be the part of the plane $z=1-x-y$ with $x\geq0$, $y\geq0$, and $z\geq0$. Evaluate the surface integral of $f\left(x,y,z\right)=xy$ over $S$.
    8. 訣竅利用曲面積分的定義計算求解。
    解法設 $D=\left\{\left(x,y\right)\in\mathbb{R}^2:\,x,y\geq0,\,x+y\leq1\right\}$，如此所求之曲面積分可寫為

    $\displaystyle\begin{aligned}\iint_Sfd\sigma&=\iint_Df\left(x,y\right)\cdot\sqrt3dA=\sqrt3\int_0^1\int_0^{1-x}xydydx=\frac{\sqrt3}2\int_0^1x\left(1-x\right)^2dx\\&=\frac{\sqrt3}2\int_0^1\left(x^3-2x^2+x\right)dx=\left.\frac{\sqrt3}2\left(\frac{x^4}4-\frac{2x^3}3+\frac{x^2}2\right)\right|_0^1=\frac{\sqrt3}{24}\end{aligned}$

    其中 $\sqrt3=\sqrt{1+z_x^2+z_y^2}=\sqrt{1+\left(-1\right)^2+\left(-1\right)^2}$。

  8. (15 pts) Let $C$ be the curve which consists of a line segment from $\left(1,1\right)$ to $\left(4,4\right)$, a semicircular arc from $\left(4,4\right)$ to $\left(6,4\right)$, and then a line segment from $\left(6,4\right)$ to $\left(8,4\right)$. Let ${\bf F}=\left(\exp\left(x-2y\right),-2\exp\left(x-2y\right)\right)$.
    1. (3 pts) Plot curve $C$.
    2. (12 pts) Evaluate $\displaystyle\int_C{\bf F}\cdot d{\bf r}$.
    9. 訣竅按照題意描繪各線段或圓弧，隨後找出對應的位能函數並使用線積分基本定理求解即可。
    解法
    1. 直接分段描繪如下
    2. 容易觀察發現可取位能函數 $f\left(x,y\right)=\exp\left(x-2y\right)$ 滿足 $\nabla f={\bf F}$，如此由線積分基本定理可知

      $\displaystyle\int_C{\bf F}\cdot d{\bf r}=f\left(8,4\right)-f\left(1,1\right)=1-e^{-1}$

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