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成功大學 96 年度 微積分

本頁整理這份微積分考卷的題目、解答與詳解步驟，可直接對照每題內容與答案重點。

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這份試題整理

依本站已整理題目自動彙整題數、分數與題型，方便先判斷這份考卷的出題結構。

PastExamLab Summary
題數
10
題
總分
100
分
已整理詳解
0%
0 / 10 題

微積分 其他年度考古題

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第 1 題10 分
Show that tan⁡−1x+1x+tan⁡−1x=C\tan^{-1} \frac{x+1}{x} + \tan^{-1} x = C, where CC is a constant. And find the value of CC. (Note: tan⁡−1x=arctan⁡x\tan^{-1} x = \arctan x)
第 2 題10 分
Find the convergence set for the power series
∑n=0∞(−1)n(x−4)nn+1,\sum\limits_{n=0}^{\infty} \frac{(-1)^n(x-4)^n}{n+1},
and also find its sum.
第 3 題10 分
Find the average value of f(x)=tan⁡−1x2f(x) = \tan^{-1} \frac{x}{2} on the interval [0,2][0,2].
第 4 題10 分
For ∫π8π4tan⁡x(ln⁡cos⁡x)2dx\int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{\tan x}{(\ln \cos x)^2} dx,
(a)5 分
show that its an improper integral;
(b)5 分
evaluate or show that its diverges.
第 5 題10 分
The position vector of a particle at time t≥0t \geq 0 is
r⃗(t)=(cos⁡t+tsin⁡t)i⃗+(sin⁡t−tcos⁡t)j⃗.\vec{r}(t) = (\cos t + t \sin t)\vec{i} + (\sin t - t \cos t)\vec{j}.
(a)5 分
Show that the speed ds/dt=tds/dt = t.
(b)5 分
Show that the tangential and normal components of acceleration, aT=1a_T = 1 and aN=ta_N = t, respectively.
第 6 題10 分
Determine whether there are any points on the surface z2+xy−2x−y2=1z^2 + xy - 2x - y^2 = 1 at which the tangent plane is parallel to z=2z = 2.
第 7 題10 分
The function f(x,y)=6x2−8x+2y2−5f(x,y) = 6x^2 - 8x + 2y^2 - 5 is continuous on the closed region RR defined by x2+y2≤1x^2 + y^2 \leq 1. Find its absolute extrema over RR.
第 8 題10 分
The graph of x23+y23=1x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1, is called a hypocycloid,
(a)5 分
find equations of the tangent lines to the graph at the points corresponding to x=18x = \frac{1}{8};
(b)5 分
also find d2ydx2\frac{d^2y}{dx^2} for the equation in (a) for those points corresponding to x=18x = \frac{1}{8}
第 9 題10 分
Compute ∬Rcos⁡12(x−y)3x+ydA\iint_R \frac{\cos \frac{1}{2}(x-y)}{3x+y} dA, where RR is the region bounded by the graphs of y=xy = x, y=x−πy = x - \pi, y=−3x+3y = -3x + 3, and y=−3x+6y = -3x + 6.
第 10 題10 分
Find the surface area of the portions of the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4 that are within the cylinder (x−1)2+y2=1(x-1)^2 + y^2 = 1.