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# 偏微分の問題演習

1. 次の関数を偏微分せよ．
 $3\displaystyle{z=x- 2y}$ 解答 $3\displaystyle{z={x}^{2}+{y}^{2}}$ 解答 $3\displaystyle{z={x}^{2}- 3xy+2{y}^{2}}$ 解答 $3\displaystyle{z=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$ 解答 $3\displaystyle{z=\frac{\hspace{2} x- y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ 解答 $3\displaystyle{z=\sqrt{ 3x- 4y }}$ 解答 $3\displaystyle{z={e}^{ xy }}$ 解答 $3\displaystyle{z=\log \( x- y \) }$ 解答 $3\displaystyle{z=2\sin \sqrt{ xy }}$ 解答 $3\displaystyle{z={\tan }^{ - 1 }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$ 解答 $3\displaystyle{z={x}^{3}+3{x}^{2}y+2{y}^{3}}$ 解答 $3\displaystyle{z={e}^{ - ax }\text{\hspace{1}\hspace{1}} \( \sin by+\cos by \) }$ 解答 $3\displaystyle{f \( x,y \) =\frac{\hspace{2} 5x+3y \hspace{2}}{\hspace{2} 3x+2y \hspace{2}}}$ 解答 $3\displaystyle{f \( x,y \) ={x}^{y}}$ 解答
2. 次の関数について $3\displaystyle{\text{\hspace{1}}\text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f \( 1,- 2 \) \text{\hspace{1}}\text{\hspace{1}}}$$3\displaystyle{\text{\hspace{1}}\text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f \( - 1,2 \) \text{\hspace{1}}\text{\hspace{1}}}$ を求めよ．
 $3\displaystyle{f \( x,y \) ={x}^{2}+xy- {y}^{2}}$ 解答 $3\displaystyle{f \( x,y \) =\sqrt{ {x}^{2}- xy }}$ 解答 $3\displaystyle{ f \( x,y \) ={ \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }$ 解答
3. 次のことを証明せよ．
 $3\displaystyle{z=f \( \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} \) \text{\hspace{1}}\text{\hspace{1}}}$ ならば $3\displaystyle{\text{\hspace{1}}\text{\hspace{1}}x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=0\text{\hspace{1}}\text{\hspace{1}}}$ である． 解答 $3\displaystyle{z=f \( {x}^{2}- {y}^{2} \) \text{\hspace{1}\hspace{1}}}$ ならば $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=0\text{\hspace{1} }\text{\hspace{1} }}$ である． 解答 $3\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}f \( \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} \) \text{\hspace{1} }\text{\hspace{1} }}$ ならば $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}+z=0\text{\hspace{1} }\text{\hspace{1} }}$ である． 解答 $3\displaystyle{z=\log \sqrt{ {x}^{2}+{y}^{2} }\text{\hspace{1} }\text{\hspace{1} }}$ ならば $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} \) }^{2}+{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} \) }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} {e}^{ 2z } \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$ である． 解答
4. 次の関数の $3\displaystyle{\frac{\hspace{2} dz \hspace{2}}{\hspace{2} dt \hspace{2}}}$ を求めよ．
 $3\displaystyle{z=f \( x,y \) \text{\hspace{1} },}$ $3\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=a+ht\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=b+kt\text{\hspace{1} }\text{\hspace{1} }}$ 解答 $3\displaystyle{z={x}^{2}+{y}^{2}\text{\hspace{1} },}$ $3\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=t- \sin t\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=1- \cos t\text{\hspace{1} }\text{\hspace{1} }}$ 解答 $3\displaystyle{z=xy\text{\hspace{1} },}$ $3\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=2{t}^{2}+1\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y={t}^{2}+3t+1\text{\hspace{1} }\text{\hspace{1} }}$ 解答 $3\displaystyle{z=x\tan y\text{\hspace{1} },}$ $3\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x={\sin }^{ - 1 }2t\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y={\cos }^{ - 1 }2t\text{\hspace{1} }\text{\hspace{1} }}$ 解答
