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台灣大學 · 生物環境系統工程學系 · 轉學考考古題 · 民國111年（2022年）

111 年度 微積分(B)

台灣大學 · 生物環境系統工程學系 · 轉學考

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Section A. Fill in the blanks.

Only answers will be graded. Label clearly your answer to each blank with the number of each blank on the answer sheet. 5 points are assigned to each blank.
第 1 題5 分
(a) lim⁡x→05cot⁡(x)+6sin⁡1x7sec⁡(x)−8sin⁡1x=\lim\limits_{x \to 0} \frac{5\cot(x) + 6\sin \frac{1}{x}}{7\sec(x) - 8\sin \frac{1}{x}} = __(1)__
(a)5 分
lim⁡x→05cot⁡(x)+6sin⁡1x7sec⁡(x)−8sin⁡1x=\lim\limits_{x \to 0} \frac{5\cot(x) + 6\sin \frac{1}{x}}{7\sec(x) - 8\sin \frac{1}{x}} = __(1)__
(b)5 分
lim⁡x→0(e2x−2x−2x2)12x=\lim\limits_{x \to 0} (e^{2x} - 2x - 2x^2)^{\frac{1}{2x}} = __(2)__
(c)5 分
lim⁡n→∞∑k=1nkn+n2(n2+kn+n2)2=\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{kn + n^2}{(n^2 + kn + n^2)^2} = __(3)__
第 2 題10 分
(a) Let f(x,y)=(x+y)x2f(x,y) = (x + y)^{x^2}. Then d2dy2f(1,t)∣t=0=\frac{d^2}{dy^2} f(1,t)|_{t=0} = __(4)__ (b) Let g(x)=1x2+2x+5g(x) = \frac{1}{\sqrt{x^2 + 2x + 5}}. Then g(5)(−1)=g^{(5)}(-1) = __(5)__
(a)5 分
Let f(x,y)=(x+y)x2f(x,y) = (x + y)^{x^2}. Then d2dy2f(1,t)∣t=0=\frac{d^2}{dy^2} f(1,t)|_{t=0} = __(4)__
(b)5 分
Let g(x)=1x2+2x+5g(x) = \frac{1}{\sqrt{x^2 + 2x + 5}}. Then g(5)(−1)=g^{(5)}(-1) = __(5)__
第 3 題5 分
Consider a function f:(−π,π)→Rf : (-\pi, \pi) \to \mathbb{R} defined by
f(x)={sin⁡x+axsin⁡xif −π<x<0x3+bx+5if 0≤x<πf(x) = \begin{cases} \sin x + \frac{ax}{\sin x} & \text{if } -\pi < x < 0 \\ x^3 + bx + 5 & \text{if } 0 \leq x < \pi \end{cases}
If ff is differentiable on (−π,π)(-\pi, \pi), then (a,b)=(a,b) = __(6)__
第 4 題5 分
Consider the parametric curve x=2t2+1x = 2t^2 + 1, y=4ty = 4t. Let PP be the point (2p2+1,4p)(2p^2 + 1, 4p). The greatest value of pp such that the normal to the curve at PP passes through (31,−24)(31, -24) is p=p = __(7)__
第 5 題5 分
Let f(x,y,z)=∫zx−yett2+1dtf(x, y, z) = \int_z^{x-y} \frac{e^t}{t^2 + 1} dt. The linearization of f(x,y,z)f(x, y, z) at (1,1,0)(1, 1, 0) is L(x,y,z)=L(x, y, z) = __(8)__
第 6 題10 分
(a) ∫01x(sin⁡x+sin⁡−1x)dx=\int_0^1 x(\sin x + \sin^{-1} x) dx = __(9)__ (b) Let DD be the region enclosed by the curve y=(10x−x2−21)12y = (10x - x^2 - 21)^{\frac{1}{2}} and the xx-axis. The volume of the solid obtained by revolving DD about the xx-axis is __(10)__
(a)5 分
∫01x(sin⁡x+sin⁡−1x)dx=\int_0^1 x(\sin x + \sin^{-1} x) dx = __(9)__
(b)5 分
Let DD be the region enclosed by the curve y=(10x−x2−21)12y = (10x - x^2 - 21)^{\frac{1}{2}} and the xx-axis. The volume of the solid obtained by revolving DD about the xx-axis is __(10)__
第 7 題5 分
Let f(x,y)=2x3−12xy+y3+13f(x, y) = 2x^3 - 12xy + y^3 + 13. Let P=(p,q)P = (p, q) be the point on R2\mathbb{R}^2 at which the rate of change of f(x,y)f(x, y) in the direction i+j\mathbf{i} + \mathbf{j} is the smallest. Then (p,q)=(p, q) = __(11)__
第 8 題10 分
(a) ∫14∫ln⁡z11(ey−y)2dydz=\int_1^4 \int_{\ln z}^1 \frac{1}{(e^y - y)^2} dy dz = __(12)__ (b) Let R={(x,y)∈R2:1≤x2+y2<4,x≥0,0≤y≤1}R = \{(x, y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 < 4, x \geq 0, 0 \leq y \leq 1\}. Then ∬Rxyx2+y2dA=\iint_R \frac{xy}{x^2 + y^2} dA = __(13)__
(a)5 分
∫14∫ln⁡z11(ey−y)2dydz=\int_1^4 \int_{\ln z}^1 \frac{1}{(e^y - y)^2} dy dz = __(12)__
(b)5 分
Let R={(x,y)∈R2:1≤x2+y2<4,x≥0,0≤y≤1}R = \{(x, y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 < 4, x \geq 0, 0 \leq y \leq 1\}. Then ∬Rxyx2+y2dA=\iint_R \frac{xy}{x^2 + y^2} dA = __(13)__
第 9 題15 分
(a) The work done by the force field F(x,y)=(1+x2)i+(xy)j\mathbf{F}(x, y) = (\sqrt{1 + x^2})\mathbf{i} + (xy)\mathbf{j} in moving a particle along a triangular path with vertices (0,0)(0, 0), (1,0)(1, 0), (2,2)(2, 2) counter-clockwise is __(14)__ (b) Let SS be part of the cone z=2x2+2y2z = \sqrt{2x^2 + 2y^2} that lies below the plane x+z=1x + z = 1. Then ∬SzdS=\iint_S z dS = __(15)__ (c) Let DD be a closed surface in R3\mathbb{R}^3, oriented outward. The maximum flux of the vector field
F(x,y,z)=(x+2x3)i−y(x2+z2)j−(3x2z2+4y2z)k\mathbf{F}(x, y, z) = (x + 2x^3)\mathbf{i} - y(x^2 + z^2)\mathbf{j} - (3x^2z^2 + 4y^2z)\mathbf{k}
among all possible choices of DD is __(16)__
(a)5 分
The work done by the force field F(x,y)=(1+x2)i+(xy)j\mathbf{F}(x, y) = (\sqrt{1 + x^2})\mathbf{i} + (xy)\mathbf{j} in moving a particle along a triangular path with vertices (0,0)(0, 0), (1,0)(1, 0), (2,2)(2, 2) counter-clockwise is __(14)__
(b)5 分
Let SS be part of the cone z=2x2+2y2z = \sqrt{2x^2 + 2y^2} that lies below the plane x+z=1x + z = 1. Then ∬SzdS=\iint_S z dS = __(15)__
(c)5 分
Let DD be a closed surface in R3\mathbb{R}^3, oriented outward. The maximum flux of the vector field
F(x,y,z)=(x+2x3)i−y(x2+z2)j−(3x2z2+4y2z)k\mathbf{F}(x, y, z) = (x + 2x^3)\mathbf{i} - y(x^2 + z^2)\mathbf{j} - (3x^2z^2 + 4y^2z)\mathbf{k}
among all possible choices of DD is __(16)__
第 10 題5 分
The greatest value of pp such that the series ∑n=1∞(−1)n⋅tan⁡(1np)⋅ln⁡(1+1np)\sum_{n=1}^{\infty} (-1)^n \cdot \tan\left(\frac{1}{\sqrt{np}}\right) \cdot \ln\left(1 + \frac{1}{n^p}\right) converges conditionally is p=p = __(17)__

