PastExamLabPastExamLab

臺灣綜合大學系統 108 年度 微積分A

原始 PDF
第 1 題10 分極限
Find the following limits.
(a)5 分
lim⁡x→3x2+x−12x−3\lim\limits_{x \to 3} \frac{x^2 + x - 12}{x - 3}.
(b)5 分
lim⁡x→0x2sec⁡x−1\lim\limits_{x \to 0} \frac{x^2}{\sec x - 1}.
第 2 題10 分偏微分
Evaluate ∂f∂x∣(0,0)\frac{\partial f}{\partial x}|_{(0,0)} and ∂f∂y∣(0,0)\frac{\partial f}{\partial y}|_{(0,0)} for
f(x,y)={3x4+xy22x3+4xy+y,(x,y)≠(0,0)0,(x,y)=(0,0).f(x,y) = \begin{cases} \frac{3x^4 + xy^2}{2x^3 + 4xy + y}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0). \end{cases}
第 3 題10 分最佳化
Given
f(t)={1;t≤01−t;t>0f(t) = \begin{cases} 1; & t \leq 0 \\ 1 - t; & t > 0 \end{cases}
and
F(x)=∫−12ax+2f(t) dtF(x) = \int_{-1}^{2ax+2} f(t) \, dt
with a>0a > 0, find aa so that FF is maximum at x=−2ax = -2a.
第 4 題10 分最佳化
Find the largest possible area of a triangle with vertices (0,2)(0,2), (1,0)(1,0) and the third vertex on the ellipse
x2+y24=1.x^2 + \frac{y^2}{4} = 1.
第 5 題10 分積分
Evaluate
∫ln⁡14ln⁡12ex1−4e2x dx.\int_{\ln \frac{1}{4}}^{\ln \frac{1}{2}} \frac{e^x}{\sqrt{1 - 4e^{2x}}} \, dx.
第 6 題10 分多重積分
Evaluate
∫022∫y1−y2ex2+y2 dxdy.\int_0^{\frac{\sqrt{2}}{2}} \int_y^{\sqrt{1-y^2}} e^{x^2+y^2} \, dx dy.
第 7 題10 分級數
Derive the complete Taylor series expansion for
ln⁡(1+2x1−2x)\ln \left( \frac{1 + 2x}{1 - 2x} \right)
about x=0x = 0. (In the form ∑k=0∞akx2k−1\sum\limits_{k=0}^{\infty} a_k x^{2k-1} with a general formula for aka_k.)
第 8 題10 分微分
Given function T(x,y)=1+x2−y2T(x,y) = 1+x^2-y^2, find the curve γ(t)=(x(t),y(t))\gamma(t) = (x(t), y(t)) so that γ(0)=(1,4)\gamma(0) = (1,4) and γ′(t)=−∇T(γ(t))\gamma'(t) = -\nabla T(\gamma(t)).
第 9 題10 分偏微分
Find (a,b)(a,b) with −12≤b≤12-\frac{1}{2} \leq b \leq \frac{1}{2} so that • The point P=(1,a,b)P = (1,a,b) is on the surface EE defined by
x2−y4+sin⁡(2z)4=0\frac{x}{2} - \frac{y}{4} + \frac{\sin(2z)}{4} = 0
• The tangent plane to EE at PP contains lines
l1(t)=P+t(12,1,0)andl2(t)=P+t(0,2,1)l_1(t) = P + t \left( \frac{1}{2}, 1, 0 \right) \quad \text{and} \quad l_2(t) = P + t(0, 2, 1)
第 10 題10 分向量微積分
Evaluate ∫LF⃗⋅dr⃗\int_L \vec{F} \cdot d\vec{r}, where
F⃗=(4x+5y,ecos⁡y+7x)\vec{F} = (4x + 5y, e^{\cos y} + 7x)
and LL is the path from AA to BB, to CC to AA and to DD as shown below:
第 10 題圖表