國立臺灣大學一百零四學年度研究所碩士班入學考試試題：微積分(A)

1. (10 points) A rectangle with corners at $\left(-x,0\right)$, $\left(x,0\right)$, $\left(x,y\right)$, $\left(-x,y\right)$ is inscribed in a half circle $x^2+y^2=1$ where $y\geq0$. Note that rectangle is in the upper half plane. Assume we move $x$ on the half circle, $x^2+y^2=1$, as $x\left(t\right)=t^2$ and $t$ goes from $0$ to $1$.
   1. (7 points) Find the rate of change of $y\left(t\right)$.
   2. (3 points) Find the rate of change of the area $A\left(t\right)=2x\left(t\right)y\left(t\right)$ of the rectangle.
訣竅
根據題意獲得變量之間的關係並使用微分公式求導即可。
解法

1. 由於 $y=\sqrt{1-x^2}$，故 $y\left(t\right)=\sqrt{1-\left(t^2\right)^2}=\sqrt{1-t^4}$，因此所求為

   $\displaystyle y'\left(t\right)=\frac{-4t^3}{2\sqrt{1-t^4}}=\frac{-2t^3}{\sqrt{1-t^4}}$

2. 直接求導可知

   $\displaystyle A'\left(t\right)=2x'\left(t\right)y\left(t\right)+2x\left(t\right)y'\left(t\right)=4t\sqrt{1-t^4}-\frac{4t^5}{\sqrt{1-t^4}}$



2. (10 points)
   1. (5 points) Analyze the local extrema of the function

      $\displaystyle f\left(x\right)=\frac{x}{1+x^2}$

      on the real axes using the second derivative test.
   2. (5 points) Are there any global extrema?
訣竅
利用求導找出極值候選點，隨後使用二階導函數判定其是否為局部極值；再者觀察無窮遠處的行為以瞭解是否為全域極值。
解法
1. 運用微分公式求導可知

   $\displaystyle f'\left(x\right)=\frac1{1+x^2}-\frac{2x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2}$

   為了找出局部極值，我們解方程式 $f'\left(x\right)=0$，易得 $x=\pm1$。又進一步求二階導函數可知

   $\displaystyle f''\left(x\right)=\frac{-2x}{\left(1+x^2\right)^2}-\frac{2\left(1-x^2\right)\cdot2x}{\left(1+x^2\right)^3}=\frac{-6x+2x^3}{\left(1+x^2\right)^3}$

   故有 $\displaystyle f''\left(1\right)=-\frac12<0$ 且 $\displaystyle f''\left(-1\right)=\frac12>0$，因此 $f$ 在 $x=1$ 處達到極大值而在 $x=-1$ 處達到極小值。
2. 又由前一小題的結果可以知道當 $x\in\left(-\infty,-1\right)\cup\left(1,\infty\right)$ 有 $f'\left(x\right)<0$ 而在 $x\in\left(-1,1\right)$ 有 $f'\left(x\right)>0$，且因 $\displaystyle\lim_{x\to\pm\infty}f\left(x\right)=0$，因此 $f$ 在 $x=-1$ 與 $x=1$ 處分別達到全域極小值與全域極大值。


3. (10 points) Suppose that $x^2y+xz^2=5$, and let $w=x^3y$. Express $\displaystyle\left(\frac{\partial w}{\partial z}\right)_y$ as a function of $x$, $y$, and $z$, and evaluate it numerically when $\left(x,y,z\right)=\left(1,1,2\right)$.
訣竅
運用多變數連鎖律求解即可，其中應準確理解記號之意義。
解法
首先注意到題目意指將 $w$ 視為 $\left(y,z\right)$ 的雙變數函數，故
$\displaystyle\frac{\partial w}{\partial z}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial z}=3x^2y\frac{\partial x}{\partial z}$
而利用題目原先給定的條件可知
$\displaystyle 2xy\frac{\partial x}{\partial z}+2z^2\frac{\partial x}{\partial z}+2xz=0$
即有 $\displaystyle\frac{\partial x}{\partial z}=-\frac{xz}{xy+z^2}$，故所求為 $\displaystyle\frac{\partial w}{\partial z}=-\frac{3x^3yz}{xy+z^2}$。


4. (15 points) Consider the region $R$ in the first quadrant bounded by the curves $y=x^2$, $y=x^2/5$, $xy=2$, and $xy=4$.
   1. (7 points) Compute $dxdy$ in terms of $dudv$ if $u=x^2/y$ and $v=xy$.
   2. (8 points) Find a double integral for the area of $R$ in $uv$ coordinates and evaluate it.
訣竅
按照 Jacobian 行列式的意義計算即可；運用變數變換求出平面上的區域面積。
解法

1. 按照 Jacobian 行列式的定義可計算如下

   $\displaystyle\begin{aligned}dxdy&=\Big|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\Big|dudv=\Big|\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u}&\displaystyle\frac{\partial x}{\partial v}\\\displaystyle\frac{\partial y}{\partial u}&\displaystyle\frac{\partial y}{\partial v}\end{vmatrix}\Big|dudv=\Big|\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x}&\displaystyle\frac{\partial u}{\partial y}\\\displaystyle\frac{\partial v}{\partial x}&\displaystyle\frac{\partial v}{\partial y}\end{vmatrix}\Big|^{-1}dudv\\&=\Big|\begin{vmatrix}2x/y&-x^2/y^2\\y&x\end{vmatrix}\Big|^{-1}dudv=\frac{y}{3x^2}dudv=\frac1{3u}dudv\end{aligned}$

