PastExamLabPastExamLab

台灣大學 · 工程科學及海洋工程學系 · 轉學考考古題 · 民國109年（2020年）

109 年度 微積分 (B)

台灣大學 · 工程科學及海洋工程學系 · 轉學考

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PART 1 : Fill in the blanks.

• Only answers will be graded. • Each answer must be clearly labeled on the answer sheet. • 4 points are assigned to each blank.
第 1 題12 分
Fill in the blanks (4 points are assigned to each blank):
(a)4 分
lim⁡x→∞(1+3x)x=\lim\limits_{x \to \infty} (1 + \frac{3}{x})^x = __(1)__, where [x][x] is the greatest integer which is less than or equal to xx.
(b)4 分
lim⁡x→0(cos⁡ax)11−cos⁡bx=\lim\limits_{x \to 0} (\cos ax)^{\frac{1}{1 - \cos bx}} = __(2)__, where a,ba, b are constants and ab≠0ab \neq 0.
(c)4 分
lim⁡n→∞∑i=1n12n1−(i2n)2=\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{1}{2n\sqrt{1 - (\frac{i}{2n})^2}} = __(3)__.
第 2 題16 分
Fill in the blanks (4 points are assigned to each blank):
(a)4 分
f(x)=[ln⁡(1+x2)]2,ddxf(x)=f(x) = [\ln(1 + x^2)]^2, \frac{d}{dx}f(x) = __(4)__.
(b)4 分
x3−y2+y3=xx^3 - y^2 + y^3 = x. At (x,y)=(0,1),d2ydx2=(x,y) = (0,1), \frac{d^2y}{dx^2} = __(5)__.
(c)4 分
f(x,y,z)=∫zxyetdt,∇f=f(x,y,z) = \int_z^{x^y} e^{\sqrt{t}} dt, \nabla f = __(6)__.
(d)4 分
f(x,y)=sin⁡(x2y)x2+y2f(x,y) = \frac{\sin(x^2y)}{x^2 + y^2} for (x,y)≠(0,0)(x,y) \neq (0,0) and f(0,0)=0f(0,0) = 0. The directional derivative of ff along u=(cos⁡θ,sin⁡θ)u = (\cos\theta, \sin\theta) at (0,0)(0,0) is __(7)__.
第 3 題12 分
f(x)={∣x∣−1,for−1≤x≤3(x−6)2−4,for 3<x≤10f(x) = \begin{cases} |x| - 1, & \text{for} -1\leq x \leq 3 \\ (x-6)^2 - 4, & \text{for } 3 < x \leq 10 \end{cases}, g(x)=∫0xf(t)dtg(x) = \int_0^x f(t)dt, for −1≤x≤10-1 \leq x \leq 10.
(a)4 分
Choose correct statement(s) about g(x)g(x): __(8)__ i. g(x)g(x) is discontinuous at x=3x = 3. ii. g(x)g(x) is not differentiable at x=3x = 3. iii. g(x)g(x) is differentiable at x=0x = 0 and g′(0)=−1g'(0) = -1. iv. g(x)<0g(x) < 0 for −1≤x<0-1 \leq x < 0.
(b)4 分
The local minimum values of g(x)g(x) occur at x=x = __(9)__.
(c)4 分
The xx-coordinates of points of inflection for y=g(x)y = g(x) are __(10)__.
第 4 題4 分
A vertical fence 2 m high is located 1 m away from a wall. The length of the shortest ladder that can extend from the wall over the fence to a point on the ground is __(11)__.
第 5 題32 分
Compute the following integrals:
(a)4 分
∫0∞2x−3(x−1)(x2+4)dx=\int_0^\infty \frac{2x-3}{(x-1)(x^2+4)} dx = __(12)__.
(b)4 分
∫dxx2−6x=\int \frac{dx}{\sqrt{x^2-6x}} = __(13)__.
(c)4 分
∫sin⁡−1(x)dx=\int \sin^{-1}(\sqrt{x}) dx = __(14)__.
第 6 題8 分
Double and triple integrals:
(a)4 分
∬Dy dA=\iint\limits_D y \, dA = __(15)__, where DD is the region in the first quadrant that lies between the circles x2+y2=4x^2 + y^2 = 4 and (x−1)2+y2=1(x-1)^2 + y^2 = 1.
(b)4 分
The volume of the solid RR lying below the plane z=3−2yz = 3 - 2y and above the paraboloid z=x2+y2z = x^2 + y^2 is __(16)__.
第 7 題8 分
Vector calculus:
(a)4 分
CC is a smooth curve in the upper half plane going from (1,1)(1,1) to (0,2)(0,\sqrt{2}). F(x,y)=x−3yx2+y2i+3x+yx2+y2j\mathbf{F}(x,y) = \frac{x-3y}{x^2+y^2}\mathbf{i} + \frac{3x+y}{x^2+y^2}\mathbf{j}. Then, ∫CF⋅dr=\int_C \mathbf{F} \cdot d\mathbf{r} = __(17)__.
(b)4 分
SS is the part of the sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9 that lies above the cone z=x2+y2z = \sqrt{x^2 + y^2} with upward orientation. F(x,y,z)=xi+yj+zk(x2+y2+z2)2\mathbf{F}(x,y,z) = \frac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{(x^2 + y^2 + z^2)^2}. Then, ∬SF⋅dS=\iint\limits_S \mathbf{F} \cdot d\mathbf{S} = __(18)__.
第 8 題8 分
Series:
(a)4 分
∑n=0∞3n(2n)!=\sum\limits_{n=0}^\infty \frac{3^n}{(2n)!} = __(19)__.
(b)4 分
The Maclaurin series (the Taylor series at a=0a = 0) for cos⁡−1x\cos^{-1} x is __(20)__.

PART 2 :

• Solve the following problems. You need to write down complete arguments. • 10 points are assigned to each problem.
第 1 題10 分
Find the equation of the curve on the xyxy-plane that passes through (3,1)(3,1) such that for any point PP on the curve, the midpoint of the tangent line at PP that lies in the first quadrant is PP itself.
第 2 題10 分
f(x,y)f(x,y) is a differentiable function. On the curve tan⁡−1(xy)+y=2+π4\tan^{-1}(xy) + y = 2 + \frac{\pi}{4}, ff obtains local maximum at (x,y)=(12,2)(x,y) = (\frac{1}{2}, 2). Suppose that fx(12,2)=2f_x(\frac{1}{2}, 2) = 2.
(a)5 分
Find ∇f(12,2)\nabla f(\frac{1}{2}, 2).
(b)5 分
Assume that on another curve tan⁡−1(xy)+y=1.9+π4\tan^{-1}(xy) + y = 1.9 + \frac{\pi}{4}, ff obtains local maximum at (x0,y0)(x_0, y_0) which is near (12,2)(\frac{1}{2}, 2). Use linear approximation to estimate f(x0,y0)−f(12,2)f(x_0, y_0) - f(\frac{1}{2}, 2).
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