PastExamLabPastExamLab

成功大學 114 年度 微積分C

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第 1 題10 分
Find the following limits.
(a)
lim⁡h→1∫0h2−1ex2−2dxh−1\lim\limits_{h \to 1} \frac{\int_0^{h^2-1} e^{x^2-2} dx}{h-1}.
(b)
lim⁡x→0+∣x−sin⁡(3x)∣7x\lim\limits_{x \to 0^+} \frac{|x - \sin(3x)|}{7x}.
第 2 題10 分
Find point(s) (x0,y0)(x_0, y_0) on the curve x+y=12\sqrt{x} + \sqrt{y} = 12 at which normal line(s) to this curve has(have) slope(s) 13\frac{1}{3}.
第 3 題10 分
Find the exact value of the infinite sum ∑k=3∞(k+1)(−1)kk!\sum\limits_{k=3}^{\infty} \frac{(k+1)(-1)^k}{k!}.
第 4 題10 分
Two particles, A,BA, B, initially start from origin (0,0)(0,0), move along two straight lines with angle π6\frac{\pi}{6} between them. Particle AA is moving at a speed of 2, and particle BB is moving at a speed of 6. What is the rate of change of the distance between two particles, when particle AA is 1 unit from the starting point?
第 5 題10 分
Given the function f(x,y,z)=2xey−xsin⁡zf(x, y, z) = 2xe^y - x \sin z, and point P=(1,0,0)P = (1, 0, 0), find a,ba, b so that the rate of change of ff at PP along the unit vector u=(a,0,b)\mathbf{u} = (a, 0, b) is 0. Here, a>0a > 0.
第 6 題10 分
Find the maximum possible volume for a rectangular box, with edges parallel to coordinate axes, that is inscribed in the ellipsoid x29+y24+z2=1\frac{x^2}{9} + \frac{y^2}{4} + z^2 = 1.
第 7 題10 分
Find all the saddle point(s) of f(x,y)=ey(y2−x2)f(x, y) = e^y(y^2 - x^2).
第 8 題10 分
Evaluate the following integral ∫01∫0x−x2x2+y2 dydx\int_0^1 \int_0^{\sqrt{x-x^2}} \sqrt{x^2 + y^2} \, dy dx. (The formula cos⁡3θ=cos⁡(3θ)+3cos⁡θ4\cos^3 \theta = \frac{\cos(3\theta) + 3\cos\theta}{4} may be useful.)
第 9 題10 分
Evaluate the line integral ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r}, where F=(ln⁡(y2+1)−4y+ex,2xyy2+1+x)\mathbf{F} = \left(\ln(y^2 + 1) - 4y + e^x, \frac{2xy}{y^2 + 1} + x\right) and CC is the path connecting (2,0)(2, 0) to (1,3)(1, \sqrt{3}) through the circle x2+y2=4x^2 + y^2 = 4, and then to the origin (0,0)(0, 0) by straight line. See the figure below:
第 9 題圖表
第 10 題10 分
Evaluate the upward flux ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S}, where F=(cos⁡(yz2)+5x,ln⁡(x2+2z2+1)−6y,z−cos⁡(x2+y2))\mathbf{F} = (\cos(yz^2) + 5x, \ln(x^2 + 2z^2 + 1) - 6y, z - \cos(x^2 + y^2)), and SS is the upper half (i.e. z≥0z \geq 0 part) of the torus (x2+y2−4)2+z2=4\left(\sqrt{x^2 + y^2} - 4\right)^2 + z^2 = 4 as shown below
第 10 題圖表
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