PastExamLabPastExamLab

臺灣綜合大學系統 105 年度 微積分A

原始 PDF
第 1 題10 分極限
lim⁡x→0x2+sin⁡2x1−cos⁡x=\lim\limits_{x \to 0} \frac{x^2 + \sin^2 x}{1 - \cos x} = ____1(a)____; lim⁡x→0+(ex+x)1x=\lim\limits_{x \to 0^+} (e^x + x)^{\frac{1}{x}} = ____1(b)____
第 2 題10 分微分
Let x2+xy+2y2=1x^2 + xy + 2y^2 = 1. Find dydx=\frac{dy}{dx} = ____2(a)____ and d2ydx2=\frac{d^2y}{dx^2} = ____2(b)____ at (x,y)=(1,0)(x,y) = (1,0).
第 3 題10 分級數
Find the interval = ____3(a)____ of convergence of the power series ∑n=0∞(−1)n⋅x2n+12n+1\sum\limits_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n+1}}{2n+1}, and its sum = ____3(b)____
第 4 題10 分微分
The graph of the equation y=x(x+3)2y = \frac{x}{(x+3)^2} is strictly increasing on the interval = ____4(a)____ and concave upward on the interval = ____4(b)____.
第 5 題10 分積分
Evaluate ∫0∞x2e−x2dx\displaystyle\int_0^{\infty} x^2 e^{-x^2} dx from the known integral ∫0∞e−x2dx=π2\displaystyle\int_0^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}
第 6 題10 分微分
Show that F(x)=x2a2−x2+a22arcsin⁡(xa)F(x) = \displaystyle\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \arcsin \left(\frac{x}{a}\right), a>0a > 0 is an anti-derivative for f(x)=a2−x2f(x) = \sqrt{a^2 - x^2}.
第 7 題10 分偏微分
Let z=ln⁡(1+x1+y)z = \ln \left(\frac{1+x}{1+y}\right) where x=cos⁡tx = \cos t, y=tan⁡ty = \tan t. Use the chain rule to find the value of dzdt\frac{dz}{dt} when t=0t = 0.
第 8 題10 分最佳化
An observer at (3,6)(3,6) is watching an object descend the graph y2=xy^2 = x. At what point of it's path is the object closest to the observer.
第 9 題10 分多重積分
Find the area of surface ∬Sz2ds\displaystyle\iint\limits_S z^2 ds over the hemisphere z=1−x2−y2z = \sqrt{1 - x^2 - y^2}.
第 10 題10 分多重積分
Find the volume of the solid bounded above by the cone z=2−x2+y2z = 2 - \sqrt{x^2 + y^2} and below by the disk R:x2+(y−1)2≤1R: x^2 + (y-1)^2 \leq 1.