2. 據前一小題的結果，所求之面積為

   $\displaystyle A=\iint_R1dA=\int_2^4\int_1^5\frac1{3u}dudv=\frac13\left(\int_2^4dv\right)\left(\int_1^5\frac{du}u\right)=\frac13\cdot2\cdot\ln5=\frac{2\ln5}3$



5. (15 points) Find the Lagrange multiplier equations for the point of the surface

   $x^4+y^4+z^4+xy+yz+zx=6$

   at which $x$ is largest. (Do not solve.) Explain clearly how you reach your conclusion.
訣竅
按照拉格朗日乘子法的概念列式即可。
解法

首先設定拉格朗日乘子函數如下

$F\left(x,y,z,\lambda\right)=x+\lambda\left(x^4+y^4+z^4+xy+yz+zx-6\right)$

據此能知需解的聯立方程組為

$\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=1+\lambda\left(4x^3+y+z\right)=0\\&F_y\left(x,y,z,\lambda\right)=\lambda\left(4y^3+x+z\right)=0\\&F_z\left(x,y,z,\lambda\right)=\lambda\left(4z^3+x+y\right)=0\\&F_{\lambda}\left(x,y,z,\lambda\right)=x^4+y^4+z^2+xy+yz+zx-6=0\end{aligned}\right.$



6. (10 points) Prove or disprove the following statement:

   $\displaystyle\frac{x}{1+x^2}<\tan^{-1}\left(x\right)<x$ for all $x>0$.

訣竅
利用函數的單調性進行證明。
解法

設 $\displaystyle f\left(x\right)=\tan^{-1}\left(x\right)-\frac{x}{1+x^2}$，那麼可知 $f\left(0\right)$ 且

$\displaystyle f'\left(x\right)=\frac1{1+x^2}-\left(\frac1{1+x^2}-\frac{2x^2}{\left(1+x^2\right)^2}\right)=\frac{2x^2}{\left(1+x^2\right)^2}>0$

故 $f$ 在 $\left(0,\infty\right)$ 上嚴格遞增，因此對所有 $x>0$ 恆有 $f\left(x\right)>f\left(0\right)=0$，即 $\displaystyle\tan^{-1}\left(x\right)>\frac{x}{1+x^2}$，故第一個不等式證明完畢。

現設 $g\left(x\right)=x-\tan^{-1}\left(x\right)$，那麼求導可知

$\displaystyle g'\left(x\right)=1-\frac1{1+x^2}=\frac{x^2}{1+x^2}>0$

故 $g$ 在 $\left(0,\infty\right)$ 上嚴格遞增，即對任何 $x\in\left(0,\infty\right)$ 恆有 $g\left(x\right)>g\left(0\right)=0$，故 $x>\tan^{-1}\left(x\right)$，第二個不等式證明完畢。



7. (15 points)
   1. (3 points) Find the Taylor series of $\ln\left(1+x\right)$ centered at $a=0$.
   2. (5 points) Determine the radius of convergence of this Taylor series.
   3. (2 points) Use the first two non-zero terms of the power series you found in (a) to approximate $\ln\left(1.5\right)$.
   4. (5 points) Give an upper bound on the error in your approximation in (c) using Taylor's inequality.
訣竅
首先由經典的無窮等比級數出發導出對數函數的泰勒級數，並由比值審歛法求出收斂半徑，也可由泰勒級數進行近似值與誤差的估算。
解法

1. 直接由不定積分與無窮等比級數的結論可知

   $\displaystyle\ln\left(1+x\right)=\int_0^x\frac{dt}{1+t}=\int_0^x\sum_{n=0}^{\infty}\left(-t\right)^ndt=\sum_{n=0}^{\infty}\left(-1\right)^n\int_0^xt^ndt=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n}{n+1}x^{n+1}$

2. 運用比值審歛法的概念可知收斂半徑為

   $\displaystyle R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\left(\frac{1}{n+1}\cdot\frac{n+2}1\right)=1$

3. 取 $\ln\left(1+x\right)$ 的前兩階近似為 $\displaystyle\ln\left(1+x\right)\approx f\left(x\right)=x-\frac{x^2}2$，故

   $\displaystyle\ln\left(1.5\right)\approx0.5-\frac{0.5^2}2=0.375=\frac38$

4. 由交錯級數的誤差估計可知其最大誤差為

   $\displaystyle\left|\ln\left(1.5\right)-\left(0.5-\frac{0.5^2}2\right)\right|\leq\frac{0.5^3}3=\frac1{24}$



8. (15 points) The point $x=\cos t$, $y=5+\sin t$ travels on a circle with center at $\left(0,5\right)$. Revolving that circle around the $x$ axis produces a doughnut. Find its surface area.
訣竅
運用旋轉體表面積的公式計算求解即可。
解法

使用旋轉體表面積之公式，並考慮此為參數曲線，如此所求為

$\displaystyle A=\int_0^{2\pi}2\pi y\left(t\right)\sqrt{x'^2\left(t\right)+y'^2\left(t\right)}dt=2\pi\int_0^{2\pi}\left(5+\sin t\right)\cdot1dt=2\pi\cdot5\cdot2\pi=20\pi^2$