5. 次のことを示せ．
 $3\displaystyle{z=f \( x,y \) }$ , $3\displaystyle{x=r\cos {\theta}}$ , $3\displaystyle{y=r\sin {\theta}}$ ならば $3\displaystyle{{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} \) }^{2}+{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} \) }^{2}={ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}} \) }^{2}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}} \) }^{2}}$ となる． 解答 $3\displaystyle{z=f \( x,y \) \text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }x=u\cos {\alpha}- v\sin {\alpha}\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=u\sin {\alpha}+v\cos {\alpha}\text{\hspace{1} }\text{\hspace{1} }}$ ならば $3\displaystyle{{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} \) }^{2}+{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} \) }^{2}={ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}u \hspace{2}} \) }^{2}+{ \( \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}v \hspace{2}} \) }^{2}\text{\hspace{1} }\text{\hspace{1} }}$ となる． 解答
6. 次の関数の第2次偏導関数を求めよ．
 $3\displaystyle{z=3x- 2y}$ 解答 $3\displaystyle{z={x}^{3}+{y}^{3}}$ 解答 $3\displaystyle{z=2{x}^{2}- 3xy+4{y}^{2}}$ 解答 $3\displaystyle{z=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$ 解答 $3\displaystyle{z=\frac{\hspace{2} x- y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ 解答 $3\displaystyle{z=\sqrt{ 2x- 3y }}$ 解答 $3\displaystyle{z={e}^{ xy }}$ 解答 $3\displaystyle{z=\log \( x- y \) }$ 解答 $3\displaystyle{z=\sin \sqrt{ xy }}$ 解答 $3\displaystyle{z={\tan }^{ - 1 }\text{\hspace{1} }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$ 解答
7. 次の関数の第2次偏導関数を求めよ．
 $3\displaystyle{z=f \( x,y \) \text{\hspace{1} },}$ $3\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=t- \sin t\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=1- \cos t\text{\hspace{1} }\text{\hspace{1} }}$ 解答 $3\displaystyle{z=f \( x,y \) \text{\hspace{1} },}$ $3\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} \hspace{1} }x=2{t}^{2}- 3\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y={t}^{2}+3t+7\text{\hspace{1} }\text{\hspace{1} }}$ 解答
8. 次のことを示せ．
 $3\displaystyle{z=f \( x,y \) \text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }x=u\cos {\theta}- v\sin {\theta}\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }y=u\sin {\theta}+v\cos {\theta}\text{\hspace{1} }\text{\hspace{1} }}$ のとき $3\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$ となる． 解答 $3\displaystyle{z=f \( x,y \) \text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}x=r\cos {\theta}\text{\hspace{1}},\text{\hspace{1}}\text{\hspace{1}}y=r\sin {\theta}\text{\hspace{1}}\text{\hspace{1}}}$ のとき $3\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{r}^{2} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}r\hspace{2}}\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{{\theta}}^{2} \hspace{2}}\text{\hspace{1}}\text{\hspace{1}}}$ となる． 解答
9. 次のことを示せ．
 $3\displaystyle{y=f \( x,t \) }$ において，1次元の波動方程式 $3\displaystyle{\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{t}^{2} \hspace{2}}={c}^{2}\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}}$ に $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }{\xi}=x- ct\text{\hspace{1} },\text{\hspace{1} }\text{\hspace{1} }{\eta}=x+ct\text{\hspace{1} }\text{\hspace{1} }}$ なる変換を行うと $3\displaystyle{\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{\eta}{\partial}{\xi} \hspace{2}}=0}$ となる． 