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)limx→1x3−1√x−1=limx→1(x3−1)′(√x−1)′=limx→13x212√x=limx→16x5/2=6(b)limn→∞[4√2n2−12+4√2n2−22+⋯+4√2n2−n2]=limn→∞n∑k=14√2n2−k2=limn→∞n∑k=14n√2−(k/n)2=I=∫104√2−x2dxx=√2cosθ⇒dx=−√2sinθdθ⇒I=∫π/4π/2−4√2sinθ√2sinθdθ=4⋅π4=π解答:f(x,y,z)=√xy3z+cos(xy)+zlnx=x1/2y3/2z1/2+cos(xy)+zlnx⇒fx=12x−1/2y3/2z1/2−ysin(xy)+zx⇒fxy=∂∂yfx⇒fxy=34x−1/2y1/2z1/2−sin(xy)−xycos(xy)⇒fxyz=∂∂zfxy⇒fxyz=38x−1/2y1/2z−1/2
解答:(a)y=x−3⇒dy=dx⇒I=∫∞3(x−3)2e−xdx=∫∞0y2e−(y+3)dy{u=y2dv=e−(y+3)dy⇒{du=2ydyv=−e−(y+3)⇒I=[−y2e−(y+3)]|∞0+2∫∞0ye−(y+3)dy=0+2∫∞0ye−(y+3)dy=2[−ye−(y+3)−e−(y+3)]|∞0=2e−3(b)cosx=1−tan2(x/2)1+tan2(x/2)⇒I=∫π013−cos(x)dx=∫π013−1−tan2(x/2)1+tan2(x/2)dxu=tan(x/2)⇒x=2tan−1u⇒dx=21+u2du⇒I=∫∞013−1−u21+u2⋅21+u2du=∫∞011+2u2duu=v√2⇒du=√22dv⇒I=√22∫∞011+v2dv=√22[tan−1v]|∞0=√22⋅π2=√2π4
解答:√1+x=(1+x)1/2=∞∑k=0(1/2k)xk=∞∑k=0(2kk)(−1)k+122k(2k−1)xk
解答:y′+3y=e2x⇒積分因子I(x)=e∫3dx=e3x⇒I(x)y′+3yI(x)=e2xI(x)⇒e3xy′+3e3xy=e5x⇒(e3xy)′=e5x⇒e3xy=15e5x+c1⇒y=15e2x+c1e−3xInitial value y(0)=1⇒1=15+c1⇒c1=45⇒y=15e2x+45e−3x
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+fyrcosθ⇒frr=(fxcosθ+fysinθ)r=(fxxxr+fxyyr)cosθ+(fyxxr+fyyyr)sinθ=(fxxcosθ+fxysinθ)cosθ+(fyxcosθ+fyysinθ)sinθ=fxxcos2θ+2fxycosθsinθ+fyysin2θ⇒fθθ=(−fxrsinθ+fyrcosθ)θ=(−fxxxθ−fxyyθ)rsinθ−fxrcosθ+(fyxxθ+fyyyθ)rcosθ−fyrsinθ=(fxxrsinθ−fxyrcosθ)rsinθ−fxrcosθ+(−fyxrsinθ+fyyrcosθ)rcosθ−fyrsinθ=fxxr2sin2θ−2fxyr2cosθsinθ+fyyr2cos2θ−fxrcosθ−fyrsinθ⇒frr+1r2fθθ=fxx(cos2θ+sin2θ)+fyy(sin2θ+cos2θ)−1r(fxcosθ+fysinθ)=fxx+fyy+1rfr⇒frr+1r2fθθ−1rfr=fxx+fyy=0⇒frr+1r2fθθ−1rfr=0.QED
解答:Given dVdt=120,V=43πr3⇒dVdt=dVdrdrdt=4πr2drdt⇒120=4π⋅302⋅drdt⇒drdt=130πcm/s
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