Section B. Long Question.

Solve the following problem. You need to write down a complete and correct argument to receive full credits. Your work is graded on the quality of your writing as well as the validity of the mathematics. 15 points are assigned to this question.
第 1 題15 分
Consider the function f:R2→Rf : \mathbb{R}^2 \to \mathbb{R} defined by
f(x,y)={x⋅sin⁡(y2)x2+y3if (x,y)≠(0,0)0if (x,y)=(0,0)f(x, y) = \begin{cases} \frac{x \cdot \sin(y^2)}{x^2 + y^3} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } (x, y) = (0, 0) \end{cases}
(a) Is ff continuous at (0,0)(0, 0)? Justify your answer. (b) Let u=ai+bj\mathbf{u} = a\mathbf{i} + b\mathbf{j} be a unit vector. Find the directional derivative of ff at (0,0)(0, 0) in the direction u\mathbf{u}. Express your answer in terms of aa and bb. (c) Find the direction(s) that ff changes the most rapidly at (0,0)(0, 0).
(a)5 分
Is ff continuous at (0,0)(0, 0)? Justify your answer.
(b)5 分
Let u=ai+bj\mathbf{u} = a\mathbf{i} + b\mathbf{j} be a unit vector. Find the directional derivative of ff at (0,0)(0, 0) in the direction u\mathbf{u}. Express your answer in terms of aa and bb.
(c)5 分
Find the direction(s) that ff changes the most rapidly at (0,0)(0, 0).
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