臺灣綜合大學系統107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:假設{P在y=x3/2上⇒P(t,t3/2),t∈RQ(5/2,0)⇒¯PQ=√(t−5/2)2+t3=√t3+t2−5t+25/4令f(t)=t3+t2−5t+25/4⇒f′(t)=3t2+2t−5⇒f″(t)=6t+2因此f′(t)=0⇒(3t+5)(t−1)=0⇒t=1,−5/3⇒{f″(1)>0f″(−5/3)<0⇒當t=1時,¯PQ有最小值,此時P=(1,1)
解答:{x=3t2y=2t3⇒{x′=6ty′=6t2⇒曲線長=∫20√(x′)2+(y′)2dt=∫20√36t2+36t4dt=∫106t√1+t2dt=∫213√udu(u=1+t2)=[2u3/2]|21=25/2−2
解答:(a)limn→∞|an+1an|=limn→∞|(n+1)2xn+1n2xn|=|x|<1⇒收斂半徑=1(b)令f(x)=11−x=∞∑n=0xn⇒f′(x)=1(1−x)2=∞∑n=1nxn−1⇒xf′(x)=x(1−x)2=∞∑n=1nxn⇒(xf′(x))′=1+x(1−x)3=∞∑n=1n2xn−1⇒g(x)=x(1+x)(1−x)3=∞∑n=1n2xn⇒g(1/2)=6=∞∑n=1n22n⇒∞∑n=1n22n=6
解答:x2y+exyz−2cos(xz)=0⇒→n|(1,1,0)=(2xy+yzexyz+2zsin(xz),x2+xzexyz,xyexyz+2xsin(xz))|(1,1,0)=(2,1,1)⇒切平面方程式:(2,1,1)⋅(x−1,y−1,z)=0⇒2(x−1)+(y−1)+z=0⇒2x+y+z=3
解答:依題意{x=u2−v2y=2uv⇒{∂x∂u=2u∂x∂v=−2v∂y∂u=2v∂y∂v=2u⇒{gu=∂f∂x∂x∂u+∂f∂y∂y∂u=2ufx+2vfygv=∂f∂x∂x∂v+∂f∂y∂y∂v=−2vfx+2ufy⇒{guu=2fx+2u(∂∂xfx⋅∂x∂u+∂∂yfx⋅∂y∂u)+2v(∂∂xfy⋅∂x∂u+∂∂yfy⋅∂y∂u)=2fx+4u2fxx+8uvfxy+4v2fyygvv=−2fx−2v(∂∂xfx⋅∂x∂v+∂∂yfx⋅∂y∂v)+2u(∂∂xfy⋅∂x∂v+∂∂yfy⋅∂y∂v)=−2fx+4v2fxx−8uvfxy+4u2fyy⇒guu+gvv=4(u2+v2)fxx+4(u2+v2)fyy=4(u2+v2)(fxx+fyy)=0
解答:取{u=x+yv=x−y⇒{x=(u+v)/2y=(u−v)/2⇒J=∂(x,y)∂(u,v)=|xuxvyuyv|=|1/21/21/2−1/2|=−12⇒∬Dsin(x+y)dA=∫0π∫π0−12sinududv=∫0π−1dv=π
解答:取{x=rcosθy=rsinθz=z⇒∭E√x2+y2dV=∫2π0∫10∫r21r2dzdrdθ=∫2π0∫10r4−r2drdθ=∫2π0712dθ=76π
解答:{M(x,y)=yex+sinyN(x,y)=ex+xcosy⇒{My=ex+cosyNx=ex+cosy⇒My=Nx⇒存在potential function f滿足{fx=Mfy=N⇒f=∫Mdx=∫N,dy由於{∫yex+sinydx=yex+xsiny+ϕ(y)∫ex+xcosydy=yex+xsiny+ρ(x)⇒f(x,y)=yex+xsiny+C又{p0=r(t=0)=(1,−1)p1=r(t=2)=(5,3)⇒該線積分=f(p1)−f(p0)=3e5+5sin3−(−e+sin(−1))=3e5+5sin3+e+sin1
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解題僅供參考,其他轉學考歷屆試題及詳解
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