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2024年4月19日 星期五

113年中山應數系碩士班-微積分詳解

國立中山大學113學年度碩士班招生考試

科目名稱: 微積分【應用數學系】

解答:(a)limx1x31x1=limx1(x31)(x1)=limx13x212x=limx16x5/2=6(b)limn[42n212+42n222++42n2n2]=limnnk=142n2k2=limnnk=14n2(k/n)2=I=1042x2dxx=2cosθdx=2sinθdθI=π/4π/242sinθ2sinθdθ=4π4=π
解答:f(x,y,z)=xy3z+cos(xy)+zlnx=x1/2y3/2z1/2+cos(xy)+zlnxfx=12x1/2y3/2z1/2ysin(xy)+zxfxy=yfxfxy=34x1/2y1/2z1/2sin(xy)xycos(xy)fxyz=zfxyfxyz=38x1/2y1/2z1/2
解答:(a)y=x3dy=dxI=3(x3)2exdx=0y2e(y+3)dy{u=y2dv=e(y+3)dy{du=2ydyv=e(y+3)I=[y2e(y+3)]|0+20ye(y+3)dy=0+20ye(y+3)dy=2[ye(y+3)e(y+3)]|0=2e3(b)cosx=1tan2(x/2)1+tan2(x/2)I=π013cos(x)dx=π0131tan2(x/2)1+tan2(x/2)dxu=tan(x/2)x=2tan1udx=21+u2duI=0131u21+u221+u2du=011+2u2duu=v2du=22dvI=22011+v2dv=22[tan1v]|0=22π2=2π4
解答:1+x=(1+x)1/2=k=0(1/2k)xk=k=0(2kk)(1)k+122k(2k1)xk
解答:y+3y=e2xI(x)=e3dx=e3xI(x)y+3yI(x)=e2xI(x)e3xy+3e3xy=e5x(e3xy)=e5xe3xy=15e5x+c1y=15e2x+c1e3xInitial value y(0)=11=15+c1c1=45y=15e2x+45e3x
解答:f(x(r,θ),y(r,θ)), where {x(r,θ)=rcosθy(r,θ)=rsinθ{fr=fxxr+fyyr=fxcosθ+fysinθfθ=fxxθ+fyyθ=fxrsinθ+fyrcosθfrr=(fxcosθ+fysinθ)r=(fxxxr+fxyyr)cosθ+(fyxxr+fyyyr)sinθ=(fxxcosθ+fxysinθ)cosθ+(fyxcosθ+fyysinθ)sinθ=fxxcos2θ+2fxycosθsinθ+fyysin2θfθθ=(fxrsinθ+fyrcosθ)θ=(fxxxθfxyyθ)rsinθfxrcosθ+(fyxxθ+fyyyθ)rcosθfyrsinθ=(fxxrsinθfxyrcosθ)rsinθfxrcosθ+(fyxrsinθ+fyyrcosθ)rcosθfyrsinθ=fxxr2sin2θ2fxyr2cosθsinθ+fyyr2cos2θfxrcosθfyrsinθfrr+1r2fθθ=fxx(cos2θ+sin2θ)+fyy(sin2θ+cos2θ)1r(fxcosθ+fysinθ)=fxx+fyy+1rfrfrr+1r2fθθ1rfr=fxx+fyy=0frr+1r2fθθ1rfr=0.QED
解答:Given dVdt=120,V=43πr3dVdt=dVdrdrdt=4πr2drdt120=4π302drdtdrdt=130πcm/s
 

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解題僅供參考,其他歷年試題及詳解

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