偏微分の問題演習

1. 次の関数を偏微分せよ．

   • ${\displaystyle z=x-2y}$

     ⇒ 解答

   • ${\displaystyle z=x^2+y^2}$

     ⇒ 解答

   • ${\displaystyle z=x^2-3xy+2y^2}$

     ⇒ 解答

   • ${\displaystyle z=\frac{y}{x}}$

     ⇒ 解答

   • ${\displaystyle z=\frac{x-y}{x+y}}$

     ⇒ 解答

   • ${\displaystyle z=\sqrt{3x-4y}}$

     ⇒ 解答

   • ${\displaystyle z=e^{xy}}$

     ⇒ 解答

   • ${\displaystyle z=log\left(x-y\right)}$

     ⇒ 解答

   • ${\displaystyle z=2sin\sqrt{xy}}$

     ⇒ 解答

   • ${\displaystyle z={tan}^{-1}\frac{y}{x}}$

     ⇒ 解答

   • ${\displaystyle z=x^3+3x^{2}y+2y^3}$

     ⇒ 解答

   • ${\displaystyle z=e^{-ax}\left(\sin by+\cos by\right)}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)=\frac{5x+3y}{3x+2y}}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)=x^y}$

     ⇒ 解答

   • ${\displaystyle z=3x^3-6x^{2}y+2x{y}^2+5y^4}$

     ⇒ 解答

   • ${\displaystyle z=\frac{2x^2+y^3}{x^3-3y^2}}$

     ⇒ 解答

   • ${\displaystyle z=\sqrt{5x^2+7y^3}}$

     ⇒ 解答

   • ${\displaystyle z=e^{x^3+2y^2}}$

     ⇒ 解答

   • ${\displaystyle z=\log (7x^4+5y^3)}$

     ⇒解答

   • ${\displaystyle z=\sin 4x^2+\cos 5y^3}$

     ⇒解答

   • ${\displaystyle z=3\sin \left(4x^3-7y^4\right)}$

     ⇒解答

   • ${\displaystyle z=5\cos x{y}^2}$

     ⇒解答

   • ${\displaystyle z={\sin }^{-1}xy}$

     ⇒解答

   • ${\displaystyle z={\cos }^{-1}2xy}$

     ⇒解答

   • ${\displaystyle z={\sin }^{-1}3x^2{y}^3}$

     ⇒解答

   • ${\displaystyle z={\cos }^{-1}\frac{y}{x}}$

     ⇒解答

   • ${\displaystyle z={\tan }^{-1}xy}$

     ⇒解答

   • ${\displaystyle z={\tan }^{-1}\sqrt{\frac{x}{y}}}$

     ⇒解答
2. 次の関数について ${\displaystyle \frac{\partial }{\partial x}f\left(1,-2\right)}$ と ${\displaystyle \frac{\partial }{\partial y}f\left(-1,2\right)}$ を求めよ．

   • ${\displaystyle f\left(x,y\right)=x^2+xy-y^2}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)=\sqrt{x^2-xy}}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)={sin}^{-1}\frac{x}{y}}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)=\frac{3x^{2}y+2x{y}^2}{x^2-y^2}}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)=\left(x+y\right)\left(x^2+x{y}^3\right)}$

     ⇒ 解答

   • ${\displaystyle f\left(x,y\right)={\tan }^{-1}\frac{y}{x}}$

     ⇒ 解答

3. 次のことを証明せよ．

   • ${\displaystyle z=f\left(\frac{y}{x}\right)}$ ならば${\displaystyle x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0}$ である． ⇒ 解答

   • ${\displaystyle z=f\left(x^2-y^2\right)}$ ならば ${\displaystyle y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0}$ である ⇒ 解答

   • ${\displaystyle z=\frac{1}{x}f\left(\frac{y}{x}\right)}$ ならば ${\displaystyle x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=0}$ である ⇒ 解答
4. 次の関数の ${\displaystyle \frac{dz}{dt}}$ を求めよ．

   • ${\displaystyle z=f\left(x,y\right),}$ ${\displaystyle x=a+ht,y=b+kt}$　⇒ 解答

   • ${\displaystyle z=x^2+y^2,}$ ${\displaystyle x=t-sint,y=1-cost}$　⇒ 解答

   • ${\displaystyle z=xy,}$ ${\displaystyle x=2t^2+1,y=t^2+3t+1}$　⇒ 解答

   • ${\displaystyle z=xtany,x={sin}^{-1}2t,}$    ${\displaystyle y={cos}^{-1}2t}$　⇒ 解答

5. 次のことを示せ．

   • 	${\displaystyle z=f\left(x,y\right)}$ , ${\displaystyle x=rcos\theta }$ , ${\displaystyle y=rsin\theta }$ ならば
     	${\displaystyle {\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2={\left(\frac{\partial z}{\partial r}\right)}^2+\frac{1}{r^2}{\left(\frac{\partial z}{\partial \theta }\right)}^2}$
     	となる．      ⇒ 解答

   • 	${\displaystyle z=f(x,y),x=u\cos \alpha -v\sin \alpha ,}$${\displaystyle y=u\sin \alpha +v\cos \alpha }$ ならば
     	${\displaystyle {\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2={\left(\frac{\partial z}{\partial u}\right)}^2+{\left(\frac{\partial z}{\partial v}\right)}^2}$
     	となる．      ⇒ 解答
6. 次の関数の第2次偏導関数を求めよ．