解答 $3\displaystyle{z}$ に関する方程式 $3\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{t}^{2} \hspace{2}}={c}^{2} \( \frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{r}^{2} \hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}r\hspace{2}}\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}} \) }$ において， $3\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}r\hspace{2}}u}$ とおき， $3\displaystyle{u}$ に関する方程式に変換すると $3\displaystyle{\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}{t}^{2} \hspace{2}}={c}^{2}\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}{r}^{2} \hspace{2}}}$ となる 解答
10. 次の関係で定義される陰関数 $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y={\varphi} \(x\) \text{\hspace{1} }\text{\hspace{1} }}$ について $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$ を求めよ．
 $3\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$ 解答 $3\displaystyle{{y}^{2}=4px}$ 解答 $3\displaystyle{\frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2} {a}^{2} \hspace{2}}+\frac{\hspace{2} {y}^{2} \hspace{2}}{\hspace{2} {b}^{2} \hspace{2}}=1}$ 解答 $3\displaystyle{y={e}^{ x+y }}$ 解答 $3\displaystyle{\log \text{\hspace{1} }\sqrt{ {x}^{2}+{y}^{2} }- {\tan }^{ - 1 }\text{\hspace{1} }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}=0}$ 解答 $3\displaystyle{{x}^{3}+{y}^{3}- 3axy=0}$ 解答
11. 次の関係で定義される陰関数 $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y={\varphi} \(x\) \text{\hspace{1} }\text{\hspace{1} }}$ の指定された点における接線の方程式を求めよ．
 $3\displaystyle{\text{\hspace{1} \hspace{1} }f \( x,y \) =0\text{\hspace{1} }\text{\hspace{1} }}$ の点 $3\displaystyle{\text{\hspace{1} } \( a,b \) \text{\hspace{1} }\text{\hspace{1} }}$ での接線の方程式 解答 $3\displaystyle{\text{\hspace{1} \hspace{1} }{x}^{3}+{y}^{3}- xy=0\text{\hspace{1} \hspace{1} }}$ の点 $3\displaystyle{\text{\hspace{1} } \( \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}},\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} \) \text{\hspace{1} }\text{\hspace{1} }}$ 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 $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }y={\varphi} \(x\) \text{\hspace{1} }\text{\hspace{1} }}$ の極値を調べよ．
 $3\displaystyle{{x}^{2}- 2xy+3{y}^{2}=8}$ 解答 $3\displaystyle{{x}^{2}y- x{y}^{2}+128=0}$ 解答 $3\displaystyle{{x}^{2}{y}^{2}- 2x+9{y}^{2}=0}$ 解答 $3\displaystyle{{x}^{3}- 12xy+2{y}^{3}=0}$ 解答 $3\displaystyle{{x}^{4}- 16xy+3{y}^{4}=0}$ 解答 $3\displaystyle{{x}^{4}+4{x}^{2}+3{y}^{3}- 2y=0}$ 解答 $3\displaystyle{{x}^{2}+2xy+{y}^{4}+2{y}^{2}=6}$ 解答
13. 次のことを示せ．
 $3\displaystyle{z=xf \( ax+by \) +yg \( ax+by \) \text{\hspace{1} }}$ならば $3\displaystyle{{b}^{2}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}- 2ab\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}+{a}^{2}\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=0}$ である 解答
14. 次の関数の極値を求めよ．
 $3\displaystyle{f \( x,y \) ={x}^{2}+2{y}^{2}+10x}$ 解答 $3\displaystyle{f \( x,y \) ={x}^{2}- 2xy+3{y}^{2}- 4x+5y}$ 解答 $3\displaystyle{f \( x,y \) =4{x}^{2}+2xy+{y}^{2}+4x+4y}$ 解答 $3\displaystyle{f \( x,y \) =xy \( x- 2y- 3 \) }$ 解答 $3\displaystyle{f \( x,y \) ={x}^{3}+{y}^{3}- 12x- 27y}$ 解答 $3\displaystyle{f \( x,y \) ={x}^{3}+{y}^{3}+6xy- 24}$ 解答 $3\displaystyle{f \( x,y \) =xy+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}}}$ 解答 $3\displaystyle{f \( x,y \) =\cos x+\cos y+\cos \( x+y \) }$ $3\displaystyle{\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} } \( 0< x\text{\hspace{1} },\text{\hspace{1} }y< {\pi} \) }$ 解答

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