臺灣綜合大學系統108學年度學士班轉學生聯合招生考試
科目名稱:微積分A
解答:{fx(0,0)=limh→0f(h,0)−f(0,0)h=limh→03h/2h=32fy(0,0)=limh→0f(0,h)−f(0,0)h=limh→00h=0⇒{∂f∂x|(0,0)=32∂f∂y|(0,0)=0
解答:F(x)=∫2ax+2−1f(t)dt⇒F′(x)=2af(2ax+2)={2a,2ax+2≤02a(1−2ax−2)=−2a(1+2ax),2ax>0又F′(−2a)=0⇒−2a(1−4a2)=0⇒a=12
解答:令{A(0,2)B(1,0),及P在x2+y24=1上,即P(cosθ,2sinθ)⇒{→PA=(−cosθ,2−2sinθ)→PB=(1−cosθ,−2sinθ)⇒△PAB面積=12‖
解答:u=2e^x \Rightarrow du=2e^xdx \Rightarrow \int_{\ln(1/4)}^{\ln(1/2)}{e^x \over \sqrt{1-4e^{2x}}}\,dx =\int_{1/2}^1 {1/2\over \sqrt{1-u^2}}du =\left.\left[ {1\over 2}\sin^{-1} u \right] \right|_{1/2}^1 \\={1\over 2}({\pi\over 2}-{\pi\over 6})= \bbox[red,2pt]{\pi \over 6}
解答:\int_0^{\sqrt 2/2} \int_y^{\sqrt{1-y^2}} e^{x^2+y^2}\,dxdy =\int_0^{\pi/4} \int_0^1 re^{r^2}\,drd\theta =\int_0^{\pi/4} \left. \left[{1\over 2}e^{r^2} \right] \right|_0^1 \;d\theta ={1\over 2}(e-1)\times{\pi \over 4} \\=\bbox[red,2pt]{{\pi \over 8}(e-1)}
解答:f(x)=\ln\left({1+2x\over 1-2x}\right) \Rightarrow f'(x)={4\over 1-4x^2} =4(1+4x^2+ (4x^2)^2+ (4x^2)^3+\cdots ) \\= 4+4^2 x^2+ 4^3x^4 +4^4 x^6+\cdots +4^{n+1}x^{2n}+\cdots =\sum_{n=0}^\infty 4^{n+1}x^{2n}\\\Rightarrow f''(x)= \sum_{n=1}^\infty 2n\cdot 4^{n+1}x^{2n-1} \Rightarrow f(x)=f(0)+f'(0)x +{f''(0)\over 2!}x^2 +\cdots \\ =0+4x +{16\over 3} x^3 +{64\over 5}x^5 +\cdots +{4^k\over 2k-1}x^{2k-1}+\cdots \Rightarrow f(x)= \bbox[red,2pt]{\sum_{k=1}^\infty {4^k\over 2k-1}x^{2k-1}}
解答:T(x,y)=1+x^2-y^2 \Rightarrow \nabla T=(2x,-2y) \Rightarrow -\nabla T(\gamma (t))=(-2x(t),2y(t)) =\gamma'(t)=(x'(t),y'(t))\\ \Rightarrow \cases{-2x(t)= x'(t) \\ 2y(t)= y'(t)} \Rightarrow \cases{x(t)=C_1e^{-2t}\\ y(t)= C_2e^{2t}},又\gamma(0)=(x(0),y(0))= (1,4),因此\cases{x(0)=C_1= 1\\ y(0)= C_2=4} \\ \Rightarrow \bbox[red, 2pt]{\gamma(t)= (e^{-2t}, 4e^{2t})}
解答:令F(x,y,z)={x\over 2}-{y\over 4}+{\sin(2z)\over 4} \Rightarrow \cases{F_x=1/2\\ F_y= -1/4\\ F_z=\cos(2z)/2}\\ \Rightarrow 過P(1,a,b)之切平面: F_x(P)(x-1) +F_y(P)(y-a)+ F_z(P)(z-b)=0\\ \Rightarrow {1\over 2}(x-1)-{1\over 4}(y-a)+{\cos (2b)\over 2} (z-b)=0 \Rightarrow 2(x-1)-(y-a)+2\cos(2b)(z-b)=0 \cdots(1)\\ 又\cases{\ell_1方向向量\vec u=(1/2,1,0)\\ \ell_2方向向量\vec v=(0,2,1)} \Rightarrow \vec n=\vec u\times \vec v=(1,-1/2,1)\\ \Rightarrow 切平面方程式:(x-1)-{1\over 2}(y-a)+(z-b)=0 \Rightarrow 2(x-1)-(y-a)+2(z-b)=0 \cdots(2) \\ 因此(1)=(2) \Rightarrow \cos(2b)=1 \Rightarrow b=0\Rightarrow F(1,a,0)=0 \Rightarrow {1\over 2}-{a\over 4}+0=0 \Rightarrow a=2 \\ \Rightarrow (a,b)=\bbox[red, 2pt]{(2,0)}
解答:\cases{C_1=A\to B\to C\to A為一loop \\ C_2=\overline{AD} =\{(-t,0)\mid t=0\to 1\}}\\ 依\text{Green's Theorem}: \vec F=(P(x,y),Q(x,y)) =(4x+5y,e^{\cos y}+7x) \\\Rightarrow \int_{C_1}\vec F\cdot d\vec r = \int \left({\partial \over \partial x}Q -{\partial \over \partial y}P\right)\;dA =\int (7-5)\,dA = \int 2\,dA = 2\times {\pi\over 4}={\pi \over 2}\\ 又\int_{C_2} \vec F\cdot d\vec r =\int_0^1 (-4t,1-7t)\cdot (-1,0)dt =\int_0^1 4t\,dt =2\\ 因此 \int_L \vec F\cdot d\vec r= \bbox[red,2pt]{{\pi \over 2}+2}
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