   • ${\displaystyle z=3x-2y}$

     ⇒ 解答

   • ${\displaystyle z=x^3+y^3}$

     ⇒ 解答

   • ${\displaystyle z=2x^2-3xy+4y^2}$

     ⇒ 解答

   • ${\displaystyle z=\frac{y}{x}}$

     ⇒ 解答

   • ${\displaystyle z=\frac{x-y}{x+y}}$

     ⇒ 解答

   • ${\displaystyle z=\sqrt{2x-3y}}$

     ⇒ 解答

   • ${\displaystyle z=e^{xy}}$

     ⇒ 解答

   • ${\displaystyle z=\log \left(x-y\right)}$

     ⇒ 解答

   • ${\displaystyle z=sin\sqrt{xy}}$

     ⇒ 解答

   • ${\displaystyle z={tan}^{-1}\frac{y}{x}}$

     ⇒ 解答

7. 次の関数の第2次偏導関数を求めよ．

   • ${\displaystyle z=f\left(x,y\right),}$${\displaystyle x=t-sint,y=1-cost}$⇒ 解答

   • ${\displaystyle z=f\left(x,y\right),}$ ${\displaystyle x=2t^2-3,y=t^2+3t+7}$⇒ 解答

8. 次のことを示せ．

   • 	${\displaystyle z=f(x,y),x=u\cos \theta -v\sin \theta ,}$${\displaystyle y=u\sin \theta +v\cos \theta }$のとき
     	${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}=\frac{{\partial }^{2}z}{\partial u^2}+\frac{{\partial }^{2}z}{\partial v^2}}$
     	となる．      ⇒ 解答

   • 	${\displaystyle z=f\left(x,y\right),x=r\cos \theta ,y=r\sin \theta }$のとき
     	${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}=\frac{{\partial }^{2}z}{\partial r^2}+\frac{1}{r}\frac{\partial z}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}z}{\partial {\theta }^2}}$
     	となる．      ⇒ 解答

9. 次のことを示せ．

   • 	${\displaystyle y=f\left(x,t\right)}$ において，1次元の波動方程式
     	${\displaystyle \frac{{\partial }^{2}y}{\partial t^2}=c^2\frac{{\partial }^{2}y}{\partial x^2}}$
     	に ${\displaystyle \xi =x-ct,\eta =x+ct}$ なる変換を行うと
     	${\displaystyle \frac{{\partial }^{2}y}{\partial \eta \partial \xi }=0}$
     	となる．      ⇒ 解答

   • 	${\displaystyle z}$ に関する方程式
     	${\displaystyle \frac{{\partial }^{2}z}{\partial t^2}=c^2\left(\frac{{\partial }^{2}z}{\partial r^2}+\frac{2}{r}\frac{\partial z}{\partial r}\right)}$
     	において， ${\displaystyle z=\frac{1}{r}u}$ とおき， ${\displaystyle u}$ に関する方程式に変換すると
     	${\displaystyle \frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial r^2}}$
     	となる．      ⇒ 解答
10. 次の関係で定義される陰関数 ${\displaystyle y=\varphi \left(x\right)}$ について ${\displaystyle \frac{d^{2}y}{d{x}^2}}$ を求めよ．

    • ${\displaystyle 3x^2+2xy+y^2=1}$

      ⇒ 解答

    • ${\displaystyle y^2=4px}$

      ⇒ 解答

    • ${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$

      ⇒ 解答

    • ${\displaystyle y=e^{x+y}}$

      ⇒ 解答

    • ${\displaystyle log\sqrt{x^2+y^2}-{tan}^{-1}\frac{y}{x}=0}$

      ⇒ 解答

    • ${\displaystyle x^3+y^3-3axy=0}$

      ⇒ 解答

11. 次の関係で定義される陰関数 ${\displaystyle y=\varphi \left(x\right)}$ の指定された点における接線の方程式を求めよ．

    • ${\displaystyle f\left(x,y\right)=0}$ の点 ${\displaystyle \left(a,b\right)}$ での接線の方程式

    • ⇒ 解答

    • ${\displaystyle x^3+y^3-xy=0}$ の点 ${\displaystyle \left(1/2,1/2\right)}$ における

    • 接線の方程式     ⇒ 解答

12. 次の関係で定義される陰関数 ${\displaystyle y=\varphi \left(x\right)}$ の極値を調べよ．
    

    • ${\displaystyle x^2-2xy+3y^2=8}$

      ⇒ 解答

    • ${\displaystyle x^{2}y-x{y}^2+128=0}$

      ⇒ 解答

    • ${\displaystyle x^2{y}^2-2x+9y^2=0}$

      ⇒ 解答

    • ${\displaystyle x^3-12xy+2y^3=0}$

      ⇒ 解答

    • ${\displaystyle x^4-16xy+3y^4=0}$

      ⇒ 解答

    • ${\displaystyle x^4+4x^2+3y^3-2y=0}$

      ⇒ 解答

    • ${\displaystyle x^2+2xy+y^4+2y^2=6}$

      ⇒ 解答

13. 次のことを示せ．

    • 	${\displaystyle z=xf\left(ax+by\right)+yg\left(ax+by\right)}$ならば
      	${\displaystyle b^2\frac{{\partial }^{2}z}{\partial x^2}-2ab\frac{{\partial }^{2}z}{\partial x\partial y}+a^2\frac{{\partial }^{2}z}{\partial y^2}=0}$
      	である．　⇒ 解答
14. 次の関数の極値を求めよ．

    • ${\displaystyle f\left(x,y\right)=x^2+2y^2+10x}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=x^2-2xy+3y^2-4x+5y}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=4x^2+2xy+y^2+4x+4y}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=xy\left(x-2y-3\right)}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=x^3+y^3-12x-27y}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=x^3+y^3+6xy-24}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=xy+\frac{1}{x}+\frac{1}{y}}$

      ⇒ 解答

    • ${\displaystyle f\left(x,y\right)=cosx+cosy+cos\left(x+y\right)}$ ${\displaystyle \left(0<x,y<\pi \right)}$

      ⇒ 解答

 

 

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最終更新日： 2023年1月20日